TS Inter 1st Year

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(f) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(f)

Question 1.
Verify Rolle’s theorem for the following functions. (V.S.A.Q.)
(i) x² – 1 on [-1, 1]
Answer:
Let f(x) = x² defined on [-1, 1].
Since the function f(x) is a polynomial, it is continuous on [-1, 1] and differentiable on (-1, 1)
Also f(1) = 1 – 1 = 0, f(- 1) = (- 1)² – 1 = 0
∴ f (-1) = f (1). Hence f satisfy all conditions of Rolle’s theorem. Now we have to find a point c ∈ (-1,1) such that f'(c) = 0, f'(x) = 2x and f'(c) = 0 ⇒ 2c = 0
⇒ c = 0 ∈ (-1, 1)
Hence Rolle’s theorem is verified.

(ii) sin x – sin 2x on [0, π]
Answer:
Let f(x) = sin x – sin 2x defined over [0, π] f is continuous over [0, π] and differentiable over (0, π).
Also f(0) = 0 and f(π) = sin π – sin 2π = 0
∴ f(0) = f(π) = 0.
Hence f satisfy conditions of Rolle’s theorem’.
Also f'(x) = cos x – 2 cos 2x and f’ (c) = 0
⇒ cos c – 2 cos 2c = 0
⇒ cos c = 2 (2 cos²c – 1)
⇒ 4 cos²c – cos c – 2 = 0

Hence conditions of Rolle’s theorem are verified.

(iii) log (x² + 2) – log 3; over [-1, 1]
Answer:
f(x) = log (x² + 2) – log 3
This function is continuous over [-1,1] and differentiable over (-1,1)
f (-1) = log (1 + 2) – log 3 = 0
f(1) = log 3 – log 3 = 0
f (-1) = f (1)
Hence f satisfy all the conditions of Rolle’s theorem. So we have to find c ∈ (-1,1)
such that f’ (c) = 0 and f’ (c) = 0
⇒ c = 0 ∈ (-1, 1)
So Rolle’s theorem is verified.

Question 2.
It is given that Rolle’s theorem holds for the function f(x) = x³ + bx² + ax on [1, 3] with c = 2 + \(\frac{1}{\sqrt{3}}\). Find the values of a and b.
Answer:
Given that f(x) = x³ + bx² + ax defined over [1, 3] satisfy all the conditions of Rolle’s theorem.
∴ f ’(x) = 3x² + 2bx + a

b = -6, and b² – 3a = 3
⇒ 36 – 3a = 3 ⇒ a = 11
Hence a = 11 and b = – 6

Question 3.
Show that there is no real number k, for which the equation x² – 3x + k = 0 has two distinct roots in [0, 1]. (S.A.Q.)
Answer:
Let f(x) = x² – 3x + k and suppose there exists two different roots α, β (α < β). Since f is a polynomial in x, it is continuous over [0, 1] and differentiable in (0, 1). f is continuous over [a, 3] and differentiable in (a, β).
Also f(α) = α² – 3α + k = 0 and
f(β) = β² – 3β + k = 0
f(α) = f(β) = o :
f satisfy the conditions of Rolle’s theorem.
Also f'(c) = 0 ⇒ 3c² – 3 = 0
⇒ c² = 1
⇒ c = ± 1
This is a contradiction since
0 < α < c < β < 1
Hence there does not exists roots α, β in (0, 1).
So, there is no real number k for which the equation x² – 3x + k = 0 has distinct roots in[0, 1]

Question 4.
Find a point on the graph of the curve y = (x – 3)² where the tangent is parallel to the chord joining (3, 0) and (4, 1). (S.A.Q.)
Answer:
Let the points be A (3, 0) and B (4, 1)
∴ Slope of chord AB = \(\frac{1-0}{4-3}\) = 1
Given y = (x – 3)²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2(x – 3)
Slope of the chord 2 (x – 3) = 1
⇒ 2x = 7 ⇒ x = \(\frac{7}{2}\)
∴ y = (x – 3)² = (\(\frac{7}{2}\) – 3) = \(\frac{1}{4}\)
The point on the curve = \(\left(\frac{7}{2}, \frac{1}{4}\right)\)

Question 5.
Find a point on the graph of the curve y = x³ where the tangent is parallel to the chord joining (1, 1) and (3, 27). (S.A.Q.)
Answer:
Let the points be A (1, 1) and B (3, 27)
Slope of chord AB = \(\frac{27-1}{3-1}=\frac{26}{2}\) = 13

Question 6.
Find ‘c’ so that f'(c) = \(\frac{f(b)-f(a)}{b-a}\) in the following cases. (S.A.Q)
(i) f(x) = x² – 3x – 1; a = \(\frac{-11}{7}\), b = \(\frac{13}{7}\)
Answer:

Question 7.
Verify the Rolle’s theorem for the function (x² – 1) (x – 2) on [- 1, 2]. Find a point in the interval where the derivative vanishes. (S.A.Q.)
Answer:
Let f(x) = (x² – 1) (x – 2)
= x³ – 2x² – x + 2 defined over [-1, 2] f being a polynomial in ‘x’, it is continuous over [-1, 2] and differentiable over (- 1, 2) f (- 1) = 0, f(2) = 0 f(-1) = f(2)
∴ f satisfy all the conditions of Rolle’s theorem.
f'(x) = 3x² – 4x – 1
and f'(c) = 0 ⇒ 3c² – 4c – 1 = 0

= 1.549 ∈ (-1, 2)
Rolle’s theorem is verified.

Question 8.
Verify the conditions of the Lagrange’s mean value theorem for the following functions. In each case find a point ‘c’ in the interval as stated by the theorem. (S.A.Q.)
(i) x² – 1 on [2, 3]
Answer:
Let f(x) = x² – 1 defined over [2, 3]. This being a polynomial in x, continuous on [2,3] and djfferentiable over (2,3). So by Lagrange’s mean value theorem there exists a point ‘c’ ∈ (2, 3) such that
f'(c) = \(\frac{\mathrm{f}(3)-\mathrm{f}(2)}{3-2}\)
f'(x) = 2x ⇒ f'(c) = 2c f(3) = 8, 1(2) = 3
∴ 2c = \(\frac{8-3}{1}\) = 5 ⇒ c = \(\frac{5}{2}\) ∈ (2, 3)

(ii) sin x – sin 2x on [0, π]
Answer:
Let f(x) = sin x – sin 2x defined over [0, π]. This is continuous on [0, π] and differentiable over (0, π) since
f’ (x) = cos x – 2 cos 2x exists for all x ∈ (0, π). So by Lrgrange s mean value theorem
f'(c) = \(\frac{\mathrm{f}(\pi)-\mathrm{f}(0)}{\pi-0}\), f(π) = 0, f(0) = 0
∴ cos c – 2 cos 2c = \(\frac{0}{\pi}\) =0
⇒ cos c – 2 cos 2c = 0
⇒ cos c – 2 (2 cos² c – 1) = 0
⇒ 4cos² c – cos c – 2 = 0
⇒ cos c = \(\frac{1 \pm \sqrt{1+32}}{8}=\frac{1 \pm \sqrt{33}}{8}\)
∴ c ∈ cos^-1\(\left(\frac{1+\sqrt{33}}{8}\right)\) ∈ (0, π)

(ii) log x on [1, 2]
Let f(x) = log x defined over [1,2]
This is continuous over [1, 2] and differentiable over (1, 2) since

Hence Lagrange’s mean value theorem is satisfied.

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(e) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(e)

I.
Question 1.
At time ‘t’ the distance ‘x’ of a particle moving in a straight line is given by s = – 4t² + 2t. Find the average velocity between t = 2 and t = 8 sec. (V.S.A.Q.)
Answer:
s = – 4t² + 2t
∴ v = \(\frac{\mathrm{d} s}{\mathrm{dt}}\) = -8t + 2
Velocity at t = 2 is -16 + 2 = -14 units/sec.
Velocity at t = 8 is – 64 + 2 = – 62 units/sec.
Average velocity between t = 2 and t = 8 sec
= \(\frac{-62-14}{2}\) = – 38 units/sec.

Question 2.
If y = x⁴ then find the average rate of change of y between x = 2 and x = 4. (V.S.A.Q.)
Answer:

Question 3.
A particle moving along a straight line has the relation s = t³ + 2t + 3 connecting the distance s described by a particle in time t. Find the velocity and acceleration of the particle at t = 4 sec. (V.S.A.Q.)
Answer:
s = t³ + 2t + 3

Question 4.
The distance – time formula for the motion of a particle along a straight line is s = t³ – 9t² + 24t – 18. Find when and Where the velocity is zero. (S.A.Q.)
Answer:
Given s = t³ – 9t² + 24t – 18
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 3t² – 18t + 24
v = 0 ⇒ \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 0 ⇒ 3 (t² – 6t + 8) = 0
⇒ t = 2 or 4
The velocity is zero after 2 and 4 seconds

Case – (i) :
When t = 2, we have s = t3 – 9t2 + 24t – 18
= 8 – 36 + 48 – 13
= 56 – 54
= 2

Case – (ii) :
When t = 4, we have
s = t3 – 9t2 + 24t – 18
= 64 – 144 + 96- 18
= 160 – 162
= – 2
The particle is at a distance of 2 units from the starting point 0 on either side.

Question 5.
The displacement s of a particle travelling in a straight line in t seconds is given by s = 45t + lit2 – t3. Find the time when the particle comes to rest. (S.A.Q.)
Answer:
s = 45t + 11t² – t³
v = \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 45 + 22t – 3t²
If the particle comes to rest then v = 0
⇒ 45 + 22t – 3t² = 0
⇒ 3t² – 22t – 45 = 0
⇒ 3t² – 27t + 5t – 45 = 0
⇒ (3t + 5) (t – 9) = 0
⇒ t = \(\frac{-5}{3}\) or t = 9
∴ t = 9 and the particle comes to rest after t = 9 seconds.

II.
Question 1.
The volume of a cube is increasing at the rate of 8 cm3/sec. How fast is the surface area increasing when the length of an edge is 12 cm ? (Mar. ’14) (S.A.Q.)
Answer:
Suppose ‘a’ is the edge of the cube and V be the volume of the cube.
V = a³ ………….(1)

Question 2.
A stone is dropped into a quiet lake and ripples move in circles at the speed of 5 cm/sec. At the instant when the radius of circular ripple is 8 cm, how fast is the enclosed area increases ? (S.A.Q.)
Answer:
Suppose r is the radius of the outer ripple and A be its area.
Then the area A = πr2
dA „ dr dr _
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 2πr \(\frac{\mathrm{dr}}{\mathrm{dt}}\); Given r = 8, \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5
\(\frac{\mathrm{dA}}{\mathrm{dt}}\) = 2π(8) (5) = 80 π cm² / sec

Question 3.
The radius of a circle is increasing at the rate of 0.7 cm/sec. What is the rate of Increase of its circumference ? (S.A.Q.)
Answer:
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.7 cm/sec at
Circumference C = 2πr
\(\frac{\mathrm{dC}}{\mathrm{dt}}\) = 2π\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(0.7) = 1.4π cm/sec

Question 4.
A balloon which always remains spherical on inflation is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of balloon increases when the radius is 15 cm. (S.A.Q.)
Answer:
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 900 c.c/sec and r = 15 cm
Volume of the sphere V = \(\frac{4}{3}\)πr³

Question 5.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/sec. At what rate is the volume of the bubble is increasing when the radius is 1 cm ? (S.A.Q.)
Answer:
\(\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{2}\) cm/sec end radius r = 1 cm
Given volume of sphere V = \(\frac{4}{3}\)πr³
\(\frac{\mathrm{dV}}{\mathrm{dt}}=\frac{4}{3}\)π3r² = \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4πr²\(\frac{\mathrm{dr}}{\mathrm{dt}}\)
= 4π × 1 × \(\frac{1}{2}\) = 2π cm³/sec

Question 6.
Assume that an object is launched upward at 980 m/sec. Its position would be given by s = – 4.9t2 + 9801. Find the maximum height attained by the object. (S.A.Q.)
Answer:
Given relation between s and t as s = – 4.9t² + 980 t
= – 9.8t + 980
For maximum height \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = 0
s = – 4.9 (100)² + 980 (100)
= – 49000 + 98000 = 49000 units
The maximum height attained by the object is 49000 units.

Question 7.
Let a kind of bacteria grow in such a way that at time t sec, there are t3^2 bacteria. Find the rate of growth at time t = 4 hours. (S.A.Q.)
Answer:
Let ‘s’ be the amount of growth of bacteria at time’t’. Then s(t) = t^3/2
The growth rate at time’t’ is given by
\(\frac{\mathrm{ds}}{\mathrm{dt}}=\frac{3}{2}\)
When t = 4, \(\frac{3}{2}\) (4 × 60 × 60)
= \(\frac{3}{2}\) (2 × 60) = 180 units.

Question 8.
Suppose we have a rectangular aquarium with dimensions of length 8 m, width 4 in and height 3 m. Suppose we are filling the tank with water at the rate of 0.4 m3/sec. How feist is the height of water changing when the water level is 3 m ? (S.A.Q.)
Answer:
Length of aquarium l = 8 m
Width of aquarium b = 4 m
Height of aquarium h = 3 m
\(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 0.4 m/sec dt
V = lbh = 8 (4) (3) = 96
and log V = log l + log b + log h
\(\frac{1}{V} \frac{d V}{d t}=\frac{1}{h} \frac{d h}{d t}\) (∵l, b are constants)
⇒ \(\frac{0.4}{96}=\frac{1}{3} \frac{\mathrm{dh}}{\mathrm{dt}}\)
⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}=\frac{3 \times 0.4}{96}=\frac{1}{80}\)

Question 9.
A container is in the shape of an inverted cone has height 8 m and radius 6 m at the top. If it is filled with water at the rate of 2 m³/minuite how fast is the height of the water changing when the level is 4 m ? (S.A.Q.)
Answer:
Given h = 8 m = OA
r = 6 m = AB

The above measurements are at instant of time’t’
In ΔOAB, ΔOCD we have

Question 10.
The total cost c(x) in rupees associated with the production of x units of an item is given by c(x) = 0. 007 x³ – 0.003 x² + 15x + 4000. Find the marginal cost when 17 units are produced. (S.A.Q.)
Answer:
Let m represents the marginal cost then m = \(\frac{\mathrm{dc}}{\mathrm{dx}}\)
Hence
m = \(\frac{d}{d x}\)(0.007x³ – 0.003x² + 15x + 4000)
= 0.007 (3x²) – 0.003 (2x) + 15
The marginal cost at x = 17 is
(m)_x=17 = 17 = (0.007) (3 × 172) – 0.003 (34) + 15
= (0.007) (3 × 289) – (0.003) (34) + 15
= (0.007) 867 – (0.003) (34) + 15
= 20.967

Question 11.
The total revenue in rupees received from the sale of x units of a produce is given by R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7. (S.A.Q.)
Answer:
Let m denotes the margin revenue. Then
m = \(\frac{\mathrm{d}}{\mathrm{dx}}\)
Given R(x) = 13x² + 26x + 15 m = 26x + 26
The marginal revenue when x = 7 is
(m)_x=7 = 7= 26 (7) + 26 = 208

Question 12.
A point P is moving on the curve y = 2x². The x coordinate of P is increasing at the rate of 4 units per second. Find the rate at which y coordinate is increasing when the point is at (2, 8). (S.A.Q.)
Given equation of the curve is y = 2x² ………….. (1)
Let P (x₁, y₁) be any point on the curve (1)
Let y₁ = 2x₁
x₁ = 2 and y₁ = 8
Differentiating (2) w.r.t ‘t’
\(\frac{d y_1}{d t}\) = 4x₁ \(\frac{d x_1}{d t}\)
= 4 (2) (4)
= 32 units/sec.
∴ y co-ordinate is increasing at the rate of 32 units/sec.

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(d) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)

I. Find the angle between the curves given below. (March 2014) (E.Q.)

Question 1.
x + y + 2 = 0, x² + y² – 10y = 0
Answer:
x + y + 2 = 0 ⇒ x = – (y + 2)
x² + y² – 10y = 0
⇒ (y + 2)² + y² – 10y = 0
⇒ y² + 4y + 4 + y² – 10y = 0
⇒ 2y² – 6y + 4 = 0
⇒ y² – 3y + 2 = 0
⇒ (y – 2) (y – 1) = 0
⇒ y = 1 (or) y = 2
x = – (y + 2)
y = 1 ⇒ x = -(1 + 2) = -3
y = 2 ⇒ x = -(2 + 2) = -4
The points of intersection at P (- 3, 1) and Q (- 4, 2)
Equation of curve is x + y – 10y = 0
Differentiating w.r.to ‘x’

Question 2.
y² = 4x, x² + y² = 5. (E.Q.) (May 2007)
Answer:
The given equations of curves are y² = 4x and x² + y² = 5
Eliminating y, we get x² + 4x = 5
⇒ x² + 4x – 5 = 0
⇒ (x – 1) (x + 5) = 0
⇒ x = 1 or – 5
From y² = 4x and x = 1
⇒ y² = 4 ⇒ y = ±2
But when x = -5, y is imaginary and not real
Points of Intersection are
P(1, 2) and Q(1, -2)
From the equation y² = 4x
⇒ 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 4
⇒ \(\frac{d y}{d x}=\frac{z}{y}\) …….(1)
At P(1, 2), slope of first curve from (1)
m₁ = \(\frac{2}{2}\) = 1
and m₂ = slope of second curve from (2)
= \(\frac{-1}{2}\)

Question 3.
x² + 3y = 3, x² – y² + 25 = 0 (E.Q.)
Answer:
x² + 3y = 3 ……….(1),
x² – y² + 25 = 0 ………….(2)
are the equations of given curves.
From (1) x² = 3 – 3y
From (2), 3 – 3y – y² + 25 = 0
⇒ – y² – 3y + 28 = 0
⇒ y² + 3y – 28 = 0
⇒ (y + 7)(y – 4) = 0
⇒ y = 4 or y = -7
If y = 4 then x² = 3 – 12 = -9
Values of x are not real.
If y = – 7 then x² = 3 + 21
⇒ x² = 24 = ± 2√6
∴ Points of intersection are
P (2√6 , -7) and Q (-2√6 , -7)
From (1), 2x + 3 \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{-2 x}{3}\)
Slope of the tangent at P (2√6 , -7) to the curve (1) is m₁ = \(\frac{-2(2 \sqrt{6})}{3}=\frac{-4 \sqrt{6}}{3}\)

Question 6.
y² = 8x, 4x² + y² = 32
Answer:
y = 8x ….(1)
4x² + y² = 32
⇒ 4x² + 8x – 32 = 0
⇒ x² + 2x – 8 = 0
⇒ (x + 4) (x – 2) = 0
⇒ x = 2 or x = – 4
When x = 2, y² = 16 ⇒ y = ± 4
∴ Points of intersection are P (2, 4) and Q (2, – 4)
From (1), 2y \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 8 ⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{\mathrm{y}}\)

Question 7.
x²y = 4, y(x² + 4) = 8.
Answer:
x²y = 4 ………………(1)
y(x² + 4) = 8 …………(2)
From (1), x² = \(\frac{4}{y}\)

Substitute x² value in equation (2)
∴ y(\(\frac{4}{y}\)+4) = 8 ⇒ y\(\left(\frac{4+4 y}{y}\right)\) = 8
⇒ 4 + 4y = 8 ⇒ y = 1
∴ x² = \(\frac{4}{y}\)
⇒ x² = 4
⇒ x = ±2
∴ Points of intersection are P(2, 1), Q(-2, 1)
From (1)

Question 8.
Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\). (E.Q)
Answer:
The given curves are 6x² – 5x + 2y = 0 ………..(1)
and 4x² + 8y² = 3 ………..(2)

Slopes of two curves at \(\left(\frac{1}{2}, \frac{1}{2}\right)\) are equal and hence the given curves touch each other at \(\left(\frac{1}{2}, \frac{1}{2}\right)\).

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(c) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)

I.
Question 1.
Find the length of subtangent and sub¬normal at a point of the curve y = b sin\(\frac{x}{a}\).
Answer:
Equation of the curve is y = b sin\(\left(\frac{x}{a}\right)\)
\(\frac{d y}{d x}\) = b cos\(\left(\frac{x}{a}\right) \cdot \frac{1}{a}=\frac{b}{a}\) cos\(\left(\frac{x}{a}\right)\)

Question 2.
Show that the length of the subnormal at any point in the curve xy = a² varies as the cube of the ordinate of the point. (V.S.A.Q.)
Answer:
Equation of the curve is xy = a²

∴ Length of the subnormal αy₁³ = cube of the ordinate.

Question 3.
Show that at any point (x, y) on the curve y = be^x/a, the length of the subtangent is a constant and the length of the subnormal is \(\frac{y^2}{a}\). (V.S.A.Q)
Answer:
Equation of the given curve is y = be^x/a
∴ \(\frac{d y}{d x}=\frac{b}{a}\)e^x/a = \(\frac{\mathrm{y}}{\mathrm{a}}\)
∴ Slope at any point P(x, y) = \(\frac{y}{a}\)
Length of the subtangent = |y₁/f'(x₁)|
= |y/\(\frac{y}{a}\)| = a = constant
Length of the subnormal = |y₁/f'(x₁)|
= |y.\(\frac{y}{a}\)| = \(\left|\frac{y^2}{a}\right|\)

II.
Question 1.
Find the value of k so that the length of the subnormal at any point on the curve xy^k = a^k+1 is a constant. (S.A.Q.)
Answer:
Equation of the curve is xy^k = a^k+1
Let P(xj, yO be any point on the curve then
x₁y₁^k = a^k+1 …………….(1)
Differentiating w.r. to x,

Length of the subnormal is constant at any point on the curve is independent of x₁ and y₁.
\(\frac{\mathrm{y}_1^{\mathrm{k}+2}}{\mathrm{k} \cdot \mathrm{a}^{\mathrm{k}+1}}\) is independent of x₁ y₁.
∴ k + 2 = 0
⇒ k = – 2

Question 2.
At any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, subtangent and subnormal. (S.A.Q.) (June 2004, Board Model Paper)
Answer:
Equation of the curve is
x = a (t + sin t), y = a (1 – cos t)

Question 3.
Find the length of normal and subnormal at a point on the curve y = \(\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)\) (S.AQ.) (March 2013)
Answer:
Equation of the curve is y = \(\frac{a}{2}\left(e^{\frac{x}{a}}+e^{\frac{-x}{a}}\right)\)
= a cos h\(\left(\frac{x}{a}\right)\)

Question 4.
Find the lengths of subtangent, subnormal at a point’t’ on the curve x = a (cos t + t sin t), y = a (sin t – t cos t) (May 2014) (S.A.Q.)
Answer:
Equation of the curve is
x = a (cos t + t sin t)
y = a (sin t – t cos t)
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a(-sint + tcost + sint) = at cos t
and \(\frac{\mathrm{dy}}{\mathrm{dt}}\)= a (cost + tsint – cost) = at sin t dt
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} / \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{at} \sin \mathrm{t}}{\mathrm{at} \cos \mathrm{t}}\) = tan t
= \(\left|\frac{\mathrm{a}(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})}{\tan \mathrm{t}}\right|\)
= |a cot t(sin t – t cos t)|
Length of the subnormal = |y₁. f'(x₁)|
= |a(sin t – t cos t) tan t|
= |a tan t(sin t – t cos t)|

Students must practice this TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(b) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b)

I.
Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4. (V.S.A.Q.)
Answer:
Equation of the given curve is y = 3x⁴ – 4x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 12x³ – 4
Slope of the tangent at x = 4 is 12(4)³ – 4
= 12(64) – 4
= 764

Question 2.
Find the slope of the tangent to the curve y = \(\frac{x-1}{x-2}\) x ≠ 2 at x = 10. (V.S.A.Q.)
Answer:
Equation of the curve is y = \(\frac{x-1}{x-2}\)
= 1 + \(\frac{1}{x-2}\)
∴ \(\frac{d y}{d x}=-\frac{1}{(x-2)^2}\)
and slope of the tangent at x = 10 is
= \(-\frac{1}{(10-2)^2}=-\frac{1}{64}\)

Question 3.
Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x coordinate is ‘2’. (V.S.A.Q.)
Answer:
Equation of the curve is y = x³ – x + 1
∴ \(\frac{d y}{d x}\) = 3x² – 1
Slope of the tangent at x = 2 is 3(2)² – 1 = 11

Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x co-ordinate is ‘3’. (V.S.A.Q.)
Answer:
Equation of the curve is y = x³ – 3x + 2
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3x² – 3
Slope of the tangent at x = 3 is 3 (3)² – 3 = 24

Question 5.
Find the slope of the normal to the curve
x = a cos³θ, y = a sin³θ at θ = \(\frac{\pi}{4}\). (V.S.A.Q.)
Answer:
x = a cos³θ ⇒ \(\frac{d x}{d \theta}\) = 3a cos²θ (-sin θ) dθ
and y = a sin³θ ⇒ \(\frac{d y}{d \theta}\) = 3a sin²θ (cos θ)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{d} \theta} / \frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{3 \mathrm{a} \sin ^2 \theta \cos \theta}{-3 \mathrm{a} \cos ^2 \theta \sin \theta}\) = – tan θ
∴ Slope of the tangent at θ = \(\frac{\pi}{4}\) is – tan \(\frac{\pi}{4}\) = -1
and hence the slope of the normal = 1.

Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos²θ at θ = \(\frac{\pi}{2}\). (V.S.A.Q.)
Answer:

Question 7.
Find the points at which the tangent to the curve y = x³ – 3x² – 9x + 7 Is parallel to the X-axis. (V.SA.Q.)
Answer:
Equation of the curve is y = x³ – 3x² – 9x + 7
∴ \(\frac{d y}{d x}\) = 3x² – 6x – 9

If the tangent is parallel to X – axis then
\(\frac{d y}{d x}\) = 0, ⇒ 3x² – 6x – 9 = 0
⇒ x² – 2x – 3 = 0
x = 3 or – 1
When x = 3, then
y = 3³ – 3(3)² – 9(3) + 7 = – 20
When x = – 1, then y = (- 1)³ – 3(- 1)² – 9 = (-1) + 7 = 12
∴ The required points are (3, – 20), (-1, 12)

Question 8.
Find a point on the curve y = ( x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4, 4). (V.S.A.Q.)
Answer:
Equation of the curve is y = (x – 2)²
∴ \(\frac{d y}{d x}\) = 2(x – 2)
Slope of the chord joining points (2, 0) and (4, 4) is = \(\frac{4}{2}\) = 2
The tangent is parallel to the chord 2 (x – 2) = 2
⇒ 2x = 6
⇒ x = 3
y = (x – 2)² = (3 – 2)² = 1
The required point = (3, 1)

Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11. (V.S.A.Q.)
Answer:
Equation of the curve isy = x³ – 11x + 5
∴ \(\frac{d y}{d x}\) = 3x² – 11
The tangent is y = x – 11
Slope of the tangent is 3x² -11 = 1
⇒ 3x² = 12
⇒ x² = 4
⇒ x = ± 2
y = x – 11, if x = 2 ⇒ y = – 9
The point on the curve is (2, – 9)

Question 10.
Find the equations of all lines having slope zero which are tangents to the curve y = \(\frac{1}{x^2-2 x+3}\) (V.S.A.Q)
Answer:
Equation of the given curve is y = \(\frac{1}{x^2-2 x+3}\)
∴ \(\frac{d y}{d x}=\frac{-1(2 x-2)}{\left(x^2-2 x+3\right)^2}\)
Given slope of the tangent is ‘O’ .
\(\frac{-2(x-1)}{\left(x^2-2 x+3\right)^2}\) = 0 ⇒ x = 1
At x = 1 ⇒ y = \(\frac{1}{1-2+3}=\frac{1}{2}\)
The point is (1, \(\frac{1}{2}\))
Slope of the tangent is ‘0’
∴ Equation of the tangent is
y – \(\frac{1}{2}\) = 0 (x – 1) ⇒ 2y – 1 = 0

II.

Question 1.
Find the equations of tangent and normal to the following curves at the points indicated against. (S.A.Q.)
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5)
Answer:
\(\frac{d y}{d x}\) = 4x³ – 18x² + 26x – 10
At x = 0, slope of the tangent = – 10
∴ Equation of tangent is
y – 5 = – 10 (x – 0) = – 10x
⇒ 10x + y – 5 = 0
Slope of the normal = \(\frac{1}{10}\)
∴ Equation of normal is
y – 5 = \(\frac{1}{10}\)(x – 0) ⇒ 10y – 50 = x
⇒ x – 10y + 50 = 0

(ii) y = x³ at (1, 1)
Answer:
\(\frac{d y}{d x}\) = 3x²
At (1, 1), slope of the tangent = 3 (1)2 = 3
Equation of the tangent at P (1, 1) is
y – 1 = 3 (x – 1)
⇒ 3x – y – 2 = 0
Slope of the normal = –\(\frac{1}{3}\)

∴ Equation of the normal is
y – 1 = –\(\frac{1}{3}\)(x – 1)
⇒ 3y – 3 = -x + 1
⇒ x + 3y – 4 = 0

(iii) y = x² at (0, 0)
Equation of the curve is y = x²
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 2x
At (0, 0) the slope of the tangent is ‘O’ Equation of the tangent is y – 0 = 0 (x – 0)
⇒ y = o
Let equation of normal which is perpendicular to the tangent is of the form x = k
Since the normal passes through (0, 0) we have k = 0
∴ Equation of the normal is x = 0

(iv) x = cos t, y = sin t at t = \(\frac{\pi}{4}\)
Answer:

(v) y = x² – 4x + 2 at (4, 2)
Answer:
Equation of the curve is y = x² – 4x + 2
∴ \(\frac{d y}{d x}\) = 2x – 4
Slope of tangent at (4, 2) is 2(4) – 4 = 4
Equation of the tangent is y – 2 = 4(x – 4)
⇒ 4x – y – 14 = 0
Slope of the normal = \(\frac{-1}{4}\)
∴ Equation of the normal is
y – 2= \(\frac{-1}{4}\)(x – 4)
⇒ 4y – 8 = -x + 4
⇒ x + 4y – 12 = 0

(vi) y = \(\frac{1}{1+x^2}\) at (0, 1)
Answer:
Given equation of the curve is y = \(\frac{1}{1+x^2}\)
∴ \(\frac{d y}{d x}=\frac{-2 x}{\left(1+x^2\right)^2}\)
At (0, 1), slope of the tangent = 0

Equation of the tangent is y – 1 = 0 (x – 0)
⇒ y = 1
Slope of the normal = ∞

∴ Equation of the normal is
y – 1 = \(\frac{1}{0}\) (x – 0)
⇒ x = 0

Question 2.
Find the equations of tangent and normal to the curve xy = 10 at (2, 5). (S.A.Q.)
Answer:
Given equation of the curve is xy = 10
y = \(\frac{10}{x} \Rightarrow \frac{d y}{d x}=\frac{-10}{x^2}\)

Slope of the tangent at the point (2, 5) is \(-\frac{10}{4}=-\frac{5}{2}\)
Equation of the tangent is
y – 5 = \(-\frac{5}{2}\) (x-2)
2y – 10 = -5x + 10
⇒ 5x + 2y – 20 = 0

Slope of the normal is \(\frac{2}{5}\)
∴ Equation of normal at (2, 5) is
y – 5 = \(\frac{2}{5}\)(x – 2)
⇒ 5y – 25 = 2x – 4
⇒ 2x – 5y + 21 = 0

Question 3.
Find the equations of tangent and normal to the curve y = x³ + 4x² at (-1, 3). (S.A.Q.) (May 2014)
Answer:
Equation of the curve is y = x³ + 4x²
∴ \(\frac{d y}{d x}\) = 3x² + 8x
Slope of the tangent at (- 1, 3) is
= 3 (- 1)2 + 8 (- 1) = – 5
Equation of tangent at (- 1, 3) is
y – 3 = – 5 (x + 1)
⇒ 5x + y + 2 = 0

Slope of the normal at (- 1, 3) is \(\frac{1}{5}\)
∴ Equation of normal at (- 1, 3) is
y – 3 = \(\frac{1}{5}\)(x + 1)
⇒ 5y – 15 = x + 1
⇒ x – 5y + 16 = 0

Question 4.
If the slope of the tangent to the curve x² – 2xy + 4y = 0 at a point on it is \(\frac{-3}{2}\), then find the equations of tangent and normal at that point. (S.A.Q.)
Answer:
Equation of the curve is
x² – 2xy + 4y = 0
Differentiating with respect to x

⇒ 2x – 2y = – 3x + 6
⇒ 5x – 2y – 6 = 0 (2)
⇒ 2y = 5x – 6
P (x, y) is a point on the curve and x² – x (5x – 6) + 2 (5x – 6) = 0
⇒ x² – 5x² + 6x + 10x -12 = 0
⇒ -4x² + 16x- 12 = 0
⇒ – 4 (x² – 4x + 3) = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1) (x – 3) = 0

Case – (i): x = 1, then form (1)
1 – 2y + 4y = 0 ⇒ 2y + 1 = 0 ⇒ y = –\(\frac{1}{2}\)
∴ Required point is P(1, –\(\frac{1}{2}\))

Equation of tangent is
y + \(\frac{1}{2}=\frac{-3}{2}\)(x – 1)
⇒ 2y + 1 = -3x + 3
⇒ 3x + 2y – 2 = 0

Equation of normal is
y + \(\frac{1}{2}=\frac{2}{3}\)(x – 1)
⇒ \(\frac{2 y+1}{2}=\frac{2}{3}\)(x – 1)
⇒ 6y + 3 = 4x – 4
⇒ 4x – 6y – 7 = 0

Case – (ii): When x = 3
∴ From (1); 9 – 6y + 4y = 0
⇒ y = \(\frac{9}{2}\)

Equation of the tangent is
y – \(\frac{9}{2}=\frac{-3}{2}\)(x – 33)
⇒ 2y – 9 = -3x + 9
⇒ 3x + 2y – 18 = 0

Equation of the normal is
y – \(\frac{9}{2}=\frac{2}{3}\)(x – 3)
⇒ \(\frac{2 y-9}{2}=\frac{2}{3}\)(x – 3)
⇒ 6y- 27 = 4x- 12
⇒ 4x – 6y + 15 = 0

Question 5.
If the slope of the tangent to the curve y = x log x at a point on it is \(\frac{3}{2}\), then find the equations of tangent and normal at that point. (S.A.Q.)
Answer:
Equation of the curve is y = x log x
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x . \(\frac{1}{x}\) + log x = 1 + log x

Question 6.
Find the tangent and normal to the curve y = 2e^-x/3 at the point where the curve meets the Y-axis.
Answer:
Equation of the curve is y = 2e^-x/3
Equation of Y – axis is x = 0 ⇒ y = 2
Required point = (0, 2)
Equation of tangent at P (0, 2) is
y – 2 = \(\frac{-2}{3}\)(x – 0)
⇒ 3y – 6 = – 2x
⇒ 2x + 3y – 6 = 0
Equation of the normal is
y – 2 = y(x – 0)
⇒ 2y – 4 = 3x
⇒ 3x – 2y + 4 = 0

III.

Question 1.
Show that the tangent at P(x₁, y₁) on the curve curved √x + √y = √a is yy₁^{\(\frac{-1}{2}\)} – 2 +xx₁^{\(\frac{-1}{2}\)} = a^{\(\frac{1}{2}\)} (E.Q.) [June 2004]
Answer:
Equation of the curve is √x + √y = √a
Differentiating with respect to x,

Question 2.
At what points on the curve x2 – y2 = 2, the slopes of the tangents are equal to 2 ?(E.Q.)
Answer:
Given equation of the curve
x² – y² = 2 ……….(1)
Differentiating with respect to ‘x’
2x – 2y\(\frac{d y}{d x}\) = 0
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{y}}\)

∴ Slope of the tangent at P(x₁, y₁) = 2
Also \(\frac{x_1}{y_1}\) = 2 ⇒ x₁ = 2y₁

Since P(x,, y,) is a point on the curve

Question 3.
Show that the curves x² + y² = 2 and 3x² + y² = 4x have a common tangent at the point (1, 1). (E.Q.)
Answer:
Equations of the curves are
x² + y² = 2 ……….(1)
3x² + y² = 4x ……(2)
From (1) we have 2x + 2y \(\frac{d y}{d x}\) = 0
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-x}{y}\)
Slope of the tangent to the curve (1) at (1, 1) is m₁ = \(\frac{-1}{1}\) = -1

Also from (2), 6x + 2y = 4 dy
⇒ 2y\(\frac{d y}{d x}\) = 4 – 6x
⇒ \(\frac{d y}{d x}=\frac{4-6 x}{2 y}\)

Slope of the tangent to the curve (2) at (1, 1) is m₂ = \(\frac{4-6}{2}=-\frac{2}{2}\) = -1
The slope of the tangents to both the curves at P(1, 1) are same and pass through the same point (1,1).
∴ The given curves have a common tangent at P(1, 1).

Question 4.
At a point (x₁, y₁) on the curve x³ + y³ = 3axy show that the tangent is (x₁² – ay₁)x + (y₁² – ax₁) y = ax₁y₁ (E.Q.)
Answer:
Equation of the curve is x³ + y³ = 3axy
Differentiating w.r. to ‘x’

⇒ y (y₁² – ax₁) – y₁(y₁² – ax₁) = -x(x₁² – ay₁) – x₁(x₁² – ay₁)
⇒ x(x₁² – ay₁) + y(y₁² – ax₁) = -x(x₁² – ay₁) + x₁(x₁² – ay₁)
⇒ x(x₁² – ay₁) +y₁(y₁² – ax₁)
= x₁³ – ax₁y₁ + y₁³ – ax₁y₁
= x₁³ – ax₁y₁ + y₁³ – ax₁y₁
= x₁³ + y₁³ – 2ax₁y₁
= 3ax₁y₁ – 2ax₁y₁
= ax₁y₁
(∵ (x₁, y₁) is a point on the given curve and x₁³ + y₁³ = 3ax₁y₁)

Question 5.
Show that the tangent at the point P(2, – 2) on the curve y(1 – x) = x makes intercepts of equal length on the coordinate axes and the normal at P passes through the origin. (E.Q.)
Answer:
Given equation of the curve is y(1 – x) = x
⇒ y = \(\frac{x}{1-x}\)
∴ Differentiating w.r.t to ‘x’
\(\frac{d y}{d x}=\frac{(1-x)-x(-1)}{(1-x)^2}=\frac{1}{(1-x)^2}\)
At P(2, -2), slope of the tangent = \(\frac{1}{(1-2)^2}\) = 1

Equation of tangent at P(2, – 2) is
y + 2 = 1 (x – 2)
⇒ x – y = 4
⇒ \(\frac{x}{4}+\frac{y}{(-4)}\) = 1 (Intercepts form}
The tangent makes equal intercepts on the co-ordinate axes but they make intercepts with opposite sign.
Equation of normal at P is
y – y₁ = – 1 (x – x₁)
⇒ y + 2 = – 1(x – 2)
⇒ x + y = 0
This passes through the origin.

Question 6.
If the tangpnt at any point on the curve x^2/3+ y^2/3 – a^2/3 intersects the co-ordinate axes in A and B, then show that the length AB is aconstant (E.Q.) [Mar.’14,T3, ’08, ’07, ’05]
Answer:

Equation of the curve is
x^2/3+ y^2/3 = a^2/3 ……………..(1)
Differentiating w.r. to ‘x’

Question 7.
If the tangent at any point P on the curve xm y” = am+“ (mn 5* 0) meets the co-ordinate axes in A and B then show that AP : PB is a constant. (E.Q.)
Answer:
Given curve is x^myⁿ = a^m+n, (mn ≠ 0)
Differentiating with respect to ’x’
x^mny^n-1.\(\frac{d y}{d x}\) + yⁿ. mx^m-1 = 0
⇒ \(\frac{d y}{d x}=-\frac{y^n m x^{m-1}}{x^m n y^{n-1}}=-\frac{m}{n} \cdot \frac{y}{x}\)
Slope of the tangent at (x₁, y₁) = \(\frac{-\mathrm{my}_1}{\mathrm{nx}_1}\)

∴ Equation of the tangent at (x₁, y₁) is
y – y₁ = nx₁ (x – x₁)
⇒ nx₁y – nx₁y₁ = \(\frac{-m y_1}{n x_1}\)mxy₁ + mx₁y₁
⇒ mxy₁ + nx₁y = (m + n) x₁y₁

If P divides AB in the ratio m₁ : m₂ then co-ordinate of P are

∴ Point P divides AB in the ratio n : m and AP : PB = n : m = a constant.

Telangana TSBIE TS Inter 1st Year Accountancy Study Material 1st Lesson Book Keeping and Accounting Textbook Questions and Answers.

TS Inter 1st Year Accountancy Study Materia 1st Lesson Book Keeping and Accounting

Short Answer Questions:

Question 1.
State any 5 advantages of accounting.
Answer:
Advantages of Accounting : The following are the main advantages of Accounting :

1. Net Result of Business Operations:
Accounting provides the operational result (Profit and Loss) of business for a given period of time.

2. Ascertainment of Financial Position :
The proprietor requires a full picture of his financial position to plan for the next year’s business. Balance sheet provides the financial status of the business.

3. Facilitates Comparative Study:
Accounting provides the facility of comparative study of the various aspects of the business with that of previous year and helps the business man to take necessary decisions.

4. Control over Assets :
In the course of business, the proprietor acquires various assets like building, machinery, furniture etc., which are well protected by generating records.

5. Helps Management :
Accounting helps management on important issues like ascertainment of cost and price fixation of goods and services.

6. Evidence :
Accounting records act as an approved evidence in legal matters.

Question 2.
Give limitations of accounting.
Answer:
Limitations of Accounting : The following are the limitations of Accounting :

1. Records only monetary transactions :
Accounting considers monetary transactions only, non-monetary transactions like skills of human resources, quality, organization culture, units of production sales etc., are ignored in accounting.

2. Historical in nature :
Accounting considers only historical transactions, i.e., transactions which have occurred in the past only recorded in accounting books. Transactions relating to future estimations or forecasts are not considered in accounting.

3. Price level changes are not considered :
Accounting does not consider price level changes which may occur from time to time, thus, it doesn’t reflect the current position.

4. Does not provide realistic information :
Accounting may not provide realistic information which in turn affects the overall results of the business concern.

Question 3.
State any 5 basic differences between Book-keeping and Accounting.
Answer:
The following are the differences between Book-keeping and Accounting.

Question 4.
Explain the steps involved in Accounting process.
Answer: Accounting process involves the following steps :
1. Identifying :
Identifying the business transaction from the source document.

2. Recording :
The next step of accounting process is to keep a systematic record of all business transactions, in orderly manner, soon after their occurence in the journal or subsidary books.

3. Classifying :
This is concerned with the classification of the recorded business transactions so as to group the similar transactions at one place i.e., in ledger by extracting the balance or total of accounts.

4. Summarising:
It is the process of finding the total of balances of all accounts so as to prepare trial balance.

5. Reporting :
The information from the trial balance is used to prepare profit and loss account and a balance sheet in a manner useful to uses of accounting information.

6. Analysing:
It establishes the relationship between the items of profit and loss account and balance sheet. The purpose of analysing is to identify the financial strength and weakness of. the business enterprise.

7. Interpreting :
It is concerned with explaining the meaning and significance of the relationship so established by the analysis. Interpretation should be useful to the users, so as enable them to take correct decisions.

Question 5.
State the objectives of accounting.
Answer:
The main objectives of accounting are :

1. To maintain accounting records.
2. To find out the result of operations.
3. To ascertain the financial position.
4. To communicate the information to users.

Question 6.
State the general features of IFRS.
Answer: IFRS :
1. International Financial Reporting Standards (IFRS) are the standards issued by IFRS Foundation and the International Accounting Standards Board (IASB).
2. It provides a common / uniform global language for business affairs, so that the company accounts are understood and compared across international boundaries.
3. IFRS Standards have been steadily replacing the accounting standards of many countries. At present 160 countries are implementing IFRS.
4. The General Features of IFRS include :

1. Fair presentation and compliance with IFRS.
2. Going concern.
3. Accrual basis of accounting.
4. Materiality and aggregation.
5. Off-setting allowed only under specific conditions.
6. Frequency of reporting.
7. Compare the information.
8. Consistancy of presentation.

Question 7.
Briefly explain any 5 concepts of accounting. –
Answer:
The term concept means an idea or thought. Basic Accounting concepts are the fundamental ideas or basic assumptions underlying the theory and practice of financial accounting. These concepts are termed as Generally Accepted Accounting Principles.

1. Business Entity Concept :
Business is treated separate from the proprietor. All the transactions are recorded in the books of business but not in the books of proprietor. The proprietor is also treated as creditor of the business. When he contributes capital he is treated as a person who has invested his amount in the business. Therefore, capital appears on the liabilities side of the balance sheet of the business.

2. Going Concern Concept:
The concept relates with the long life of the business. The assumption is that business will continue to exist for unlimited period unless it is dissolved due to some reason or other. When the final accounts are prepared, recording is made for outstanding expenses and prepaid expenses because of the assumption that business will continue.

3. Cost Concept:
According to this concept, an asset is recorded at cost i.e., the price which is paid at the time of acquiring it, in the books of account. In Balance Sheet, these assets appear not at cost price every year but depreciation is deducted and they appear at the amount which is cost less depreciation. Under this concept, all such events are ignored which affect the business but no cost. Ex. Death of a director.

4. Accounting Period Concept :
Every business man wants to know the result of his investment and efforts after a certain period. Usually one year period is regarded as ideal for this purpose. It may be 6 months or 2 years also. This period is called accounting period.

It depends upon the nature of business and object of the proprietor of business. From taxation point of view one year period is necessary as income tax is payable every year.

5. Duel Aspect Concept:
Under this concept, every transaction has got a two fold aspect,
i) Receiving the benefit and ii) Giving of that benefit. For instance, when a firm acquires an asset, (receiving the benefit) it must pay cash (giving of that benefit).

Therefore, two accounts are to be passed in the books of accounts, one for receiving the benefit and other for giving the benefit. Thus, there will be a double entry for every transaction – debit for receiving the benefit and credit for giving the benefit.

Question 8.
Briefly explain accounting conventions.
Answer:
Accounting conventions are customs or traditions guiding the preparation of accounts. They are adopted to make financial statements clear and meaningful.

The following are the four accounting conventions :
1. Convention of disclosure:
Accounting statements should disclosefully and completely all the significant information, based on which decisions are taken by various interest parties. It involves proper classification and explanation of accounting information which are published in financial statements.

2. Convention of materiality :
According to this conventions only those events should be recorded which have a significant bearing and insignificant things should be ignored. The avoidence of insignificant. Things will not materially affect the records of the business.

3. Convention of consistency :
The convention of consistency facilitates comparison of performance of a business unit from one accounting period to another is possible when the accounting principles followed by the firm are consistently applied over the years. Ex.: An organisation should not change the method of depreciation or valuation of stocks every year.

4. Convention of conservatism:
According to this convention, the principle of anticipate no profit but provide for all possible losses” should be applied. The principle of conservatism requires that in the situation of uncertainly of doubt, the business – transactions should be recorded in such a manner that the profits and assets are not overstated and the losses and liabilities are not understated.

Question 9.
Write a brief note on accounting standards.
Answer:

1. To promote world-wide uniformity in published accounts, the International Accounting Standards Committee (IASQ has been set up in June 1973 with nine nations as founder members. The purpose of this committee is to formulate and publish in public interest, standards to be observed in the presentation of audited financial statements and to promote their world-wide acceptance and observance.
2. In our country, the Institute of Chartered Accountants of India (ICAI) has constituted Accounting Standard Board (ASB) in 1977. The ASB has been empowered to formulate and issue accounting standards that should be followed by all business concerns in India.
3. Accounting Standard is a principle that guides and standardizes accounting practices. Accounting standards ensures uniformity in presentation of financial statements and facilitates interfirm comparison within the industry. They also ensure creditability and reliability of financial statements.,
4. At present there are 35 Indian. Accounting Standards (IAS) and are mandatory and to be complied with from the date from which they come into force in presenting audited financial statements.

Very Short Answer Questions:

Question 1.
What is transaction ?
Answer:
Transactions :

1. Transactions are those activities of a business, which involve transfer of money or goods or services between two persons or two accounts.
2. For example, purchase of goods, sale of goods, borrowing from bank etc.
3. Transactions are of two types namely cash and credit transactions. Every transaction brings about change in the financial position of business.

Question 2.
What is Book-keeping ?
Answer:

1. Book-keeping is the art of recording business transactions in regular and systematic manner.
2. According to Carter “Book-keeping is the science and art of correctly recording the books of accounts all those business transactions that result in transfer of money or money’s worth.

Question 3.
Define Accounting.
Answer:

1. Accounting is called as the “Language of Business”.
2. The American Institute of Public Accountants defined accounting as “the art of recording, classifying and summerising in significant manner and interms of money transactions and events which are in part, atleast of financial character and interpreting the results thereof.

Question 4.
What is Accounting cycle ?
Answer:

1. An accounting cycle is a complete sequence of accounting process that begins with the recording of business transactions and ends with the preparation of final accounts.
2. These include journal, ledger, trial balance and financial statements such as trading account, profit and loss account and balance sheet.

Question 5.
What is an Accounting Standard ?
Answer:

1. Accounting Standard is a principle that guides and standardises accounting practices.
2. Accounting Standards are necessary so that the financial statements are meaningful across wide variety of businesses, otherwise, the accounting rules of different companies would make comparison almost impossible.
3. At present there are 35 Indian Accounting Standards.

Question 6.
What is IFRS ?
Answer:

1. nternational Financial Reporting Standards (IFRS) are the standards issued by IFRS Foundation and the International Accounting Standards Board (IASB).
2. It provides a common / uniform global language for business affairs, so that the company accounts are understood and compared across international boundaries.
3. IFRS Standards have been steadily replacing the accounting standards of many countries. At present 160 countries are implementing of IFRS.

Question 7.
What is GAAP ?
Answer:

1. Generally accepted Accounting Principles (GAAP) defined as those rules of action / conduct which are derived from experience and practice.
2. Generally Accepted Accounting Principle (GAAP) should be Relevance, . Reliable and should ensure feasibility.

Question 8.
What is an accounting concept ?
Answer:

1. Accounting Concepts are the necessary assumptions, conditions or postulates upon which the accounting is based.
2. They are developed to facilitate communication of the accounting and financial information to all the users of the financial statements.
3. Some of the Accounting Concepts are: Business, entity concept, dual aspect concept, going concern concept, money measurement concept etc.

Question 9.
What is an accounting convention ?
Answer:
1. Accounting conventions are the customs or traditions guiding the preparation of accounts.
2. They are adopted to make financial statements clear and meaningful.
3. Following are the four accounting conventions :

1. Convention of Disclosure.
2. Convention of Materiality.
3. Convention of Consistency; and
4. Convention of Conservatism.

Question 10.
Explain convention of conservatism.
Answer:

1. According to this convention, the principle of anticipate no profit but provide for all possible losses should be applied.
2. The principle of conservatism requires that in the situation of uncertainity of doubt, the business transactions should be recorded in such a manner that the profits and assets are not overstated and the losses and liabilities are not understated.

Question 11.
Explain convention of consistency.
Answer:

1. The convention of consistency facilitates comparison of performance of a business firm from one accounting period to another is possible when the accounting principles followed by the firm are consistently applied over the years.
2. Ex: An organisation should not change the method of depreciation or valuation of stocks every year.

Question 12.
What is Matching concept ?
Answer:

1. Matching the revenues earned during an accounting period with the cost associated with the period to ascertain the result of the business concern is called the Matching concept.
2. According to this concept, incomes are to be identified with their corresponding expenses or vice versa in a given period of time.

Question 13.
Explain business entity concept of accounting.
Answer:

1. Business is treated separate from the’proprietor. All the transactions are recorded in the books of business but not in the books of proprietor.
2. The proprietor is also treated as creditor of the business. When he contributes capital he is treated as a person who has invested his amount in the business. Therefore, capital appears on the liabilities side of the balance sheet of the business.

Question 14.
Explain money measurement concept
Answer:

1. Only those transactions are recorded in accounting which can be expressed interms of money. The transactions which cannot be expressed in money is beyond the scope of accounting.
2. Receipt of income, payment of expenses, purchase of assets etc., are monetary transactions that are recorded in the books of account. Where as in the event of breakdown of machinery is not recorded because it has no monetary value.

Additional Questions:

Question 1.
What is Double entry system and explain its features.
Answer:
Every business transaction involves a transfer and as such consists of two aspects.

1. The receiving aspect
2. The giving aspect. It is necessary to note that these aspects go together, because receiving necessarily implies giving and vice-versa. The record of any business transaction will be complete only when both these aspects are recorded. The recording of the two aspects of each transaction is known as “Double entry system of book-keeping”.

Features :

1. Every transaction has two aspects i.e., receiving the benefit and giving the benefit.
2. Every transaction affects two accounts.
3. Double entry system is based upon the principles and concepts of accounting.
4. It helps in the preparation of trial balance which is a test arithmetical accuracy in accounts.
5. If facilitates the preparation of final accounts with the help of trial balance.

Question 2.
Explain the advantages of Double entry system.
Answer:
The following are the advantages of Double entry system of book-keeping.

1. The system maintains a complete record of all the business transactions, as it records both the aspects of the transaction.
2. It provides a check on the arithmatical accuracy of accounts with the help of trial balance.
3. Errors and frauds can be detected under this system. So, it reduces the chance of committing errors and frauds.
4. It reveals the results of the operations i.e., Profit or loss, by preparing and loss a/c.
5. The financial position of the business can be ascertained through the preparation of the balance sheet.
6. It is a scientific system and permits the accounts to be kept in a detailed form and provides sufficient information for the purpose of control.
7. The results of one year can be compared with the results of the previous year and reasons for the changes are known.
8. It provides accounting information readily to the management for making decisions.

Question 3.
Explain Double entry system.
Answer:

1. The procedure of recording both the receiving and giving aspects of the transaction is called Double entry system of book-keeping.
2. The fundamental rule in double entry is that for every debit there must be a corresponding value of credit.

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Direction Cosines and Direction Ratios Important Questions

Question 1.
Find the direction cosines of two lines that are connected by the relations l + m + n = 0 and mn – 2nl – 2lm = 0. [Mar. ’17 (TS), ’11, ’04; May ’15 (TS)]
Solution:
Given that, l + m + n = 0 …….(1)
mn – 2nl – 2lm = 0 ……….(2)
From (1), l = -m – n
Substituting (l) in (2),
mn – 2n(-m – n) – 2(-m – n)m = 0
mn + 2mn + 2n² + 2m² + 2mn = 0
2m² + 5mn + 2n² = 0
2m² + 4mn + mn + 2n² = 0
2m(m + 2n) + n(m + 2n) = 0
(m + 2n)(2m + n) = 0
m + 2n = 0 ……..(3)
2m + n = 0 ………(4)
Solving (1) & (3)

Question 2.
Find the direction cosines of two lines that are connected by the relations l – 5m + 3n = 0 and 7l² + 5m² – 3n² = 0. [Mar. ’18 (AP); Mar. ’16 (TS); May ’09]
Solution:
Given that,
l – 5m + 3n = 0 ……….(1)
7l² + 5m² – 3n² = 0 …………(2)
From (1), l = 5m – 3n
Substituting ‘l’ in (2)
7(5m – 3n)² + 5m² – 3n² = 0
7(25m² + 9n² – 30mn) + 5m² – 3n² = 0
175m² + 63n² – 210mn + 5m² – 3n² = 0
180m² – 210mn + 60n² = 0
6m² – 7mn + 2n² = 0
2m(3m – 2n) – n(3m – 2n) = 0
(3m – 2n) (2m – n) = 0
3m – 2n = 0 ………(3)
2m – n = 0 ………. (4)
Solving (1) & (3)

Question 3.
Show that the lines whose direction cosines are given by l + m + n = 0, 2mn + 3nl – 5lm = 0 are perpendicular to each other. [Mar. ’16 (AP); ’12]
Solution:
Given that, l + m + n = 0 ……….(1)
2mn + 3nl – 5lm = 0 ……….(2)
From (1), l = -m – n
Substituting in (2)
2mn + 3n(-m – n) – 5(-m – n)m = 0
2mn – 3mn – 3n² + 5m² + 5mn = 0
5m² + 4mn – 3n² = 0
If ax² + 2hxy + by² = 0 represents a pair of straight lines then the two lines are


Direction ratio’s of 2nd line are (a2, b2, c2) = (-3 – √19, -2 + √19, 5)
Now, a₁a₂ + b₁b₂ + c₁c₂ = (-3 + √19) (-3 – √19) + (-2 – √19)(-2 + √19) + 5(5)
= (-3)² – (√19)² + (-2)² – (√19)² + 25
= 9 – 19 + 4 – 19 + 25
= 38 – 38
= 0
∴ The two lines are perpendicular to each other.

Question 4.
Find the angle between the lines whose direction cosines satisfy the equations l + m + n = 0, l² + m² – n² = 0. [Mar. ’17 (AP), ’13, ’07 ; May ’14, ’13 (Old), ’11, ’07, ’04; Mar. ’19 (AP&TS)]
Solution:
Given that
l + m + n = 0 ……….(1)
l² + m² – n² = 0 ……….(2)
From (1), l = -m – n
Substituting in (2)
(-m – n)² + m² – n² = 0
m² + n² + 2mn + m² – n² = 0
2m² + 2mn = 0
2m(m + n) = 0
2m = 0
m = 0 ………(3)
m + n = 0 ……….(4)
Solving (1) & (3)

∴ The direction ratios of 1st line are (a₁, b₁, c₁) = (-1, 0, 1).
Solving (1) & (4)

∴ The direction ratios of the 2nd line are (a₂, b₂, c₂) = (0, -1, 1).
Now, if ‘θ’ is the angle between these two lines then

∴ θ = 60°

Question 5.
Find the angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0. [Mar. ’12, ’10, ’06; May ’15 (AP); ’10]
Solution:
Given that,
3l + m + 5n = 0 ……….(1)
6mn – 2nl + 5lm = 0 ………..(2)
From (1), m = -3l – 5n
Substituting in (2)
6n(-3l – 5n) – 2ln + 5l(-3l – 5n) = 0
-18ln – 30n² – 2ln – 15l² – 25ln = 0
-15l² – 45ln – 30n² = 0
l² + 3ln + 2n² = 0
l² + 2ln + ln + 2n² = 0
(l + n)(l + 2n) = 0
l + 2n = 0 ………(3)
l + n = 0 ……….(4)
Solving (1) & (3)

∴ Direction ratio s of 1st line are (a₁, b₁, c₁) = (2, -1, -1)
Solving (1) & (4)

∴ Direction ratio’s of 2nd line are (a₂, b₂, c₂) = (1, 2, -1)
Now, if ‘θ’ is the angle between these two lines then

Question 6.
If a ray makes angles α, β, γ, δ with the four diagonals of a cube, find cos²α + cos²β + cos²γ + cos²δ. [Mar. ’08, ’05; May ’05; B.P.]
Solution:

Let one vertex of the cube as origin ‘O’ and 3 co-terminus edges OA, OB, OC as the coordinate axes.
Let ‘A’ be the edge of the cube. Then the coordinates of the vertices of the cube are O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), D(a, 0, a), E (a, a, 0), F(a, a, a), G (0, a, a)
The four diagonals are \(\overline{\mathrm{OF}}, \overline{\mathrm{CE}}, \overline{\mathrm{AG}}, \overline{\mathrm{BD}}\).
Direction Ratios of \(\overline{\mathrm{OF}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (a – 0, a – 0, a – 0)
= (a, a, a)
Direction cosines of \(\overline{\mathrm{OF}}\) are

Direction ratios of \(\overline{\mathrm{CE}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (a – 0, a – 0, 0 – a)
= (a, a, -a)
Direction cosines of \(\overline{\mathrm{CE}}\) are

Direction ratios of \(\overline{\mathrm{AG}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (0 – a, a – 0, a – 0)
= (-a, a, a)
Direction cosines of \(\overline{\mathrm{AG}}\) are

Direction ratios of \(\overline{\mathrm{BD}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (a – 0, 0 – a, a – 0)
= (a, -a, a)
Direction cosines of \(\overline{\mathrm{BD}}\) are

Let (l, m, n) be the direction cosines of the line which makes angles α, β, γ, δ with the four diagonals \(\overline{\mathrm{OF}}, \overline{\mathrm{CE}}, \overline{\mathrm{AG}}, \overline{\mathrm{BD}}\) of the cube.
Similarly,

Question 7.
Find the angle between two diagonals of a cube. [Mar. ’18 (TS); Mar. ’15 (AP); May ’13]
Solution:

Let one vertex of the cube as origin ‘O’ and 3 co-terminus edges OA, OB, OC as the coordinate axes.
Let ‘A’ be the edge of the cube. Then the coordinates of the vertices of the cube are O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), D(a, 0, a), E (a, a, 0), F(a, a, a), G (0, a, a)
Direction ratios of \(\overline{\mathrm{OF}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (a – 0, a – 0, a – 0)
= (a, a, a)
Direction ratios of \(\overline{\mathrm{BD}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (a – 0, 0 – a, a – 0)
= (a, -a, a)
If ‘θ’ is the angle between diagonals \(\overline{\mathrm{OF}}\) & \(\overline{\mathrm{BD}}\) then,
cos θ = \(\frac{\left|\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2\right|}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2+\mathrm{c}_1^2} \sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)

Similarly, the angle between any pair of diagonals can be found to be \(\cos ^{-1}\left(\frac{1}{3}\right)\).

Question 8.
The vertices of a triangle are A(1, 4, 2), B(-2, 1, 2), C(2, 3, -4). Find ∠A, ∠B, ∠C.
Solution:
A(1, 4, 2), B(-2, 1, 2), C(2, 3, -4) are the given vertices of a triangle.

Direction ratios of \(\overline{\mathrm{AB}}\) are (-2 – 1, 1 – 4, 2 – 2) = (-3, -3, 0) = (1, 1, 0)
Direction ratios of \(\overline{\mathrm{BC}}\) are (2 + 2, 3 – 1, -4 – 2) = (4, 2, -6) = (2, 1, -3)
Direction ratios of \(\overline{\mathrm{CA}}\) are (1 – 2, 4 – 3, 2 + 4) = (-1, 1, 6)
If A is the angle between the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{CA}}\) then,

∴ A = 90°
If B is the angle between the sides \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{BC}}\) then,
cos B = \(\left|\frac{1(2)+1(1)+0(-3)}{\sqrt{1^2+1^2+0^2} \sqrt{2^2+1^2+(-3)^2}}\right|\)

If c is the angle between the sides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\) then,

Question 9.
If P(2, 3, -6), Q(3, -4, 5) are two points, find the direction cosines of \(\overline{\mathbf{O P}}, \overline{\mathbf{Q O}}\) and \(\overline{\mathbf{P Q}}\) where ‘O’ is the origin. [Mar. ’00]
Solution:
O(0, 0, 0), P(2, 3, -6), Q(3, -4, 5) are the given points.
Direction ratios of \(\overline{\mathbf{OP}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (2 – 0, 3 – 0, -6 – 0)
= (2, 3, -6)
= (a, b, c)
Direction cosines of \(\overline{\mathbf{OP}}\) are

Direction ratios of \(\overline{\mathbf{OQ}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (-3 + 0, +4 + 0, -5 + 0)
= (-3, +4, -5)
= (a, b, c)
Direction cosines of \(\overline{\mathbf{OQ}}\) are

Direction ratios of \(\overline{\mathbf{PQ}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (3 – 2, -4 – 3, 5 + 6)
= (1, -7, 11)
= (a, b, c)
Direction cosines of \(\overline{\mathbf{PQ}}\) are

Question 10.
A ray makes angles \(\frac{\pi}{3}, \frac{\pi}{3}\) with \(\overline{\mathbf{OX}}\) and \(\overline{\mathbf{OY}}\) respectively. Find the angle made by it with \(\overline{\mathbf{OZ}}\). [May ’02]
Solution:
Let α, β, γ be the angles made by the line with the positive x, y, z axis respectively.
Given that, α = 60°, β = 60°
But, cos²α + cos²β + cos²γ = 1
cos²60° + cos²60° + cos²60° = 1

γ = 45° (or) γ = 135°

Question 11.
Find the direction cosines of the line joining the points (-4, 1, 7) and (2, -3, 2). [May ’02]
Solution:
Let A = (-4, 1, 7), B = (2, -3, 2) are the given points.
Direction ratios of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (2 + 4, -3 – 1, 2 – 7)
= (6, -4, -5)
= (a, b, c)
Direction cosines of \(\overline{\mathrm{AB}}\) are

Question 12.
Find the direction ratios and direction cosines of the line joining the points (4, -7, 3), (6, -5, 2). [Mar’ 00]
Solution:
Let A = (4, -7, 3), B = (6, -5, 2) are the given points.
Direction ratios of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (6 – 4, -5 + 7, 2 – 3)
= (2, 2, -1)
= (a, b, c)
Direction cosines of \(\overline{\mathrm{AB}}\) are

Question 13.
Show that the line joining the points A(2, 3, -1) and B(3, 5, -3) is perpendicular to the lines joining C(1, 2, 3) and D(3, 5, 7). [Mar. ’03]
Solution:
A = (2, 3, -1), B = (3, 5, -3), C = (1, 2, 3), D = (3, 5, 7) are the given points.
Direction ratios of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (3 – 2, 5 – 3, -3 + 1)
= (1, 2, -2)
= (a₁, b₁, c₁)
Direction ratios of \(\overline{\mathrm{CD}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (3 – 1, 5 – 2, 7 – 3)
= (2, 3, 4)
= (a₂, b₂, c₂)
Now, a₁a₂ + b₁b₂ + c₁c₂ = 1(2) + 2(3) + (-2) 4
= 2 + 6 – 8
= 8 – 8
= 0
∴ \(\overline{\mathrm{AB}}\) is perpendicular to \(\overline{\mathrm{CD}}\).

Question 14.
Find the angle between the lines whose direction ratio’s are (1, 1, 2) (√3, -√3, 0).
Solution:
Given that, the direction ratios of two lines are (a₁, b₁, c₁) = (1, 1, 2) and (a₂, b₂, c₂) = (√3, -√3, 0)
If ‘θ’ is the angle between these two lines then
cos θ = \(\frac{\left|a_1 a_2+b_1 b_2+c_1 c_2\right|}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\)

Question 15.
Find the angle between \(\overline{\mathrm{DC}}\) and \(\overline{\mathrm{AB}}\) where A = (3, 4, 5), B = (4, 6, 3), C = (-1, 2, 4) and D = (1, 0, 5). [Mar. ’06]
Solution:
A = (3, 4, 5), B = (4, 6, 3), C = (-1, 2, 4), D = (1, 0, 5) are the given points.
Direction ratios of \(\overline{\mathrm{AB}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (4 – 3, 6 – 4, 3 – 5)
= (1, 2, -2)
= (a₁, b₁, c₁)
Direction ratios of \(\overline{\mathrm{DC}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (1 + 1, 0 – 2, 5 – 4)
= (2, -2, 1)
= (a₂, b₂, c₂)
If ‘θ’ is the angle between the lines \(\overline{\mathrm{AB}}\) & \(\overline{\mathrm{DC}}\) then,

Some More Maths 1B Direction Cosines and Direction Ratios Important Questions

Question 1.
Find the direction cosines of a line that makes equal angles with the axes.
Solution:
If α, β, γ be the angles made by the line with the axis.
Since the line is equally inclined to the axis, then α = β = γ then
cos α = cos β = cos γ
We know that,
cos²α + cos²β + cos²γ = 1
cos²α + cos²α + cos²α = 1
3 cos²α = 1
cos²α = \(\frac{1}{3}\)
cos α = \(\pm \frac{1}{\sqrt{3}}\)
∴ The direction cosines of the lines are \(\left(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right)\)
The number of lines that are equally inclined to the axes is equal to the no.of different combinations of signs with l, m, n are possible = 4.

Question 2.
If the directions of a line are \(\left(\frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}\right)\), find ‘c’.
Solution:
Given that, the direction cosines of a line are (l, m, n) = \(\left(\frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}, \frac{1}{\mathrm{c}}\right)\)
We know that,
l² + m² + n² = 1
\(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}\)
\(\frac{3}{\mathrm{c}^2}\) = 1
c² = 3
c = ±√3

Question 3.
If a line makes angles α, β, γ with the positive directions of x, y, and z-axes, what is the value of sin²α + sin²β + sin²γ?
Solution:
Since α, β, γ are the angles made by a line with the positive direction of co-ordinates axes, then
cos²α + cos²β + cos²γ = 1
Now, sin²α + sin²β + sin²γ
= (1 – cos²α) + (1 – cos²β) + (1 – cos²γ)
= 3 – (cos²α + cos²β + cos²γ)
= 3 – 1
= 2

Question 4.
Show that the line joining the points P(0, 1, 2) and Q(3, 4, 8) is parallel to the line joining the points R(-2, \(\frac{3}{2}\), -3) and S(\(\frac{5}{2}\), 6, 6).
Solution:
P = (0, 1, 2), Q = (3, 4, 8), R = (-2, \(\frac{3}{2}\), -3), S = (\(\frac{5}{2}\), 6, 6) are the given points.
Direction ratios of \(\overline{\mathrm{PQ}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (3 – 0, 4 – 1, 8 – 2)
= (3, 3, 6)
= (a₁, b₁, c₁)
Direction ratios of \(\overline{\mathrm{RS}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)

∴ \(\overline{\mathrm{PQ}}\) is parallel to \(\overline{\mathrm{RS}}\).

Question 5.
A(1, 8, 4), B(0, -11, 4), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the coordinates of D.
Solution:
A(1, 8, 4), B(0, -11, 4), C(2, -3, 1) are the given points.
D is the foot of the perpendicular from A to BC.
Suppose D divides \(\overline{\mathrm{BC}}\) in the ratio m : n then the co-ordinate of D.

Question 6.
‘O’ is the origin, P(2, 3, 4) and Q(1, k, 1) are points such that \(\overline{\mathrm{OP}} \perp \overline{\mathbf{O Q}}\). Find k.
Solution:
O(0, 0, 0), P(2, 3, 4), Q(1, k, 1) are the given points.
Direction ratio’s of \(\overline{\mathrm{OP}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (2 – 0, 3 – 0, 4 – 0)
= (2, 3, 4)
= (a₁, b₁, c₁)
Direction ratio’s of \(\overline{\mathrm{OQ}}\) are (x₂ – x₁, y₂ – y₁, z₂ – z₁)
= (1 – 0, k – 0, 1 – 0)
= (1, k, 1)
= (a₂, b₂, c₂)
Since \(\overline{\mathrm{OP}} \perp \overline{\mathbf{O Q}}\) then,
a₁a₂ + b₁b₂ + c₁c₂ = 0
2(1) + 3(k) + 4(1) = 0
2 + 3k + 4 = 0
6 + 3k = 0
3k = -6
k = -2

Question 7.
If (l₁, m₁, n₁), (l₂, m₂, n₂) are the direction cosines of two intersecting lines, show that the direction cosines of two lines bisecting the angles between them are proportional to l₁ ± l₂, m₁ ± m₂, n₁ ± n₂.
Solution:

Let the two lines intersect at the origin ‘O’.
Let A and C be two different points on one line and B be a point on the other line such that
OA = OB = OC = 1
The co-ordinates of A, B, C are A(l₁, m₁, n₁), B(l₂, m₂, n₂), C(-l₁, -m₁, n₁)
Let P, Q be the midpoints of \(\overline{\mathrm{AB}} \& \overline{\mathrm{BC}}\) respectively.
Then OP & OQ are the required bisectors.
Since, P is the midpoint of \(\overline{\mathrm{AB}}\) then


∴The directions of the bisectors are proportional to (l₁ ± l₂, m₁ ± m₂, n₁ ± n₂)

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Pair of Straight Lines Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Pair of Straight Lines Important Questions

Question 1.
If ‘θ’ is the acute angle between the lines represented by ax² + 2hxy + by² = 0 then, show that cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^2+4 h^2}}\). [Mar. ’18 (TS); Mar. ’06; May ’97]
Solution:
Let ax² + 2hxy + by² = 0 represent the lines
l₁x + m₁y = 0 ……..(1)
l₂x + m₂y = 0 ……..(2)

The combined equation of lines (1) & (2) is
ax² + 2hxy + by² = (l₁x + m₁y)(l₂x + m₂y)
= l₁l₂x² + m₁m₂xy + l₂m₁xy + m₁m₂y²
= l₁l₂x² + (l₁m₂ + l₂m₁)xy + m₁m₂y²
Comparing on both sides we get l₁l₂ = a, l₁m₂ + l₂m₁ = 2h, m₁m₂ = b
If θ is an angle between the lines (1) & (2)

Question 2.
If the equation ax² + 2hxy + by² = 0 represents a pair of lines then prove that the equation of the pair of angular bisectors is h[x² – y²] = (a – b) xy. [Mar. ’18 (AP); May ’13, ’09, ’96, ’91; Mar. ’13(old), ’09, ’00, ’95. ’92, ’90]
Solution:
Let ax² + 2hxy + by² = 0 represent a pair of straight lines
l₁x + m₁y = 0 ……(1)
l₂x + m₂y = 0 ……(2)

The combined equation of (1) & (2) is
ax² + 2hxy + by² = (l₁x + m₁y)(l₂x + m₂y)
= l₁l₂x² + l₁m₂xy + l₂m₁xy + m₁m₂y²
= l₁l₂x² + (l₁m₂ + l₂m₁)xy + m₁m₂y²
Comparing on both sides we get a = l₁l₂, 2h = l₁m₂ + l₂m₁, b = m₁m₂
The equation for the bisectors of angles between (1) & (2) is

Question 3.
Show that the product of the perpendicular distances from a point (α, β) to the pair of straight lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^2+2 h \alpha \beta+b \beta^2\right|}{\sqrt{(a-b)^2+4 h^2}}\). [May ’15 (AP); May ’14, ’13 (Old); ’11, ’08, ’07: Mar. ’08, ’07, ’04, ’01]
Solution:
Let ax² + 2hxy + by² = 0 represents two lines (1)&(2)
l₁x + m₁y = 0 …..(1), l₂x + m₂y = 0 ……(2)

The combined equation of (1) & (2) is ax² + 2hxy + by²
= (l₁x + m₁y)(l₂x + m₂y)
= l₁l₂x² + l₁m₂xy + l₂m₁xy + m₁m₂y²
= l₁l₂x² + (l₁m₂ + l₂m₁)xy + m₁m₂y²
Comparing on both sides we get a = l₁l₂, 2h = l₁m₂ + l₂m₁, b = m₁m₂
The length of the perpendicular from (α, β) to line (1) is

Question 4.
Show that the area of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^2 \sqrt{h^2-a b}}{\left|a m^2-2 h l m+b l^2\right|}\). [Mar. ’13, ’02; May ’15(TS), ’10, ’98, ’94, ’92; B.P; Mar. ’17 (AP & TS); Mar. ’19 (TS)]]
Solution:
Let ax² + 2hxy + by² = 0 represents two lines l₁x + m₁y = 0 …….(1), l₂x + m₂y = 0 ……..(2)

The combined equation of (1) & (2) is
ax² + 2hxy + by² = (l₁x + m₁y)(l₂x + m₂y)
= l₁l₂x² + l₁m₂xy + l₂m₁xy + m₁m₂y²
= l₁l₂x² + (l₁m₂ + l₂m₁)xy + m₁m₂y²
Comparing on both sides we get a = l₁l₂, 2h = l₁m₂ + l₂m₁, b = m₁m₂
Let the given line be lx + my + n = 0 ……..(3)
Clearly, the origin O is the point of intersection of (1) & (2)
∴ O = (0, 0)
Let A be the point of intersection of (1) & (3)

Question 5.
Find the centroid and area of the triangle formed by the lines 12x² – 20xy + 7y² – 0 and 2x – 3y + 4 = 0. [Mar. ’05, ’90, ’83; May ’87]
Solution:
Given the equation of the pair of lines is
12x² – 20xy + 7y² = 0
12x² – 6xy – 14xy + 7y² = 0
6x(2x – y) – 7y(2x – y) = 0
(2x – y)(6x – 7y) = 0
2x – y = 0 …….(1), 6x – 7y = 0 ……..(2)
The third equation is 2x – 3y + 4 = 0 ……(3)
∴ Vertex O: The point of intersection of (1) & (2) is O = (0, 0)

Question 6.
Show that the lines represented by (lx + my)² – 3(mx – ly)² = 0 and lx + my + n = 0 form an equilateral triangle with area \(\frac{n^2}{\sqrt{3}\left(l^2+m^2\right)}\). [Mar. ’15 (TS), ’91]
Solution:
Given equation of the pair of lines is (lx + my)² – 3(mx – ly)² = 0
(lx + my)² – [√3(mx – ly)²] = 0
(lx + my + √3(mx – ly)) (lx + my – √3(mx – ly)) = 0
(lx + my + √3mx – √3ly) (lx + my – √3mx + √3ly) = 0
[(l + √3m)x + (m – √3l)y] [(l – √3m)x + (m + √3l)y] = 0
(l + √3m)x + (m – √3l)y = 0
(l – √3m)x + (m + √3l)y = 0
∴ This equation represents the lines
(l + √3m)x + (m – √3l)y = 0 ……..(1)
(l – √3m)x + (m + √3l)y = 0 ……(2)
Let the given equation of the straight line is lx + my + n = 0 ………(3)
If A is an angle between lines (1) & (3), then


B = 60°
If O is the third angle then O = 180° – (60 + 60)
= 180° – 120°
= 60°
∴ A = B = O = 60°
The lines (1), (2), (3) form an equilateral triangle.
Now h = the perpendicular distance from the origin to the straight line lx + my + n = 0

Question 7.
Prove that the lines represented by the x² – 4xy + y² = 0 and x + y = 3 form an equilateral triangle. [May ’00]
Solution:
Given the equation of the pair of lines is x² – 4xy + y² = 0
Comparing with ax² + 2hxy + by² = 0, we get a = 1, b = 1, h = -2.

∴ Given equation represents the two lines are \(a x+\left(h \pm \sqrt{h^2-a b}\right) y=0\)
\(1 x+\left(-2 \pm \sqrt{(-2)^2-1 \times 1}\right) y=0\)
x + (-2 ± \(\sqrt{4-1}\))y = 0
x + (-2 ± √3)y = 0
x + (-2 + √3)y = 0, x + (-2 – √3)y = 0
This equation represents the lines
x + (-2 + √3)y = 0 ………(1)
x + (-2 – √3)y = 0 ………(2)
Let the given equation of the straight line is x + y – 3 = 0 ……(3)
Let A is an angle between (1) & (3) then

B = 60°
If O is the third angle then
O = 180° – (A + B)
= 180° – (60 + 60)
= 180° – 120°
= 60°
∴ O = A = B = 60°
∴ The lines (1), (2), (3) form an equilateral triangle.

Question 8.
If (α, β) is the centroid of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my = 1 prove that \(\frac{\alpha}{b l-h m}=\frac{\beta}{a m-h l}=\frac{2}{3\left(b l^2-2 h l m+a m^2\right)}\). [Mar. ’08]
Solution:
Let ax² + 2hxy + by² = 0 represents the lines l₁x + m₁y = 0 …….(1), l₂x + m₂y = 0 ……(2)
∴ ax² + 2hxy + by² = (l₁x + m₁y)(l₂x + m₂y)

Comparing on both sides we get l₁l₂ = a, (l₁m₂ + l₂m₁) = 2h, m₁m₂ = b
Given equation of the straight line is lx + my – 1 = 0 ………(3)
Vertex O: Clearly the origin O is the point of intersection of (1) & (2)
∴ O = (0, 0)
Vertex A: Solving (2) & (3)

Question 9.
Find the equation of the pair of lines intersecting at (2, -1) and (i) perpendicular to the pair 6x² – 13xy – 5y² = 0 and (ii) parallel to the pair 6x² – 13xy – 5y² = 0. [May ’98]
Solution:
Given the equation of the pair of lines is 6x² – 13xy – 5y² = 0.
Comparing with ax² + 2hxy + by² = 0, we get a = 6, h = \(\frac{-13}{2}\), b = -5
Let the given point A(x₁, y₁) = (2, -1)
(i) Equation to the pair of lines perpendicular to 6x² – 13xy – 5y² = 0 and passing through (2, -1) is
b(x – x₁)² – 2h(x – x₁)(y – y₁) + a(y – y₁)² = 0
-5(x – 2)² – 2 × \(\frac{-13}{2}\) (x – 2)(y + 1) + 6(y + 1)² = 0
-5(x² + 4 – 4x) + 13(xy + x – 2y – 2) + 6(y² + 1 + 2y) = 0
-5x² – 20 + 20x + 13xy + 13x – 26y – 26 + 6y² + 6 + 12y = 0
-5x² + 13xy + 6y² + 33x – 14y – 40 = 0
5x² – 13xy – 6y² – 33x + 14y + 40 = 0
(ii) Equation to the pair of a line parallel to 6x² – 13xy – 5y² = 0 and passing through (2, -1) is
a(x – x₁)² + 2h(x – x₁)(y – y₁) + b(y – y₁)² = 0
6(x – 2)² + 2 × \(\frac{-13}{2}\) (x – 2) (y + 1) – 5(y + 1)² = 0
6(x² + 4 – 4x) – 13(xy + x – 2y – 2) – 5(y² + 1 + 2y) = 0
6x² + 24 – 24x – 13xy – 13x + 26y + 26 – 5y² – 5 – 10y = 0
6x² – 5y² – 13xy – 37x + 16y + 45 = 0

Question 10.
If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines then prove that
(i) abc + 2fgh – af² – bg² – ch² = 0
(ii) h² ≥ ab, g² ≥ ac and f² ≥ bc. [Mar. ’16 (AP & TS), ’14, ’11, ’96, ’83; May ’95, ’90]
Solution:
Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the two lines
l₁x + m₁y + n₁ = 0 ……….(1)
l₂x + m₂y + n₂ = 0 ……….(2)

Question 11.
If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents two parallel lines, then prove that (i) h² = ab (ii) af² = bg² and (iii) the distance between the parallel lines = \(2 \sqrt{\frac{g^2-a c}{a(a+b)}}=2 \sqrt{\frac{f^2-b c}{b(a+b)}}\). [Mar. ’12, ’10, ’98; May ’06, ’01, ’97, ’95, ’91; Mar. ’19 (AP)]
Solution:
Let S = 0 represent the lines
lx + my + n₁ = 0 …….(1)
lx + my + n₂ = 0 …….(2)
∴ ax² + 2hxy + by² + 2gx + 2fy + c = (lx + my + n₁)(lx + my + n₂)
= l₂x² + lmxy + ln₂x + lmxy + m₂y² + mn₂y + ln₁x + mn₁y + n₁n₂
= l₂x² + 2lmxy + m₂y² + (ln₂ + ln₁)x + (mn₂ + mn₁)y + n₁n₂
Comparing both sides we get
l₂ = a, 2lm = 2h, m₂ = b, ln₁ + ln₂ = 2g, mn₂ + mn₁ = 2f, n₁n₂ = c, lm = h, l(n₁ + n₂) = 2g
m(n₁ + n₂) = 2f
(i) h² = (lm)² = l²m² = ab = R.H.S.
∴ h² = ab
(ii) \(\frac{2g}{2f}\) = \(\frac{l\left(n_1+n_2\right)}{m\left(n_1+n_2\right)}\)
\(\frac{\mathrm{g}}{\mathrm{f}}=\frac{l}{\mathrm{~m}}\)
gm = lf
Squaring on both sides
g²m² = l²f²
∴ af² = bg²
(iii) The distance between two parallel lines

Question 12.
Show that the equation 2x² – 13xy – 7y² + x + 23y – 6 = 0 represents a pair of straight lines. Also, find the angle between them and the coordinates of the point of intersection of the lines. [May ’12, ’00; Mar. ’03]
Solution:
Given equation is 2x² – 13xy – 7y² + x + 23y – 6 = 0.
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 2, h = \(\frac{-13}{2}\), b = -7, g = \(\frac{1}{2}\), f = \(\frac{23}{2}\), c = -6
Now,
(i) abc + 2fgh – af² – bg² – ch²

Question 13.
Find the value of λ for which the equation λx² – 10xy + 12y² + 5x – 16y – 3 = 0 represents a pair of straight lines. [May ’09]
Solution:
Given equation is λx² – 10xy + 12y² + 5x – 16y – 3 = 0.
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = λ, h = -5, b = 12, g = \(\frac{5}{2}\), f = -8, c = -3
Since the given equation represents a pair of straight lines then

Question 14.
Show that the equation 8x² – 24xy + 18y² – 6x + 9y – 5 = 0 represents a pair of parallel straight lines and find the distance between them. [Mar. ’93]
Solution:
Given, equation is 8x² – 24xy + 18y² – 6x + 9y – 5 = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 8, h = -12, b = 18, g = -3, f = \(\frac{9}{2}\), c = -5
(i) h² = ab
⇒ h² = (-12)² = 144
ab = 8 × 18 = 144
∴ h² = ab
(ii) af² = \(8\left(\frac{9}{2}\right)^2\)
= 8 × \(\frac{81}{4}\)
= 162
bg² = 18(-3)²
= 18 × 9
= 162
∴ af² = bg²
∴ The given equation represents a pair of parallel straight lines.
Now the distance between the parallel lines

Question 15.
Show that the pairs of straight lines 6x² – 5xy – 6y² = 0 and 6x² – 5xy – 6y² + x + 5y – 1 = 0 form a square. [May ’02, ’98, ’91, ’86: Mar. ’02]
Solution:
Given equations of the pair of lines are
6x² – 5xy – 6y² = 0 ……..(1)
6x² – 5xy – 6y² + x + 5y – 1 = 0 ………(2)

Now, 6x² – 5xy – 6y² = 0
6x² – 9xy + 4xy – 6y² = 0
3x(2x – 3y) + 2y(2x – 3y) = 0
(2x – 3y)(3x + 2y) = 0
2x – 3y = 0 …….(3) 3x + 2y = 0 ……..(4)
Equation (1) represents the two lines are 2x – 3y = 0 ……(3), 3x + 2y = 0 …….(4)
Now, 6x² – 5xy – 6y² + x + 5y – 1 = (2x – 3y + k) (3x + 2y + l)
Comparing the coefficient of x on both sides we get 2l + 3k = 1
Comparing the coefficient of y on both sides we get -3l + 2k = 5
Solving these two equations

Equation (2) represents the lines that are
2x – 3y + 1 = 0
3x + 2y – 1 = 0
∴ The four lines are
2x – 3y = 0 ……(3)
3x + 2y = 0 ……..(4)
2x – 3y + 1 = 0 ……..(5)
3x + 2y – 1 = 0 ………(6)
The equations (3) & (5); (4) & (6) are parallel.
The equations (3) & (4); (5) & (6) are perpendicular.
∴ The four lines form a rectangle.
The distance between the parallel lines (3) & (5) is
\(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|0-1|}{\sqrt{2^2+(-3)^2}}=\frac{|-1|}{\sqrt{4+9}}=\frac{1}{\sqrt{13}}\)
The distance between the parallel lines (4) & (6) is
\(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|0+1|}{\sqrt{3^2+2^2}}=\frac{1}{\sqrt{9+4}}=\frac{1}{\sqrt{13}}\)
∴ Given lines form a square.

Question 16.
Show that the straight lines y² – 4y + 3 = 0 and x² + 4xy + 4y² + 5x + 10y + 4 = 0 form a parallelogram and find the lengths of its sides. [Mar. ’92]
Solution:
Given equations of the straight lines are
y² – 4y + 3 = 0 …….(1)
x² + 4xy + 4y² + 5x + 10y + 4 = 0 …….(2)
(1) ⇒ y² – 4y + 3 = 0
y² – 3y – y + 3 = 0
y(y – 3) – 1(y – 3) = 0
(y – 3) (y – 1) = 0
y – 1 = o ……(3)
y – 3 = 0 ……..(4)
Equation (1) represents the lines y – 1 = 0 & y – 3 = 0.
Now, x² + 4xy + 4y² = 0
x² + 2xy + 2xy + 4y² = 0
x(x + 2y) + 2y(x + 2y) = 0
(x + 2y) (x + 2y) = 0
x + 2y = 0, x + 2y = 0
x² + 4xy + 4y² + 5x + 10y + 4 = (x + 2y + k) (x + 2y + l)
Comparing the coefficients of x on both sides l + k = 5
Comparing the coefficients of y on both sides 2l + 2k = 10
⇒ l + k = 5
Comparing constant terms on both sides
lk = 4
l = \(\frac{4}{k}\)
\(\frac{4}{k}\) + k = 5
4 + k² = 5k
k² – 5k + 4 = 0
k² – 4k – k + 4 = 0
k(k – 4) – 1(k – 4) = 0
(k – 4)(k – 1) = 0
k = 4; k = 1
If k = 4; l = \(\frac{4}{4}\) = 1
If k = 1; l = \(\frac{4}{1}\) = 4
Equations (2) represent the lines that are
x + 2y + 4 = 0 …….(5)
x + 2y + 1 = 0 …….(6)
The equations of the four lines are
y – 1 = 0 ……(3)
y – 3 = 0 ……..(4)
x + 2y + 4 = 0 ……..(5)
x + 2y + 1 = 0 ……..(6)
Clearly, lines (3), (4) & (5), (6) are parallel.

Vertex A: Solving (3) & (6)
From (3), y = 1
From (6), x + 2(1) + 1 = 0
⇒ x + 3 = 0
⇒ x = -3
Vertex A = (-3, 1)
Vertex B: Solving (3) & (5)
From (3), y = 1
From (5), x + 2(1) + 4 = 0
⇒ x + 6 = 0
⇒ x = -6
Vertex B = (-6, 1)
Vertex C: Solving (4) & (5)
From (4), y = 3
From (5), x + 2(3) + 4 = 0
⇒ x + 10 = 0
⇒ x = -10
Vertex C = (-10, 3)
Vertex D: Solving (4) & (6)
From (4), y = 3
From (6), x + 2(3) + 1 = 0
⇒ x + 7 = 0
⇒ x = -7
Vertex D = (-7, 3)
Vertices of a parallelogram are A(-3, 1), B(-6, 1), C(-10, 3) & D(-7, 3)
Now,

Question 17.
Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 is \(\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}\). [May ’98, ’93, ’90]
Solution:
ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents two lines
l₁x + m₁y + n₁ = 0 ……(1)
l₂x + m₂y + n₂ = 0 …….(2)
ax² + 2hxy + by² + 2gx + 2fy + c = (l₁x + m₁y + n₁)(l₂x + m₂y + n₂)
Comparing the coefficients on both sides we get
a = l₁l₂, 2h = l₁m₂ + l₂m₁, 2g = l₁n₂ + l₂n₁, 2f = m₁n₁ + m₂n₂
The perpendicular distance from the origin to the straight line (1) is

The perpendicular distance from the origin to the straight line (2) is

The product of the perpendicular distances

Question 18.
Find the angle between the lines joining the origin to the points of intersection of the curve x² + 2xy + y² + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0. [Mar. ’16 (AP), ’13, ’09, ’08, ’07; May ’14, ’13, ’11, ’04, ’80]
Solution:
Given equation of the curve is x² + 2xy + y² + 2x + 2y – 5 = 0 …….(1)

The equation of a straight line is
3x – y + 1 = 0
3x – y = -1
-3x + y = 1 ….(2)
Let, A, B are the points of intersection of the given line and the given curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
x² + 2xy + y² + 2x(1) + 2y(1) – 5(1)² = 0
x² + 2xy + y² + 2x(-3x + y) + 2y(-3x + y) – 5(-3x + y)² = 0
x² + 2xy + y² – 6x² + 2xy – 6xy + 2y² – 5(9x² + y² – 6xy) = 0
x² + 2xy + y² – 6x² + 2xy – 6xy + 2y² – 45x² – 5y² + 30xy = 0
x²(-5 – 45) + xy(4 – 6 + 30) + y²(1 + 2 – 5) = 0
x²(-50) + xy (+28) + y²(-2) = 0
-50x² + 28xy – 2y² = 0
50x² – 28xy + 2y² = 0
25x² – 14xy + y² = 0
which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\).
If ‘θ’ is the angle between the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\)

Question 19.
Find the value of k, if the lines joining the origin to the point of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular. [Mar. ’17 (AP), ’15 (TS), ’13 (Old) ’11, ’05, ’01; Mar. ’19 (AP & TS); May ’10, ’07, ’06; B.P.]
Solution:
Given the equation of the curve is 2x² – 2xy + 3y² + 2x – y – 1 = 0 ………(1)

The equation of a straight line is x + 2y = k
\(\frac{x+2 y}{k}=1\) ……(1)
Let, A B be the points of intersection of the given line and the given curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
2x² – 2xy + 3y² + 2x(1) – y(1) – 1(1)² = 0

2k²x² – 2k²xy + 3k²y² + 2kx² + 4kxy – kxy – 2ky² – x² – 4y² – 4xy = 0
x²(2k² + 2k – 1) + xy(-2k² + 3k – 4) + y²(3k² – 2k – 4) = 0
which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\).
Given that the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) are perpendicular
then, a + b = 0
2k² + 2k – 1 + 3k² – 2k – 4 = 0
5k² – 5 = 0
5k² = 5
k² = 1
k = ±1

Question 20.
Show that the lines joining the origin to the points of intersection of the curve x² – xy + y² + 3x + 3y – 2 = 0 and the straight line x – y – √2 = 0 are mutually perpendicular. [May ’15 (TS), ’12; Mar. ’15 (AP), ’12, ’08, ’03; Mar. ’18 (TS)]
Solution:
Given equation of the curve is x² – xy + y² + 3x + 3y – 2 = 0 ……..(1)

Equation of straight line is x – y – √2 = 0
x – y = √2
\(\frac{x-y}{\sqrt{2}}=1\) …….(2)
Let, A, B be the points of intersection of the given line and the given curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
x² – xy + y² + 3x(1) + 3y(1) – 2(1)² = 0

which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\).
Now a + b = 3 – 3 = 0
Since a + b = 0, then the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) are mutually perpendicular.

Question 21.
Find the condition for the chord lx + my = 1 of the circle x² + y² = a² (whose centre is the origin) to subtend a right angle at the origin. [Mar. ’14, ’13]
Solution:
Given equation of the curve is x² + y² = a² ………(1)
Equation of the straight line is lx + my = 1 ……..(2)

Let A, B be the point of intersection of the given line and the given curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
x² + y² = a²(1)
x² + y² = a²(lx + my)²
x² + y² = a²(l²x² + m²y² + 2lmxy)
x² + y² = a²l²x² + a²m²y² + 2a²lmxy
a²l²x² + a²m²y² + 2a²lmxy – x² – y² = 0
x²(a²l² – 1) + xy(2a²lm) + y²(a²m² – 1) = 0
which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\).
Given that, the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) are perpendicular.
Then, a + b = 0
a²l² – 1 + a²m² – 1 = 0
a²(l² + m²) – 2 = 0
a²(l² + m²) = 2
which is the required condition.

Question 22.
Find the condition for the lines joining the origin to the points of intersection of the circle x² + y² = a² and the line lx + my = 1 to coincide. [Mar. ’17 (TS); May ’03]
Solution:
Given, equation of the curve is x² + y² = a² ……….(1)
Equation of the straight line is lx + my = 1 ……….(2)

Let A and B be the point of intersection of the given line and the given curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
x² + y² = a²(1)
x² + y² = a²(lx + my)²
x² + y² = a²(l²x² + m²y² + 2lmxy)
x² + y² = a²l²x² + a²m²y² + 2a²lmxy
a²l²x² + a²m²y² + 2a²lmxy – x² – y² = 0
x²(a²l² – 1) + xy(2a²lm) + y²(a²m² – 1) = 0
Which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\)
Given that the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) coincide.
Then, h² = ab
(a²lm)² = (a²l² – 1)(a²m² – 1)
a⁴l²m² = a⁴l²m² – a²l² – a²m² + 1
a²l² + a²m² = 1
a²(l² + m²) = 1
Which is the required condition.

Question 23.
Find the equations of the straight lines bisecting the angles between the lines 7x + y + 3 = 0 and x – y + 1 = 0. [May ’05]
Solution:
Given the equation of the straight lines are
7x + y + 3 = 0 ……..(1)
x – y + 1 = 0 ……..(2)
Equations of the bisectors of the angles between the lines (1) & (2) are

7x + y + 3 + 5(x – y + 1) = 0
7x + y + 3 + 5x – 5y + 5 = 0
2x + 6y – 2 = 0
x + 3y – 1 = 0
7x + y + 3 – 5(x – y + 1) = 0
7x + y + 3 – 5x + 5y – 5 = 0
12x – 4y + 8 = 0
3x – y + 2 = 0
∴ The equation of the bisectors of the angle between the lines (1) & (2) is x + 3y – 1 = 0, 3x – y + 2 = 0.

Question 24.
If ax² + 2hxy + by² = 0 represents two straight lines such that the slope of one line is twice the slope of the other, prove that 8h² = 9ab. [May ’96]
Solution:
The given equation of the pair of lines is ax² + 2hxy + by² = 0.
Since the slope of one line is twice the slope of the other, then the slopes of the two lines represented by (1) are m, 2m.

Question 25.
If one line of the pair of lines ax² + 2hxy + by² = 0 bisects the angle between the coordinate axes, prove that (a + b)² = 4h². [May ’04]
Solution:
Given, equation of the pair of lines is ax² + 2hxy + by² = 0 ………(1)
The equation of the X-axis is y = 0
The equation of the Y-axis is x = 0
∴ Equations of the bisectors of the angles between the coordinate axes are
\(\frac{\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1}{\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2}}=\pm \frac{\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2}}\)
\(\frac{y}{\sqrt{1^2}}=\pm \frac{x}{\sqrt{1^2}}\)
y = ±x
y = x and y = -x
∴ Equations of the bisectors of the angles between the coordinate axes are y = x and y = -x.
Case 1: If one line of the pair of lines ax² + 2hxy + hy² = 0 is y = x
Substitute y = x in equation (1), and we get
ax² + 2hx(x) + b(x)² = 0
⇒ a + 2h + b = 0
⇒ a + b = -2h ………(2)
Case 2: If one line of the pair of lines is y = -x
Substitute y = -x in equation (1), and we get
ax² + 2hx(-x) + b(-x)² = 0
⇒ ax² – 2hx² + bx² = 0
⇒ a – 2h + b = 0
⇒ a + b = 2h ……..(2)
From (2) & (3)
a + b = ±2h
Squaring on both sides
(a + b)² = 4h²

Some More Maths 1B Pair of Straight Lines Important Questions

Question 26.
Find the centroid and the area of the triangle formed by the lines 2y² – xy – 6x² = 0, x + y + 4 = 0. [May ’03]
Solution:
Given the equation of the pair of lines is 2y² – xy – 6x² = 0
2y² – 4xy + 3xy – 6x² = 0
2y(y – 2x) + 3x(y – 2x) = 0
(y – 2x) (2y + 3x) = 0
y – 2x = 0 (or) 2y + 3x = 0
2x – y = 0 …….(1), 3x + 2y = 0 ……..(2)
The third equation is x + y + 4 = 0 ………(3)

∴ Vertex O: The point of intersection of (1) & (2) is O = (0, 0)
Vertex A: Solving (1) & (3)

Vertex B: Solving (2) & (3)

∴ Vertex B = (8, -12)
∴ Vertices of a triangle OAB are O = (0, 0), A(\(\frac{-4}{3}, \frac{-8}{3}\)), B = (8, -12)
Centroid of triangle

Question 27.
Find the centroid and area of the triangle formed by the lines 3x² – 4xy + y² = 0, 2x – y = 6.
Solution:
Given the equation of the pair of lines is
3x² – 4xy + y² = 0
3x² – 3xy – xy + y² = 0
3x(x – y) – y(x – y) = 0
(x – y) (3x – y) = 0
x – y = 0 ……..(1), 3x – y = 0 ………(2)

The third equation is 2x – y – 6 = 0 …….(3)
∴ Vertex O: The point of intersection of (1) & (2) is O = (0, 0)
Vertex A: Solving (1) & (3)

x = 6; y = 6
∴ Vertex A = (6, 6)
Vertex B: Solving (2) & (3)

x = -6; y = -18
Vertex B = (-6, -18)
∴ Vertices of a triangle OAB are O = (0, 0), A = (6, 6), B = (-6, -18).
Centroid of triangle

Question 28.
Show that the straight lines represented by (x + 2a)² – 3y² = 0 and x = a form an equilateral triangle. [Mar. ’03]
Solution:

Given equation of the pair of lines is (x + 2a)² – 3y² = 0
⇒ (x + 2a)² – (√3y)² = 0
⇒ (x + 2a + √3y) (x + 2a – √3y) = 0
⇒ x + 2a + √3y = 0, x – √3y + 2a = 0
⇒ x + √3y + 2a = 0, x – √3y + 2a = 0
∴ This equation represents the lines
x + √3y + 2a = 0 ……(1)
x – √3y + 2a = 0 ……….(2)
Let the given equation of the straight line is x = a ………(3)
If A is an angle between (1) & (2) then

\(\left|\frac{-2 \sqrt{3}}{-2}\right|=|\sqrt{3}|\)
A = 60°
If B is an angle between lines (2) & (3) then

B = 60°
If C is the third angle then C = 180° – (60° + 60°)
= 180° – 120°
= 60°
∴ A = B = C = 60°
∴ The lines (1), (2), (3) form an equilateral triangle.

Question 29.
Show that the straight lines represented by 3x² + 48xy + 23y² = 0 and 3x – 2y + 13 = 0 form an equilateral triangle of area \(\frac{13}{\sqrt{3}}\) sq. units.
Solution:
Given the equation of the pair of lines is 3x² + 48xy + 23y² = 0
Comparing with ax² + 2hxy + by² = 0,
we get a = 3, 2h = 48 ⇒ h = 24, b = 23

The given equation represents the two lines that are

3x + (24 + √507)y = 0, 3x + (24 – √507)y = 0
∴ This equation represents the lines
3x + (24 + √507)y = 0 ……….(1)
3x + (24 – √507)y = 0 ………..(2)
Let the given equation of the straight line is 3x – 2y + 13 = 0 …….(3)
Let A is an angle between (1) & (3) then

∴ A = 60°
If B is an angle between lines (2) & (3) then

If O is the third angle then
O = 180° – (A + B)
= 180° – (60° + 60°)
= 180° – 120°
= 60°
∴ O = A = B = 60°

∴ The lines (1), (2), (3) form an equilateral triangle.
The length of the altitude of the triangle h = the perpendicular distance from the origin O to the line (3)

Question 30.
Prove that the equation 3x² + 7xy + 2y² + 5x + 5y + 2 = 0 represents a pair of straight lines and find the coordinates of the point of intersection.
Solution:
Given equation is 3x² + 7xy + 2y² + 5x + 5y + 2 = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get

Question 31.
If x² + xy – 2y² + 4x – y + k = 0 represents a pair of straight lines, find k.
Solution:
Given equation is x² + xy – 2y² + 4x – y + k = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 1, h = \(\frac{1}{2}\), b = -2, g = 2, f = \(\frac{-1}{2}\), c = k
Since the given equation represents a pair of lines then abc + 2fgh – af² – bg² – ch² = 0

Question 32.
Find the distance between the pairs of parallel straight lines 9x² – 6xy + y² + 18x – 6y + 8 = 0. [May ’03]
Solution:
Given, equation is 9x² – 6xy + y² + 18x – 6y + 8 = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 9, h = -3, b = 1, g = 9, f = -3, c = 8
The distance between the parallel lines

Question 33.
Find the distance between the pairs of parallel straight lines x² + 2√3xy + 3y² – 3x – 3√3y – 4 = 0. [May ’03]
Solution:
Given, equation is x² + 2√3xy + 3y² – 3x – 3√3y – 4 = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 1, h = √3, b = 3, g = \(\frac{-3}{2}\), f = \(\frac{-3 \sqrt{3}}{2}\), c = -4
The distance between the parallel lines

Question 34.
Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4y – 1 = 0 form a square.
Solution:
Given equations of the pair of lines are
3x² + 8xy – 3y² = 0 ……..(1)
3x² + 8xy – 3y² + 2x – 4y – 1 = 0 ………(2)

Now, 3x² + 8xy – 3y² = 0
3x² + 9xy – xy – 3y² = 0
3x(x + 3y) – y(x + 3y) = 0
(x + 3y)(3x – y) = 0
x + 3y = 0 (or) 3x – y = 0
Equation (1) represents the two lines that are
x + 3y = 0 ………(3), 3x – y = 0 ………(4)
Now 3x² + 8xy – 3y² + 2x – 4y – 1 = (x + 3y + k) (3x – y + l)
Comparing the coefficient of x, on both sides we get l + 3k = 2
Comparing the coefficient of y on both sides we get 3l – k = -4
Solving these two equations we get


∴ Equation (2) represents the lines that are
x + 3y + 1 = 0 ……..(5)
3x – y – 1 = 0 ……….(6)
∴ The equations of the four lines are
x + 3y = 0 ……….(3)
3x – y = 0 ………..(4)
x + 3y + 1 = 0 ……….(5)
3x – y – 1 = 0 ……….(6)
The equations (3) & (5); (4) & (6) are parallel.
The equations (3) & (4); (5) & (6) are perpendicular.
∴ The four lines form a rectangle.
The distance between the parallel lines (3) & (5) is
\(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|0-1|}{\sqrt{1^2+3^2}}=\frac{|-1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}\)
The distance between the parallel lines (4) & (6) is
\(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|0+1|}{\sqrt{3^2+(-1)^2}}=\frac{1}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}\)
∴ Given lines form a square.

Question 35.
Find the lines joining the origin to the points of intersection of the curve 7x² – 4xy + 8y² + 2x – 4y – 8 = 0 with the straight line 3x – y = 2 and also the angle between them. [Mar. ’18 (AP); May ’01, ’98; Mar. ’00]
Solution:
Given the equation of the curve is
7x² – 4xy + 8y² + 2x – 4y – 8 = 0 ………(1)
The equation of a straight line is 3x – y = 2
\(\frac{3 x-y}{2}\) = 1 ……..(2)

Let A and B be the points of intersection of the given line and curve.
Now, Homogenising the equation (1) with the help of (2)
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
7x² – 4xy + 8y² + 2x(1) – 4y(1) – 8(1)² = 0

7x² – 4xy + 8y² + 3x² – xy – 6xy + 2y² – 18x² – 2y² + 12xy = 0
x²(7 + 3 – 18) + xy(-4 – 1 – 6 + 12) + y²(8 + 2 – 2) = 0
x²(-8) + xy(1) + y²(8) = 0
-8x² + xy + 8y² = 0
8x² – xy – 8y² = 0
which is the equation of the pairs of lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\).
Here, a = 8, b = -8
Now, a + b = 8 – 8 = 0
Since, a + b = 0, the lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) are mutually perpendicular.
∴ Angle between the lines = 90°

Question 36.
Find the equation of the bisector of the acute angle between the lines 3x – 4y + 7 = 0 and 12x + 5y – 2 = 0.
Solution:
Given equations of the straight lines are
3x – 4y + 7 = 0 ……….(1)
12x + 5y – 2 = 0 ………(2)
Equations of bisectors of the angles between the lines (1) & (2) are

13(3x – 4y + 7) = 5(12x + 5y – 2)
39x – 52y + 91 = 60x + 25y – 10
21x + 77y – 101 = 0
\(\frac{3 x-4 y+7}{5}=\frac{-12 x-5 y+2}{13}\)
39x – 52y + 91 + 60x + 25y – 10 = 0
99x – 27y + 81 = 0
11x – 3y + 9 = 0
∴ The equations of the bisectors of the angles between lines (1) & (2) are
21x + 77y – 101 = 0 ……….(3)
11x – 3y + 9 = 0 ……….(4)
Consider the lines
3x – 4y + 7 = 0 ……..(1)
11x – 3y + 9 = 0 ……..(4)
If θ is the angle between (1) & (4) then

∴ 11x – 3y + 9 = 0 is the acute angle bisector.

Question 37.
Show that the equation of the pair of lines bisecting the angle between the pair of bisectors of the angles between the pair of lines ax² + 2hxy + by² = 0 is (a – b) (x² – y²) + 4hxy = 0.
Solution:
Given, equation of the pair of lines is ax² + 2hxy + by² = 0 ……(1)
The equation to the pair of bisectors of angles between (1) is
h(x² – y²) = (a – b)xy
hx² – hy² = (a – b)xy
hx² – (a – b)xy – hy² = 0 ……….(2)
Now comparing (2) with ax² + 2hxy + by² = 0, we get
a = h, h = \(\frac{-(a-b)}{2}\), b = -h
The equation to the pair of bisectors of angles between (2) is
h(x² – y²) = (a – b)(xy)
\(\frac{-(a-b)}{2}\)(x² – y²) = (h + h)xy
-(a – b)(x² – y²) = 4hxy
∴ (a – b)(x² – y²) + 4hxy = 0

Question 38.
If the pairs of lines represented by ax² + 2hxy + by² = 0 and ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus, prove that (a – b)fg + h(f² – g²) = 0.
Solution:
Given equations o the pair of lines are
ax² + 2hxy + by² = 0 ………..(1)
ax² + 2hxy + by² + 2gx + 2fy + c = 0 ……..(2)
Let \(\overline{\mathrm{OA}}, \overline{\mathrm{OB}}\) be the pair of straight lines given by (1).
\(\overline{\mathrm{AC}}, \overline{\mathrm{BC}}\) be the pair of lines given by (2).
Since the lines represented by (1) are parallel to the lines represented by (2), then OACB is a parallelogram.
Now, ‘C’ is the point of intersection of (2).

(gh – af)x – (hf – bg)y = 0
Since, ‘A’ is a point on the locus (1) & (2),
the coordinates of A satisfy the equation (1) – (2) = 0.
Similarly, the coordinates of B also satisfy the equation (1) – (2) = 0.
Now, (1) – (2) = 0
⇒ ax² + 2hxy + by² – ax² – 2hxy – by² – 2gx – 2fy – c = 0
⇒ -(2gx + 2fy + c) = 0
⇒ 2gx + 2fy + c = 0
This is the linear equation in which x and y represent a line.
Hence (1) – (2) is the equation of diagonal \(\overline{\mathrm{AB}}\).
Since OACB is a rhombus, then the diagonals \(\overline{\mathrm{OC}}\) and \(\overline{\mathrm{AB}}\) are perpendicular to each other.
i.e., slope of \(\overline{\mathrm{OC}}\) × slope of \(\overline{\mathrm{AB}}\) = -1
\(\frac{-(g h-a f)}{-(h f-b g)} \times \frac{- 2 g}{2 f}=-1\)
\(\frac{g(g h-a f)}{f(h f-b g)}\) = 1
g(gh – af) = f(hf – bg)
g²h – afg = f²h – bgf
f²h – bfg – g²h + afg = 0
fg(a – b) + h(f² – g²) which is the required condition.

Question 39.
Find the value of k, if the equation 2x² + kxy – 6y² + 3x + y + 1 = 0 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k.
Solution:
Given equation is 2x² + kxy – 6y² + 3x + y + 1 = 0.
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get a = 2, h = \(\frac{k}{2}\), b = -6, g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), c = 1
Since the given equation represents a pair of straight lines then
abc + 2fgh – af² – bg² – ch² = 0

-k² + 3k + 4 = 0
k² – 3k – 4 = 0
k² – 4k + k – 4 = 0
k(k – 4) + 1(k – 4) = 0
(k – 4)(k + 1) = 0
k = 4 or -1
For k = 4, then a = 2, h = 2, b = -6, g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), c = 1

Question 40.
Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 and 5x + 2y – 8 = 0 are concurrent.
Solution:
Given equation is x² + 2xy – 35y² – 4x + 44y – 12 = 0
Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get
a = 1, h = 1, b = -35, g = -2, f = 22, c = -12

⇒ 0 = 0
∴ The given lines are concurrent.

Question 41.
Write down the equation of the pair of straight lines joining the origin to the points of intersection of line 6x – y + 8 = 0 with the pair of straight lines 3x² + 4xy – 4y² – 11x + 2y + 6 = 0. Show that the lines so obtained make equal angles with the coordinate axes. [May ’15 (AP)]
Solution:
Given the equation of the curve is
3x² + 4xy – 4y² – 11x + 2y + 6 = 0 ………(1)
Equation of straight line is 6x – y + 8 = 0
⇒ 6x – y = -8
⇒ \(\frac{6 x-y}{-8}\) = 1 ………(2)

Let, A, B are the points of intersection of the given line and the given curve.
Now, Homogenising equation (1) with the help of (2).
The combined equation of \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is
3x² + 4xy – 4y² – 11x(1) + 2y(1) + 6(1)² = 0


x²(468) + y²(-117) = 0
x²(468) – y²(117) = 0
4x² – y² = 0
which is the equation of the pair of lines \(\overline{\mathrm{OB}}\) and \(\overline{\mathrm{OB}}\).
Comparing this equation with ax² + 2hxy + by² = 0,
a = 4, h = 0, b = -1
The equation to the pair of bisectors of angles between 4x² – y = 0 is h(x² – y²) = (a – b)(xy)
0(x² – y²) – (4 + 1)xy
5xy = 0
xy = 0
x = 0, y = 0
which are the equations of the coordinate axes.
The lines \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) makes equal angles with the coordinate axes.

Students must practice these Maths 1B Important Questions TS Inter 1st Year Maths 1B Three-Dimensional Coordinates Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1B Three-Dimensional Coordinates Important Questions

Question 1.
Find the distance between the points (3, 4, -2) and (1, 0, 7). [May ’00]
Solution:
Let A = (3, 4, -2), B = (1, 0, 7) are the given points.
Now,

Question 2.
Show that the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) form an equilateral triangle. [Mar. ’18 (AP); (B.P.)]
Solution:
Let A = (1, 2, 3), B = (2, 3, 1), C = (3, 1, 2) are the given points.
Now,


∴ Given points form an equilateral triangle.

Question 3.
Show that the points (1, 2, 3), (7, 0, 1) and (-2, 3, 4) are collinear. [Mar. ’16 (TS); Mar. ’13, May ’19]
Solution:
Let A = (1, 2, 3), B = (7, 0, 1), C = (-2, 3, 4) are the given points.
Now,

AB + CA = 2√11 + √11 = 3√11 = BC
∴ A, B, C are collinear.

Question 4.
Find the ratio in which yz-plane divides the line joining A(2, 4, 5) and B(3, 5, -4). Also, find the point of intersection. [May ’10]
Solution:
A(2, 4, 5), B(3, 5, -4) are the given points.
The ratio in which the yz-plane divides
\(\overline{\mathrm{AB}}\) = -x₁ : x₂ = -2 : 3
The point divides \(\overline{\mathrm{AB}}\) in the ratio -2 : 3,
then co-ordinates of a point = \(\left[\frac{\mathrm{mx}_2+\mathrm{nx_{1 }}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{my_{2 }}+\mathrm{ny_{1 }}}{\mathrm{m}+\mathrm{n}}, \frac{\mathrm{mz_{2 }}+\mathrm{nz_{1 }}}{\mathrm{m}+\mathrm{n}}\right]\)

∴ The point of intersection = (0, 2, 23).

Question 5.
Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, -1), (3, 6, -1) and (4, 5, 1). [Mar. ’17 (TS), ’11; May ’03]
Solution:
A(2, 4, -1), B(3, 6, -1), C(4, 5, 1) are the given three vertices.
Let the fourth vertex be D(x, y, z)

3 + x = 6; 6 + y = 9; -1 + z = 0
x = 6 – 3; y = 9 – 6; z = 1
∴ x = 3, y = 3, z = 1
D = (3, 3, 1)
∴ Fourth Vertex D = (3, 3, 1)

Question 6.
Find the coordinates of the vertex ‘c’ of ∆ABC if its centroid is the origin and the vertices A, B are (1, 1, 1) and (-2, 4, 1) respectively. [Mar. ’16 (AP); ’15 (TS); May ’13, ’06]
Solution:
Let A(1, 1, 1), B(-2, 4, 1) are the given points.
Given that, Centroid G = (0, 0, 0)
Let third vertex, C = (x, y, z)
Now, the Centroid of ∆ABC is,

∴ The third vertex C = (1, -5, -2).

Question 7.
If (3, 2, -1), (4, 1, 1) and (6, 2, 5) are three vertices and (4, 2, 2) are the centroid of a tetrahedron, find the fourth vertex. [Mar. ’17, ’15 (AP), ’14, ’13 (old), ’09; May ’15 (AP), ’13, ’11, ’05]
Solution:
A(3, 2, -1), B(4, 1, 1), C(6, 2, 5) are the given points.
Given that, Centroid G = (4, 2, 2)
Let the fourth vertex, D = (x, y, z)
Now, the Centroid of tetrahedron ABCD is,


∴ The fourth vertex D = (3, 3, 3).

Question 8.
Show that the points A(3, 2, -4), B(5, 4, -6) and C(9, 8, -10) are collinear and find the ratio in which B divides \(\overline{\mathbf{A C}}\). [Mar. ’04]
Solution:
A = (3, 2, -4), B = (5, 4, -6), C = (9, 8, -10) are the given points.
Now,

Now, AB + BC = √12 + 2√12 = 3√12 = CA
∴ A, B, C are collinear.

The ratio in which ‘B’ divides \(\overline{\mathbf{A C}}\) = x₁ – x : x – x₂
= 3 – 5 : 5 – 9
= -2 : -4
= 2 : 4
= 1 : 2

Question 9.
If A(4, 8, 12), B(2, 4, 6), C(3, 5, 4) and D(5, 8, 5) are four points, show that the lines \(\overline{\mathbf{A B}}\) and \(\overline{\mathbf{CD}}\) intersect. [May ’04]
Solution:
A(4, 8, 12), B(2, 4, 6), C(3, 5, 4) and D(5, 8, 5) are the given points.
The equation of the line passing through A, B is

x – 4 = -2t; y – 8 = -4t; z – 12 = -6t
x = -2t + 4; y = -4t + 8; z = -6t + 12
∴ (x, y, z) = (-2t + 4, -4t + 8, -6t + 12) ……(1)
The equation of the line passing through C, D is

x – 3 = 2s; y – 5 = 3s; z – 4 = s
x = 2s + 3; y = 3s + 5; z = s + 4
∴ (x, y, z) = (2s + 3, 3s + 5, s + 4) ………(2)
From (1) &(2)
-2t + 4 = 2s + 3 ⇒ 2t + 2s – 1 = 0 ………(3)
-4t + 8 = 3s + 5 ⇒ 4t + 3s – 3 = 0 ………(4)
-6t + 12 = s + 4 ⇒ 6t + s – 8 = 0 ……….(5)
Solving (3) & (4)


Substitute the values of (t, s) in equation (5)
6(\(\frac{3}{2}\)) + (-1) – 8 = 0
9 – 9 = 0
0 = 0
∴ Lines \(\overline{\mathrm{AB}} \& \overline{\mathrm{CD}}\) are intersecting lines.
Substitute, t = \(\frac{3}{2}\) in equation (1) or s = -1 in equation (2)
∴ Point of intersection = [-2(\(\frac{3}{2}\)) + 4, -4(\(\frac{3}{2}\)) + 8, -6(\(\frac{3}{2}\)) + 12]
= [-3 + 4, -6 + 8, -9 + 12]
= (1, 2, 3)

Question 10.
Find the point which divides the line joining the points A(2, -3, 1), B(3, 4, -5) in the ratio 1 : 3.
Solution:
A(2, -3, 1), B(3, 4, -5) are the given points.
Let C(x, y, z) be the point which divides the line joining the points A(2, -3, 1), B(3, 4, -5) in the ratio 1 : 3.

Question 11.
Find the centroid of the triangle whose vertices are (5, 4, 6), (1, -1, 3) and (4, 3, 2).
Solution:
A(5, 4, 6), B(1, -1, 3), C(4, 3, 2) are the given vertices.
The centroid of the triangle ABC is

Question 12.
Find the centroid of the tetrahedron whose vertices are (2, 3, -4), (-3, 3, -2), (-1, 4, 2), (3, 5, 1).
Solution:
A(2, 3, -4), B(-3, 3, -2), C(-1, 4, 2) and D(3, 5, 1) are the given points.
The centroid of the tetrahedron is

Some More Maths 1B Three-Dimensional Coordinates Important Questions

Question 13.
Find the distance of P(3, -2, 4) from the origin.
Solution:
Let O(0, 0, 0), P(3, -2, 4) are the given points.
∴ Distance from origin to P(3, -2, 4) is

Question 14.
Show that the points A(3, -2, 4), B(1, 1, 1) and C(-1, 4, -2) are collinear.
Solution:
Let A = (3, -2, 4), B = (1, 1, 1), C = (-1, 4, -2) are the given points.
Now,


AB + BC = √22 + √22 = 2√22 = AC
∴ A, B, C are collinear.

Question 15.
Show that the points (5, 4, 2), (6, 2, -1) and (8, -2, -7) are collinear. [May ’07]
Solution:
Let A = (5, 4, 2), B = (6, 2, -1), C = (8, -2, -7) are the given points.

Now, AB + BC = √14 + 2√14 = 3√14 = CA
∴ A, B, C are collinear.

Question 16.
Find the ratio in which the xz-plane divides the line joining A(-2, 3, 4) and B(1, 2, 3).
Solution:
A(-2, 3, 4), B(1, 2, 3) are the given points.
xz-plane divides \(\overline{\mathrm{AB}}\) in the ratio = -y₁ : y₂
= -3 : 2 (3 : 2 externally)

Question 17.
Find the point of intersection of the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) where A = (7, -6, 1), B = (17, -18, -3), C = (1, 4, -5) and D = (3, -4, 11).
Solution:
A(7, -6, 1), B(17, -18, -3), C(1, 4, -5) and D(3, -4, 11) are the given points.
The equation of the line passing through A, B is

x – 7 = 10t; y + 6 = -12t; z – 1 = 4t
x = 10t + 7; y = -12t – 6; z = -4t + 1
∴ (x, y, z) = (10t + 7, -12t – 6, -4t + 1) ……..(1)
The equation of the line passing through C, D is

x – 1 = 2s; y – 4 = -8s; z + 5 = 16s
x = 2s + 1; y = -8s + 4; z = 16s – 5
∴ (x, y, z) = (2s + 1, -8s + 4, 16s – 5) ……….(2)
From (1) & (2)
10t + 7 = 2s + 1
⇒ 10t – 2s + 6 = 0
⇒ 5t – s + 3 = 0 ……..(3)
-12t – 6 = -8s + 4
⇒ 12t – 8s + 10 = 0
⇒ 6t – 4s + 5 = 0 ……..(4)
-4t + 1 = 16s – 5
⇒ 4t + 16s – 6 = 0
⇒ 2t + 8s – 3 = 0 ………(5)
Solving (3) & (4)

Substitute the values of t, s in equation (5)
\(2\left(\frac{-1}{2}\right)+8\left(\frac{1}{2}\right)-3=0\)
-1 + 4 – 3 = 0
⇒ 4 – 4 = 0
⇒ 0 = 0
∴ Lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are intersecting lines.
Substitute t = \(\frac{-1}{2}\) in equation (1) & s = \(\frac{1}{2}\) in equation (2)
∴ Point of intersection = \(\left[10\left(\frac{-1}{2}\right)+7,-12\left(\frac{-1}{2}\right)-6,-4\left(\frac{-1}{2}\right)+1\right]\)
= [-5 + 7, 6 – 6, 2 + 1]
= (2, 0, 3)

Question 18.
Show that the points A(-4, 9, 6), B(-1, 6, 6) and C(0, 7, 10) form a right angled isoscele.
Solution:
Let A = (-4, 9, 6), B = (-1, 6, 6), C = (0, 7, 10) are the given points triangle.
Now,

AB = CA then triangle ABC is isosceles.
AB² + BC² = (√18)² + (√18)²
= 18 + 18
= (√36)²
= AC²
∴ AB² + BC² = AC² then ΔABC is right-angled.
∴ ΔABC is a right-angled isosceles triangle.

Question 19.
Find ‘x’ if the distance between (5, -1, 7) and (x, 5, 1) is 9 units. [Mar. ’19 (AP)]
Solution:
Let A = (5, -1, 7), B = (x, 5, 1) are the given points
Now, Given that, AB = 9

Squaring on both sides
x² – 10x + 97 = 81
⇒ x² – 10x + 16 = 0
⇒ x² – 8x – 2x + 16 = 0
⇒ x(x – 8) – 2(x – 8) = 0
⇒ (x – 8) (x – 2) = 0
⇒ x = 8 or 2

Question 20.
Show that ABCD is a square where A, B, C, D are the points(0, 4, 1), (2, 3, -1), (4, 5, 0), and (2, 6, 2) respectively.
Solution:
A = (0, 4, 1), B = (2, 3, -1), C = (4, 5, 0), and D = (2, 6, 2) are the given points.
Now,


∴ Given points from a square.
∴ AB = BC = CD = DA & AC = BD

Question 21.
If (x₁, y₁, z₁) and (x₂, y₂, z₂) are two vertices and (α, β, γ) is the centroid of a triangle, find the third vertex of the triangle.
Solution:
Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be the two vertices and C(x, y, z) be the third vertex.
Given G = (α, β, γ) we have
\(\frac{x+x_1+x_2}{3}\) = α, \(\frac{y+y_1+y_2}{3}\) = β, \(\frac{z+z_1+z_2}{3}\) = γ
∴ x = 3α – x₁ – x₂, y = 3β – y₁ – y₂, z = 3γ – z₁ – z₂
∴ The third vertex = (3α – x₁ – x₂, 3β – y₁ – y₂, 3γ – z₁ – z₂)

Question 22.
If M(α, β, γ) is the midpoint of the line segment joining the points A(x₁, y₁, z₁) and B then find B.
Solution:
Let B = (x, y, z) then coordinates of midpoint = \(\left(\frac{x_1+x}{2}, \frac{y_1+y}{2}, \frac{z_1+z}{2}\right)\)
Given (α, β, γ) = \(\left(\frac{x_1+x}{2}, \frac{y_1+y}{2}, \frac{z_1+z}{2}\right)\)
∴ x = 2α – x₁, y = 2β – y₁, z = 2γ – z₁
∴ B = (2α – x₁, 2β – y₁, 2γ – z₁)

Question 23.
If H, G, S, and I respectively denote the orthocentre, centroid, circumcentre, and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1), and (3, 1, 2) then find H, G, S, I.
Solution:
Let A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) be three given points.

∴ AB = BC = CA, the triangle formed will be an equilateral triangle.
In this triangle, all the centres H, G, I and S coincide.
∴ Centroid of the triangle (G) = \(\left(\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}\right)\) = (2, 2, 2)
Hence G = H = S = I = (2, 2, 2)

Question 24.
Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0), and (0, 4, 0).
Solution:
If A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) are the vertices of a triangle
and a = BC, b = CA and c = AB are the sides of the triangle then the incentre of the triangle,

Question 25.
Find the distance between the midpoint of the line segment \(\overline{\mathrm{AB}}\) and the point (3, -1, 2) where A = (6, 3, -4) and B = (-2, -1, 2)?
Solution:
Let A = (6, 3, -4), B = (-2, -1, 2) are the given points.
Midpoint of \(\overline{\mathrm{AB}}\) is

Let D = (3, -1, 2) be given point
Now, the distance between C & D

Students must practice these TS Intermediate Maths 1B Solutions Chapter 2 Transformation of Axes Ex 2(a) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Transformation of Axes Solutions Exercise 2(a)

I.
Question 1.
When the origin is shifted to (4, – 5) by the translation of axes, find the coordinates of the following points with reference to new axes. (V.S.A.Q.)
(i) (0, 3)
(ii) (-2, 4)
(iii) (4, -5)
Answer:
(i) (0, 3)
New origin = (4, -5) = (h, k)
Old co-ordinates are (0, 3) = (x, y)
x’ = x – h = 0 – 4 = – 4 and y’ = y – k = 3 + 5 = 8
∴ New coordinates = (- 4, 8)

(ii) (- 2, 4)
New origin = (4, -5) = (h, k)
Old coordinates are (- 2, 4) = (x, y)
x’ = x – h = – 2 – 4 = – 6
y’ = y – k = 4 + 5 = 9
∴ New coordinates = (- 6, 9)

(iii) (4, – 5)
New origin = (4, -5) = (h, k)
Old coordinates are (4, -5) = (x, y)
x’ = x – h = 4 – 4 = 0
y’ = y – k = – 5 + 5 = 0
∴ New coordinates = (0, 0)

Question 2.
The origin is shifted to (2,3) by the translation of axes. If the co-ordinates of a point P change as follows, find the co-ordinates of P in the original system. (V.S.A.Q.)
(i) (4, 5)
(ii) (-4, 3)
(iii)(0, 0)
Answer:
(i) (4, 5)
New coordinates (x’,y’) = (4, 5)
origin (h, k) = (2, 3)
Old coordinates of P are x = x’ + h = 4 + 2 = 6
y = y’ + k = 5 + 3 = 8
∴ Old coordinates = (6, 8)

(ii) (- 4, 3)
New coordinates (x’, y’) = (-4, 3)
origin (h, k) = (2, 3)
Old coordinates of P are x = x’ + h = – 4 + 2 = – 2
y = y’ + k = 3 + 3 = 6
∴ Old coordinates = (- 2, 6)

(iii) (0, 0)
New coordinates (x’,y’) = (0, 0)
origin(h, k) = (2, 3)
Old coordinates of P are x = x’ + h = 0 + 2 = 2
y = y’ + k = 0 + 3 = 3
∴ Old co-ordinates = (2, 3)

Question 3.
Find the point to which the origin is to be shifted so that the point (3,0) may change to (2, -3). (V.S.A.Q.)
Answer:
(x, y) = (3, 0)
(x’,y’) = (2,-3)
Let ( h, k ) be the shifted origin,
h = x – x’ = 3 – 2 = 1
k = y – y’ = 0 + 3 = 3
∴ (h, k) = (1, 3)

Question 4.
When the origin is shifted to (-1, 2) by the translation of axes, find the transformed equations of the following.
(i) x² + y² + 2x – 4y + 1 = 0
(ii) 2x² + y² – 4x + 4y = 0 (V.S.A.Q.)
Answer:
(i) x² + y² + 2x – 4y + 1 = 0
The given equation is
x² + y² + 2x – 4y + 1 = 0
origin is shifted to (-1, 2)
∴ h = – 1, k = 2
Transformed equations are
x = x’ + h, y = y’+ k
⇒ x = x’ – 1, y = y’ + 2
∴ The new equation is
(x’ – 1)² + (y’ + 2)² + 2( x’ – 1) – 4(y’ + 2) + 1 = 0
x’² + 1 – 2x’ + y’² + 4y’ + 4 + 2x ‘ – 2 – 4y’- 8 + 1 = 0
⇒ x’² + y’² – 4 = 0

(ii) 2x² + y² – 4x + 4y = 0
The given equation is
2x² + y² – 4x + 4y = 0
By the above transformation we have
x = x’ – 1, y = y’ + 2
∴ 2 (x’ – 1)² + (y’ + 2)² – 4(x’ – 1) + 4 (y’ + 2) = 0
⇒ 2 [ x’² – 2x’ + 1] + y’² + 4y’ + 4 – 4x’ + 4 + 4 y’ + 8 = 0
⇒ 2 x’² + y’² – 8x’ + 8y’ + 18 = 0

Question 5.
The point to which the origin is shifted and the transformed equations are given below. Find the original equation.
(i) ( 3, – 4 ) ; x² + y² = 4
(ii) (-1, 2 ) ; x² + 2y² + 16 = 0 (V.S.A.Q.)
Answer:
(i) ( 3, – 4 ) ; x² + y² = 4
Shifted origin = (h, k) = ( 3, – 4)
x’ = x – h
= x – 3
y’ = y – k
= y + 4
The given equation is x² + y² = 4
The original equation of 3x’² + y’² = 4 is
(x – 3)² + (y + 4)² = 4
⇒ x² – 6x + 9 + y² + 8y + 16 = 4
⇒ x² + y² – 6x + 8y + 21 = 0

(ii) (-1, 2) ; x² + 2y² + 16 = 0
Given shifted origin = (h, k) = (- 1, 2)
x’=x + h
= x + 1
y’ = y + k
= y – 2
The original equation of x’² +2 y’² + 16 = 0 is
( x + 1)² + 2 (y – 2)² + 16 = 0
⇒ x² + 2x + 1 + 2 ( y² – 4y + 4) + 16 = 0
⇒ x² + 2y² + 2x – 8y + 25 = 0

Question 6.
The point to which the origin is shifted and the transformed equations are given below.
Find the original equation.
(i) (3, – 4); x² + y² = 4
(ii) (- 1, 2); x² + 2y² + 16 = O (V.S.A.Q.)
Answer:
(3, – 4); x² + y² = 4
Shifted origin (h, k) = (3, – 4)
x’ = x – h
= x – 3
y’ = y – k
= y + 4
The given equation is x² + y² = 4
The original equation of 3x’² + y’² = 4 is
(x – 3)² + (y + 4)² = 4
⇒ x² – 6x + 9 + y² + 8y + 16 = 4
⇒ x² + y² – 6x + 8y + 21 = 0

(ii) (- 1, 2) ; x² + 2y² + 16 = 0
Given shifted origin = (h, k) = (- 1, 2)
x’ = x + h
= x + 1

y’ = y + k
= y – 2
The origin equation of x’² + 2y’² + 16 = 0 is
(x + 1)² + 2(y – 2)² + 16 = 0
⇒ x² + 2x + 1 + 2 (y² – 4y + 4) + 16 = 0
⇒ x² + 2y² + 2x – 8y + 25 = 0

Question 7.
Find the point to which the origin is to be shifted so as to remove the first degree terms from the equation 4x² + 9y² – 8x + 36y + 4 =0. (V.S.A.Q.)
Answer:
The given equation is
4x² + 9y² – 8x + 36y + 4 = 0
To make first degree terms missing in the equation origin should be shifted to \(\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)\)
Comparing with
ax² + 2hxy + by² + 2gx + 2fy + c = 0 we have
a = 4, b = 9, h = 0, g = – 4, f = 18, c = 4

Question 8.
When the axes are rotated through an angle 30°, find the new coordinates of the following points. (V.S.A.Q.)
(i) (0, 5)
(ii) (- 2, 4)
(iii) (0, 0)
Answer:
(i) ( 0, 5 )
Given θ = 30° and (x, y) = (0, 5)
We have x’ = x cos θ + y sin θ
= 0 (cos 30°) + 5 ( sin 30°) = 5\(\left(\frac{1}{2}\right)\) = \(\frac{5}{2}\)
y’ = – x sin θ + y cos θ
= – 0 (sin 30°) + 5 cos 30° = \(\frac{5 \sqrt{3}}{2}\)
∴ New coordinates = \(\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)\)

(ii) (-2, 4)
θ = 30° and (x, y) = (-2, 4)
We have x’= x cos θ + y sin θ
= – 2 cos 30° + 4 sin 30°
= – 2 \(\left(\frac{\sqrt{3}}{2}\right)\) + 4\(\left(\frac{1}{2}\right)\) = – √3 + 2
y’ = – x sin θ + y cos θ
= – (- 2) sin 30° + 4 cos 30°
= 2 sin 30° + 4 cos 30°
= 2 \(\left(\frac{1}{2}\right)\) + 4 \(\left(\frac{\sqrt{3}}{2}\right)\) = 1 + 2√3
∴ New coordinates = (- √3 + 2, 1 + 2√3)

(iii) (0, 0)
θ = 30° and (x, y) = (0, 0)
x’= x cos θ + y sin θ
= 0 cos 30° + 0 sin 30° = 0
y’ = – x sin 0 + y cos 0
= – (0) sin 30° + 0 cos (30°) = 0
∴ New coordinates = (0, 0)

Question 9.
When the axes are rotated through an angle 60°, the new coordinates of three points are the following
(i) (3, 4)
(ii) (- 7, 2)
(iii) (2, 0)
Find their original coordinates. (V.S.A.Q.)
Answer:
Given θ = 60°, (x’, y’) = (3, 4)
We have x = x’ cos θ – y’ sin θ
= 3 cos 60° – 4 sin 60°

(ii) Given θ = 60° and (x’, y’) = (-7, 2)
We have x = x’ cos θ – y’ sin θ
= (- 7) cos 60° – 2 sin 60°

(iii) Given θ = 60°; (x’, y’) = (2, 0)
We have x = x’ cos θ – y’ sin θ
= 2 cos 60° – 0 sin 60°
= 2\(\left(\frac{1}{2}\right)\) – 0 \(\left(\frac{\sqrt{3}}{2}\right)\) = 1
Also y = x’ sin θ + y’ cos θ
= 2 sin 60° + 0 cos 60° = √3
= 2\(\left(\frac{\sqrt{3}}{2}\right)\) + 0\(\left(\frac{1}{2}\right)\) = √3
∴ Original coordinates of P = (1, √3)

Question 10
Find the angle through which the axes are to be rotated so as to remove the xy term in the equation x² + 4xy + y² – 2x + 2y – 6 = 0. (V.S.A.Q.) (June 2004)
Answer:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0
We have a = 1, 2h = 4 ⇒ h = 2,
b = 1, 2g = – 2 ⇒ g = – 1
2f = 2 ⇒ f = 1, c = – 6
Let θ be the angle of rotation of axes, then

II.
Question 1.
When the origin is shifted to the point (2, 3), the transformed equation of a curve is x² + 3xy – 2y² + 17x – 7y – 11 = 0. Find the original equation of the curve. (S.A.Q.) (March 2011)
Answer:
Transformed equations are

Transformed equation is
x’² + 3x’y’ – 2 y’² + 17x’ – 7y’ – 11 = 0
Original equation is
(x – 2)² + 3 (x – 2) (y – 3) – 2 (y – 3)² – 7 (y – 3) – 11 = 0
⇒ x² – 4x + 4 + 3 (xy – 2y – 3x + 6) – 2 (y² – 6y + 9) – 7y + 21 – 11 = 0
⇒ x² + 3xy – 2y² + 4x – y – 20 = 0
∴ The original equation of the curve is
x² + 3xy – 2y² + 4x – y – 20 = 0

Question 2.
When the axes are rotated through an angle 45°, the transformed equation of a curve is 17x² – 16xy + 17y² = 225. Find the original equation of the curve. (S.A.Q.) (May 2012)
Answer:
Angle of rotation is θ = 45°
x’ = x cos θ + y sin θ
= x cos 45° + y sin 45°
= x\(\left(\frac{1}{\sqrt{2}}\right)\) + y\(\left(\frac{1}{\sqrt{2}}\right)\) = \(\frac{x+y}{\sqrt{2}}\)
y’ = – x sin θ + y cos θ
= – x sin 45° + y cos 45°
= – \(\frac{x+y}{\sqrt{2}}\)
The original equation of the curve
⇒ 17x² – 16x’y + 17y’² = 225 is

⇒ 17x² + 17y² + 34xy – 16y² + 16x² + 17x² + 17y² – 34xy = 450
⇒ 50x² + 18y² = 450
⇒ 25x² + 9y² = 225

Question 3.
When the axes are rotated through an angle a, find the transformed equation of x cos α + y sin α = p. (S.A.Q.) (Mar. ’14, May 2007)
Answer:
Given equation is x cos α + y sin α = p and axes are rotated through an angle α.
x = x’ cos α – y’ sin α
y = x’ sin α + y’ cos α
The given equation transformed to
(x’ cos α – y’ sin α) cos α + ( x’ sin α + y’ cos α) sin α = p
⇒ x’ (cos²α + sin²α) = p
⇒ x’ = p
∴ Transformed equation is x = p.

Question 4.
When the axes are rotated through an angle \(\frac{\pi}{6}\), find the transformed equation of x² + 2 √3 xy – y² = 2a². (S.A.Q.) (March 2012, ’07, 04; May 2006 )
Answer:

⇒ 3x² – 2√3 + y² + 6x² – 2√3 xy + 6√3 xy – 6y² – x² – 2√3 xy – 3y² = 8a²
⇒ 8x² – 8y² = 8a²
⇒ x² – y² = a²
∴ Required transformed equation is x² – y² = a²

Question 5.
When the axes are rotated through an angle \(\frac{\pi}{4}\), find the transformed equation of
3x² + 10xy + 3y² = 9. (S.A.Q.) (May’14, ’11)
Answer:
Given equation is
3x² + 10xy + 3y² = 9 …………… (1)
and angle of rotation θ = \(\frac{\pi}{4}\)
Let (X ,Y) be the new coordinates of (x, y) then x = X cos θ – Y sin θ

⇒ 3X² – 6XY + 3Y² + 10X² – 10Y² + 3X² + 6XY + 3Y² = 18
⇒ 16X² – 4Y² – 18 = 0
⇒ 8X² – 2Y² = 9
∴ The transformed equation of the given equation is
8X² – 2Y² = 9

Telangana TSBIE TS Inter 1st Year Sanskrit Study Material Grammar धातुरूपाणि Questions and Answers.

TS Inter 1st Year Sanskrit Grammar धातुरूपाणि

धातवः कालवाचकाः अवस्थावाचकाश्चेति द्वेधा वर्तन्ते । ये धातवः भूत-भविष्यत्-वर्तमानादिकं कालं बोधयन्ति ते कालवाचकाः । ये धातवः विधि- आशिष् निमन्त्रण- आमन्त्रणाद्यवस्थां बोधयन्ति ते अवस्थावाचकाः । संस्कृते कालवाचकाः अवस्थावाचकाः धातवः आहत्य दश विद्यन्ते । ते च लकारा इति नाम्ना व्यवह्रियन्ते । प्रकृते च अस्माकं दशसु लकारेषु पञ्चैव पाठयभागे निर्दिष्टाः । ते च क्रमेण –

1. वर्तमाने लट्
2. विध्यादिषु लोट्
3. अनद्यतनभूते लङ्
4. विधिलिङ्
5. भविष्यति लृट्

एतेषु पञ्चसु वर्तमाने लट् अनद्यतनभूते लङ् भविष्यति लृट् इति त्रयः कालवाचकाः । विध्यादिषु लोट् विधि लिङ् इति द्वौ प्रकारबोधक । सर्वेषु लकारेषु त्रयः पुरुषाः भवन्ति । ते च –

1. प्रथमपुरुषः
2. मध्यमपुरुषः
3. उत्तमपुरुषः

1. एकवचनम्
2. द्विवचनम्
3. बहुवचनम्
पुरुषास्त्रयः वचनानि त्रीणि आहत्य लकारे नव रूपाणि भवन्ति ।

In Sanskrit there are six Tenses (काला) and four Moods (अर्थाः:). They were given a special technical name by the master grammarian Panini. All these ten verbs together are called लकाण: (Lakaras).

‘సుప్’ అనే ప్రత్యాహారము నామ విభక్తులకు సంజ్ఞయైన విధంగా, తిజ్ అనే ప్రత్యాహారము క్రియావిభక్తులకు సంజ్ఞ. తిఙంతమనగా క్రియారూపమును పొందిన ‘ధాతువు’ అని పేరు. ‘సుప్’ ప్రత్యయము చేరని నామము కానీ, తిజ్ ప్రత్యయము చేరని ధాతువు కానీ వాక్యములో ప్రయోగమునకనర్హములు.

సంస్కృత భాషలో కాలబోధకములు ఆరు, అర్థబోధకములు నాలుగు కలవు. వీటినన్నింటిని కలిపి ‘లకారము’లని అందురు.

వర్తమాన కాలము ఒకటి (లట్), భవిష్యత్ కాలములు రెండు (లుట్, లట్), భూతకాలములు మూడు (లజ్, లిట్, లుజ్)

ఇవి మొత్తం ఆరు కాలబోధకములు.

The tenses and moods are as follows:

1. वर्तमानकालः – (Present) వర్తమానకాలము – लट् (technical name)
2. भूतकालः – (1^st Past) భూతకాలము – लुङ्
3. अनध्यतन भूतकालः – (Imperfect 2^ndPast) అనద్యతన భూతకాలము – लङ्
4. परोक्ष भूतकालः – (Perfect 3^rdPast) పరోక్ష భూతకాలము – लिट्
5. अनध्यतन भविष्यत् कालः – (1^st Future) అనద్యతన భవిష్యత్కాలము – लुट्
6. भविष्यत् कालः – (2^nd) భవిష్యత్ కాలము – लृट्

అర్ధబోధకములు (Moods)

1. आज्ञा – ఆజ్ఞార్థకము – लेट्
2. विधि – విధ్యర్థకము – विधि लिड्
3. आशीः – ఆశీరర్ధకము – आशीर्लिङ्
4. संकेत – సంకేతార్థకము – लृङ्

The लेट् or Subjunctive is used in the Veda only.

In Sanskrit the verbs are classified as आत्मनेपदी and परस्मैपदी based on whether the fruit of the action rests with the doer (आत्मने) or others (परस्मै) sometime it is called (उभयपदी) as conjugated with both terminations.

All the roots in Sanskrit are grouped into ten classes called गण s. They are called by the name of the first root of that class. For example, भू + आदि भू – etc. roots are found in भ्वादि class, अद् etc.
roots in अदादि and likewise.

As with Sabdas, here also tables are prepared as models. Almost all the roots of a cluster that matter but there are found many variations and irregular conjugations, which we need not know now.

ఈ పది లకారములు కాక లేట్ (लोट्) అని ఇంకొక అర్థకము వేదభాషలో కలదు. సంస్కృతంలో ధాతువులు పరస్మైపదులు, ఆత్మనేపదులు, ఉభయపదులు అని మూడు విధములుగా విభజింపబడినవి. క్రియాఫలము కర్తకు కాక ఇతరులకు చెందుతూ ఉంటే అది పరస్మైపది ధాతువు. క్రియాఫలము కర్తకే చెందుతూ ఉంటే అది ఆత్మనేపదము. కొన్ని ధాతువులు పరస్మైపది మరియు ఆత్మనేపదులలో ఉండును. వానిని ఉభయపదులు అని అందురు.

ఈ ధాతువులన్నియునూ 10 గణములుగా విభజింపబడినవి. అవన్నియునూ గణములోని మొదటి ధాతువుచే చెప్పబడుచుండును. భ్వాది గణము అనగా (भू + आदि) भू – ధాతువు మొదటిగాయున్న ధాతువులు గణము. అదే విధంగా అదాదులు (अद् + अदादि) మొదలైనవి.

శబ్దములకు వలెనే దీనియందు కూడా పట్టికలు తెలియజేయబడినవి. దాదాపు పరస్మైపది ధాతువులు (ఒకే గణములోని) ఒక విధముగాను, ఆత్మనేపది ధాతువులన్నియు ఒక విధముగాను ఉండును.

ధాతువులకు కేవలము అయిదు లకారములు మాత్రమే పరీక్షకు నిర్దేశింపబడినవి.

1. వర్తమానకాలము – लट्
2. అనద్యనత భూతకాలము – लङ्
3. ఆజ్ఞార్ధకము – लोट्
4. విధ్యర్ధకము – लिङ्
5. భవిష్యత్ కాలము – लृट्

ధాతువులు – వాని అర్థములు (Meanings of the Conjucations)

Ex : पठ् to read – परस्मैपद root.

ఉదా : ఆర్ధే (చదువుట) – పరస్మైపది ధాతువు

1. लट् – (Present tense) (వర్తమాన కాలము)

2. लङ् (భూతకాలము)

Similarly in 2. लङ् – (Past tense)
अपठत् (He) read अपठताम् (they two) read etc.

3. लोट् – (Impetative) (ఆశీరర్థకము)
पठतु He must read etc.
पठतु – చదువును గాక
(మిగిలినవన్నియు ఇదే విధముగా అర్థము చేసికొనవలయును)

4. विधिलिड – (Potential) (విధ్యర్ధకము)
पठेतु He may read etc.
पठेतु – చదువవలయును

5. लृट् – (Future) (భవిష్యత్ కాలము)
पठिष्यति He will read etc.
पठिष्यति – చదువును

At this stage we study these five varieties only. For roots also the meanings should be understood in the same way.

ఈ విధముగానే అన్ని పట్టికలకు అర్థమును చెప్పుకొనవలయును. ఈ పట్టికలను కూడా విద్యార్థులు కంఠస్థము చేయవలయును.

परस्मैपदिधातवः

1. भू – सत्तायाम् (to be) భూ – సత్తాయామ్

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విధి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

2. अस् – भुवि (to finish) అస్ – భూవి

3. पठ् – पठने (to read) పఠ్ – పఠనే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విధి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

4. लिख – लेखने (to write) లిఖ్ – లేఖనే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

5. गम्ल – गती (to go) గమ్ల – గతీ

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

6. दा – दाने (to give) దా – దానే

7. खाद् खादने (to eat) ఖాద్ – ఖాదనే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

8. पा पाने (to drink) పా – పానే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

9. हस् – हसने (to smile) హస్ – హసనే

10. धाव धावने (to run) ధావ – ధావనే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

11. दुशिर – प्रक्षण ताने (to see) దృశిర – ప్రక్షణ తానే

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

12. डुकृञ् – करणे (to do)

13. शृ – क्षवणे (to listan)

14. कथ् – वाक्यप्रबन्धे (to tell)

आत्मनेपदधातवः (ఆత్మనేపధాతవః:)

15. बन्द – अभिवादनस्तुतयो (to salute, to praise) వన్ద – అభివాదనస్తుత్యో

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

16. लभ्- प्राप्त (to get) లభ్ – ప్రాప్త

वर्तमाने लट् – వర్తమానే లట్

विध्यादिषु लोट् – విధ్యాదిషు లోట్

अनद्यतनभूते लङ् – అనధ్యతనభూతే లజ్

विधि लिङ् – విథి లిజ్

भविष्यति लृट् – భవిష్యతి లృట్

17. वृधु – वृद्धौ (to develop)

18. सेव् – सेवने ( to serve)