TS Inter 1st Year Maths 1B Solutions Chapter 4 Pair of Straight Lines Ex 4(b)

Students must practice these TS Intermediate Maths 1B Solutions Chapter 4 Pair of Straight Lines Ex 4(b) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(b)

I.
Question 1.
Find the angle between the lines represented by 2x²
2x² + xy – 6y² + 7y – 2 = 0. (V.S.A.Q.)
Answer:
Comparing with the general equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0 …………………….. (1)
We have a = 2, h = \(\frac{1}{2}\), b = – 6, g = 0, f = \(\frac{7}{2}\), c = – 2.
Also if θ is the angle between pair of lines (1)

Question 2.
Prove that the equation 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines. (V.S.A.Q.)
Answer:
We have coefficient of x² = 2 ⇒ a = 2
and coefficient of y² = – 2 ⇒ b = – 2
Since a + b = 0; the lines are perpendicular.

II.
Question 1.
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. (S.A.Q.)
Answer:
Comparing the given equation
3x² + 7xy + 2y² + 5x + 5y + 2 = 0 with the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get a = 3, h = \(\frac{7}{2}\), b = 2, g = \(\frac{5}{2}\), f = \(\frac{5}{2}\) and c = 2.
∆ = abc + 2fgh – af² – bg² – ch²

∴ The given equation represents a pair of lines and point of intersection is

Question 2.
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. (E.Q.)
Answer:
Comparing
2x² + kxy – 6y² + 3x + y + 1 = 0 with the general equation, we get
a = 2, h = \(\frac{\mathrm{k}}{2}\), b = -6, g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), c = 1
The given equation represents a pair of straight lines if abc + 2fgh – af² – bg² – ch² = 0 (necessary condition)
⇒ – 12 + 2\(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{k}{2}\right)\) – 2\(\left(\frac{1}{4}\right)\) + 6\(\left(\frac{9}{4}\right)-\frac{k^2}{4}\) = 0
⇒ – 48 + 3k – 2 + 54 – k² = 0
⇒ – k² + 3k + 4 = 0 ⇒ k² – 3k – 4 = 0
⇒ (k – 4) (k + 1) = 0 ⇒ k = 4 or – 1

Case (i):

Case (ii):
When k = 4, then h = 2

Question 3.
Show that the equation x² – y² – x + 3y – 2 = 0 represents a pair of perpendicular lines and find their equations. (S.A.Q.)
Answer:
First we show that the given equation x² – y² – x + 3y – 2 = 0 represents a pair of lines comparing with the general equation
we get a = 1, b = -1, g = \(\frac{-1}{2}\), f = \(\frac{3}{2}\), h = 0, c = -2
∴ abc + 2fgh – af² – bg² – ch²

Hence the given equation represents a pair of lines. Coefficient of x² + coefficient of y² = 1 – 1 = 0
∴ The given equation represents a pair of perpendicular lines.
Let x² – y² – x + 3y – 2
= (x + y + c₁) (x – y + c₂)
Equating coefficients of x and y both sides
we get c₁ + c₂ = – 1 and – c₁ + c₂ = 3
Solving 2c₂ = 2 ⇒ c₂ = 1
and c₁ + c₂ = -1 ⇒ c₁ = -2
∴ Equations of lines are x + y – 2 = 0 and x – y + 1 = 0.

Question 4.
Show that the lines x² + 2xy – 35y² – 4x + 44y – 12 = 0 and 5x + 2y – 8 = 0 are concurrent. (S.A.Q.)
Answer:
Given x² + 2xy – 35y² – 4x + 44y – 12 = 0 and comparing this with general equation we get a = 1, h = 1, b = – 35, g = -2, f = 22, c = – 12
Point of intersection

∴ The point of intersection lies on the line 5x + 2y – 8 = 0.
Hence the given lines are concurrent.

Question 5.
Find the distance between the following pairs of parallel straight lines. (V.S.A.Q.)
(i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
(ii) x²+ 2√3 xy + 3y² – 3x – 3√3 y – 4 = 0
Answer:
(i) The formula for distance between parallel lines = \(\sqrt{\frac{g^2-a c}{a(a+b)}}\)

(ii) Distance between given parallel lines

Question 6.
Show that the two pairs of lines 3x² + 8xy – 3y² = 0 and 3x² + 8xy – 3y² + 2x – 4v – 1 = 0 form a square. (S.A.Q.)
Answer:

Given equations of lines represented by OA and OB is 3x² + 8xy – 3y² = 0
⇒ (x + 3y) (3x – y) = 0
⇒ x + 3y = 0, 3x – y = 0
∴ Equation of OA is 3x – y = 0 ………………… (1)
Equation of OB is x + 3y = 0 ……………… (2)
The combined equation of CA and CB is
3x² + 8xy – 3y² + 2x – 4y – 1 = 0 …………….. (3)
Let 3x² + 8xy – 3y² + 2x – 4y – 1
= (3x – y + c₁) (x + 3y + c₂)
Equating the coefficients of x and y on both sides,
and c₁ + 3C₂ = 2
3c₁ – c₂ = – 4
Solving

⇒ c₁ = – 1, c₂ = 1
Equation of BC is 3x – y – 1 = 0 ……………………. (4)
Equation of AC is x + 3y + 1 = 0 …………………… (5)
Equations OA and BC differ by a constant.
⇒ OA is parallel to BC.
Equations OB and CA differ by a constant
⇒ OB is parallel to AC.
Also from equation (3), OACB is a rectangle and a + b = 3 – 3 = 0
OA = Length of the perpendicular from O to
AC = \(\frac{|0+0+1|}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}\)
OB = Length of the perpendicular from O to
BC = \(\frac{|0+0-1|}{\sqrt{9+1}}=\frac{1}{\sqrt{10}}\)
∴ OA = OB and OACB is a rectangle
⇒ OACB is a square.

III.
Question 1.
Find the product of the length of the perpendiculars drawn from (2, 1) upon the lines (E.Q.)
12x² + 25xy + 12y² + 10x + 11y + 2 = 0
Answer:
Given equation
12x² + 25xy + 12y² + 10x + 11y + 2 = 0
represents the combined equation of AB & AC.
12x² + 25xy + 12y²
= 12x² + 16xy + 9xy + 12y²
= 4x (3x + 4y) + 3y (3x + 4y)
= (3x + 4y) (4x + 3y)
Let 12x² + 25xy + 12y² + 10x + 11y + 2 = (3x + 4y + c₁) (4x + 3y’ + C₂)
Equating the coefficients of x and y
4c₁ + 3C₂ = 10 ………………. (1)
and 3c₁ + 4c₂ = 11 ………………. (2)
Solving (1) and (2)

⇒ c₁ = 1, c₂ = 2
∴ Equation of AB is 3x + 4y + 1 = 0
and equation of AC is 4x + 3y + 2 = 0
PL = Length of the perpendicular from
P(2, 1) on AB = \(\left|\frac{6+4+1}{\sqrt{9+16}}\right|=\frac{11}{5}\)
PM = Length of the perpendicular from
P(2, 1) on AC = \(\left|\frac{8+3+2}{\sqrt{16+9}}\right|=\frac{13}{5}\)
∴ Product of lengths of perpendiculars
= PL × PM = \(\frac{11}{5} \times \frac{13}{5}=\frac{143}{25}\)

Question 2.
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. (E.Q.)
Answer:

The equation of first pair of lines is y² – 4y + 3 = 0
⇒ (y – 1) (y – 3) = 0 ⇒ y = 1 ………………….. (1)
and y = 3 ………………….. (2)
∴ AB, CD are parallel.
The equation of second pair of lines is
x² + 4xy + 4y² + 5x + 10y + 4 = 0
⇒ (x + 2y)² + 5 (x + 2y) + 4 = 0
⇒ (x + 2y)² + 4 (x + 2y) + (x + 2y) + 4 = 0
⇒ (x + 2y) [x + 2y + 4] + 1 [(x + 2y) + 4] = 0
⇒ (x + 2y + 1) (x + 2y + 4) = 0
⇒ x + 2y + 1 = 0, x + 2y + 4 = 0
Equation of AD is x + 2y + 1 = 0 ……………………. (3)
Equation of BC is x + 2y + 4 = 0 ……………………. (4)
∴ AD and BC are parallel.
Solving (1) and (3), x + 2 + 1 = 0
⇒ x = – 3
∴ Co-ordinates of A = (-3, 1)
Solving (2) and (3), x + 6 + 1 = 0 ⇒ x = -7
and coordinates of D = (-7, 3)
Solving (2) and (4) x + 6 + 4 = 0 ⇒ x = -10
∴ Coordinates of C = (-10, 3)
Solving (1) and (4), we get x + 2 + 4 = 0
⇒ x = -6
∴ Coordinates of B = (- 6, 1)
Hence A = (-3, 1), B = (-6, 1), C = (-10, 3), D = (-7, 3)

∴ AC ≠ BD
Hence a parallelogram is formed with the
lines y² – 4y + 3 = 0 and x² + 4xy + 4y² + 5x + 10y + 4 = 0.

Question 3.
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 \(\frac{|c|}{\sqrt{(a-b)^2+4 h^2}}\) (E.Q.)
Answer:
Let ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the lines
l₁x + m₁y + n₁ = 0 ……………. (1)
and l₂x + m₂y + n₂ = 0 ……………. (2)
∴ ax² + 2hxy + by² + 2gx + 2fy + c
= (l₁x + m₁y + n₁) (l₂x + m₂y + n₂)
∴ l₁l₂ = a, m₁m₂ = b, l₁m₂ + l₂m₁ = 2h,
l₁n₂ +l₂n₁ = 2g, m₁n₂ + m₂n₁ = 2f, n₁n₂ = c
∴ Perpendicular distance from origin to (1) is
= \(\frac{\left|\mathrm{n}_1\right|}{\sqrt{l_1^2+\mathrm{m}_1^2}}\)
Perpendicular distance from origin to (2) is
= \(\frac{\mathrm{n}_2}{\sqrt{l_2^2+\mathrm{m}_2^2}}\)
Product of perpendiculars

Question 4.
If the equation ax² + 2hxy + by2 + 2gx + 2fy + c = 0 represent a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is \(\). Also show that the square of this distance is \(\frac{f^2+g^2}{h^2+b^2}\) if the given lines are perpendicular. (E.Q.)
Answer:
Let the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0 represent the lines
l₁x + m₁y + n₁ = 0 ………………. (1)
and l₂x + m₂y + n₂ = 0 ……………….. (2)
∴ (l₁x + m₁y + n₁) (l₂x + m₂y + n₂)
= ax² + 2hxy + by² + 2gx + 2fy + c
Comparing coefficients
l₁l₂ = a, m₁ m₂ = b, n₁n₂ = c
l₁m₂ + l₂m₁ = 2h, l₁n₂ + l₂n₁ = 2g and
m₁n₂ + m₂n₁ = 2f
From (1) and (2) on solving