Students must practice these TS Intermediate Maths 1B Solutions Chapter 3 Straight Lines Ex 3(e) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1B Straight Lines Solutions Exercise 3(e)

Question 1.

Find the in-centre of the triangle whose vertices are ( 1, √3), ( 2, 0) and (0, 0). (S.A.Q.)

Answer:

Let 0(0, 0), A (1, √3) and B (2, 0) are the vertices of ∆ABC

∴ ABC is an equilateral triangle.

Coordinates of in-centre are

Question 2.

Find the orthocentre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0 and 2x + y – 7 = 0

Answer:

Equation of AB is x + y + 10 = 0 ………………. (1)

Equation of BC is x – y – 2 = 0 ………………. (2)

Equation of CA is 2x + y – 7 = 0 ………………… (3)

Solving (1) and (2) Coordinates of B are (-4, -6)

Solving (1) and (3) Coordinates of A are (17, -27)

Equation of BC is 2x – y – 2 = 0

AD is perpendicular to BC

Equation of AD is x + y + k = 0

AD passes through A (17, -27)

∴ 17 – 27 + k = 0 ⇒ k = 10

∴ Equation of AB is x + y + 10 = 0 ……………… (1)

Equation of AC is 2x + y – 7 = 0

BE is perpendicular to AC

Equation of DE can be taken as x – 2y = k

BE passes through D(-4 -6); – 4 + 12 = k ⇒ k = 8

Equation of BE is x – 2y = 8 ………………. (2)

x + y = – 10 ……………….. (1)

Solving (1) & (2)

– 3y = 18 ⇒ y = – 6

x + y + 10 = 0 ⇒ x – 6 + 10 = 0 ⇒ x = – 4

∴ Orthocentre of AABC is (- 4, – 6)

Question 3.

Find the orthocentre of the triangle whose sides are given by 4x – 7y + 14 = 0, x + y = 5 and 7x + 4y = 15 (V.S.A.Q.)

Answer:

Equation of sides AB and BC of a ∆ABC are

given by 4x – 7y + 10 = 0 ……………….. (1)

x + y = 5 ……………… (2)

and equation of side AC is

7x + 4y – 15 = 0 ………………. (3)

AB and AC are perpendicular and ∠A = 90°

∴ ABC is a right angle triangle with ∠A = 90°and A is the orthocentre.

Solving (1) and (3)

Question 4.

Find the circumcentre of the triangle whose sides are x = 1, y = 1 and x + y = 1 (V.S.A.Q.)

Answer:

Equation of AB is x = 1

Equation of BC is y = 1

Equation of AC is x + y = 1

AB and BC are perpendicular.

∴ ABC is a right angled triangle ∠B = 90°

Mid point of AC is the circumcentre.

Coordinates of A are (1, 0) and C are (0, 1).

∴ Circumcentre = \(\left(\frac{1}{2}, \frac{1}{2}\right)\)

Question 5.

Find the in-centre of the triangle formed by the lines x = 1, y = 1 and x + y = 1. (V.S.A.Q.)

Answer:

Equation of AB is x = 1

Equation of BC is y = 1

Equation of AC is x + y = 1

Solving these equations we get vertices as

A(1, 0), B(1, 1), C(0, 1)

Question 6.

Find the circumcentre of the triangle whose vertices are (1, 0), (-1, 2) and (3, 2).(E.Q.) (March 2012)

Answer:

Slope of SE = – 1

Equation of SE is y – 1 = – l(x – 2)

⇒ y – 1 = – x + 2

⇒ x + y – 3 = 0 ………….. (2)

Solving (1) and (2) we get y – 2 = 0 ⇒ y = 2

⇒ Circumcentre S = (1, 2).

Question 7.

Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and 3x – y + 4 = 0 is \(\frac{\pi}{4}\). (S.A.Q.)

Answer:

Equations of the given lines are

kx + y + 9 = 0

3x – y + 4 = 0

⇒ (k^{2} + 1) 10 = 2 (3k – 1)^{2}

⇒ 10k^{2} + 10 = 2(9k^{2} – 6k + 1)

⇒ 8k^{2} – 12k – 8 = 0

⇒ 4k^{2} – 6k – 4 = 0

⇒ 2k^{2} – 3k – 2 = 0

⇒ (k – 2) (2k + 1) = 0 – 1

⇒ k = 2 or \(\frac{-1}{2}\)

Question 8.

Find the equation of the straight line passing through the origin and also through the point of intersection of lines 2x – y + 5 = 0 and x + y + 1 = 0. (V.S.A.Q.)

Answer:

Equations of given lines are

L_{1} = 2x – y + 5 = 0 and L_{2} = x + y + 1= 0

Equation of any line passing through A is

L_{1} + k L_{2} = 0

⇒ (2x – y + 5) + k(x + y + 1) = 0 (1)

This line passes through (0, 0) then

5 + k = 0 ⇒ k = – 5

Substituting in (1) equation of OA is

(2x – y + 5) – 5(x + y + 1) = 0

⇒ – 3x – 6y = 0

⇒ x + 2y = 0

Question 9.

Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersection of the lines x – 2y – 3 = 0 and x + 3y – 6 = 0. (S.A.Q.)

Answer:

Let L_{1} = x + 3y – 6 = 0 and L_{2} = x – 2y – 3 = 0 be the given lines.

Equation of any line passing through the intersection is L_{1} + KL_{2} = 0

⇒ (x + 3y – 6) + k (x – 2y – 3) = 0 ….. (1)

⇒ x(1 + k) + y (3 – 2k) – (6 + 3k) = 0

This is parallel to 3x + 4y = 7

a_{1}b_{2} = a_{2}b_{1}

⇒ 4(1 + k) = 3 (3 – 2k)

⇒ 4 + 4k = 9 – 6k

⇒ 10k = 5 ⇒ k = \(\frac{1}{2}\)

∴ From (1) (x + 3y – 6) + \(\frac{1}{2}\) (x – 2y – 3) = 0

⇒ 3x + 4y – 15 = 0

Equation of the required line is 3x + 4y – 15 = 0

Question 10.

Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and passing through the point of intersection of lines x + 3y – 1 = 0 and x – 2y + 4 = 0. (S.A.Q.)

Answer:

Let L_{1} = x + 3y – 1 = 0

and L_{2} = x – 2y + 4 = 0

Equation of any line passing through the

point of intersection of L_{1} = 0, L_{2} = 0 is L_{1} + kL_{2} = 0

⇒ (x + 3y – 1) + k (x – 2y + 4) = 0 (1)

⇒ x(1 + k) + y (3 – 2k) – (1 – 4k) = 0

Slope of the line is –\(\left(\frac{1+\mathrm{k}}{3-2 \mathrm{k}}\right)\)

Slope of the given line 2x + 3y = 0 is \(\frac{-2}{3}\)

∴ Since the lines are perpendicular

\(\left(\frac{1+\mathrm{k}}{3-2 \mathrm{k}}\right)\left(\frac{2}{3}\right)\) = – 1

⇒ 2 + 2k = – 3 (3 – 2k)

= – 9 + 6k

⇒ 4k = 11 ⇒ k = \(\frac{11}{4}\)

Substituting in (1) equation of required line is

(x + 3y – 1) + \(\frac{11}{4}\)(x – 2y + 4) = 0

⇒ 4x + 12y – 4 + 11x – 22y + 44 = 0

⇒ 15x – 10y + 40 = 0

⇒ 3x – 2y + 8 = 0

Question 11.

Find the equation of the straight line making non-zero equal intercepts on the coordinate axes and passing through the point of intersection of the lines 2x – 5y + 1 = 0 and x – 3y – 4 = 0. (S.A.Q.)

Answer:

Equations of the given lines are

L_{1} = 2x – 5y + 1 = 0

and L_{2} = x – 3y – 4 = 0

Equation of any line passing through the point of intersection of L_{1} = 0, L_{2} = 0 is

L_{1} + kL_{2} = 0

⇒ (2x – 5y + 1) + k (x – 3y – 4) = 0 …. (1)

⇒ (2 + k) x – (5 + 3k) y + (1 – 4k) = 0

Intercepts on coordinate axes are equal.

⇒ 2 + k = – 5 – 3k

⇒ 4k = -7 ⇒ k = \(\frac{-7}{4}\)

∴ From (1) (2x – 5y + 1) – \(\frac{-7}{4}\) (x – 3y – 4) = 0

⇒ x + y + 32 = 0

Question 12.

Find the length of the perpendicular drawn from the point of intersection of the lines 3x + 2y + 4 = 0 and 2x + 5y – 1 = 0 to the straight line 7x + 24y – 15 = 0. (S.A.Q.)

Answer:

Equation of the given lines

3x + 2y + 4 = 0

2x + 5y – 1 = 0

Solving

⇒ x = – 2, y = 1

Coordinates of point of intersection are (-2, 1)

Equation of the given line is 7x + 24y -15 = 0

∴ Length of the perpendicular from (-2, 1) to 7x + 24y – 15 = 0 is

\(\left|\frac{-14+24-15}{\sqrt{49+576}}\right|=\frac{5}{25}=\frac{1}{5}\)

Question 13.

Find the value of ‘a’ if the distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y – 8 = 0 are equal. (S.A.Q.)

Answer:

Let A (2, 3) and B (-4, a) be the two points.

The distance from (2, 3) to the line 3x + 4y – 8 = 0 is

= \(\left|\frac{3(2)+4(3)-8}{\sqrt{3^2+4^2}}\right|\) = \(\frac{10}{5}\) = 2.

The distance from B(- 4, a) to the line 3x + 4y – 8 = 0

= \(\left|\frac{3(-4)+4 a-8}{\sqrt{3^2+4^2}}\right|\) = \(\frac{|4 a-20|}{5}\)

∴ 2 = \(\frac{|4 a-20|}{5}\)

⇒ |4a – 20| = 10 ⇒ 4a – 20 = ±10

⇒ 4a = 30 or 4a = 10

⇒ a = \(\frac{15}{2}\) or a = \(\frac{5}{2}\)

Question 14.

Find the circumcentre of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0 and x – y = 2. (E.Q.)

Answer:

Equation of AB is x + y = 0, …………………. (1)

Equation of BC is 2x + y + 5 = 0 ……………………… (2 )

Equation of CA is x – y = 2 ………………………… (3)

Solving (1) and (2), Coordinates of B are (-5, 5)

Solving (2) and (3), Coordinates of C are (-1, -3)

Solving (1) and (3), Coordinates of A are (1, -1)

Equation of SE is y + 2 = – 1 (x)

⇒ x + y + 2 = 0 ………………. ( 5 )

Solving (4) and (5) we get coordinates of circumcentre S

⇒ x = – 3, y = 1

∴ Coordinates of circumcentre S = (- 3, 1)

Question 15.

If θ is the angle between the lines \(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{x}{b}+\frac{y}{a}\) = 1, find the value of sin θ when a > b. (S.A.Q.)

Answer:

Equation of given lines \(\frac{x}{a}+\frac{y}{b}\) = 1

⇒ bx + ay – ab = 0 ……………….. (1)

II.

Question 1.

Find the equation of straight lines passing through the point (- 10, 4) and making an angle 0 with the line x – 2y = 10 such that tan θ = 2. (E.Q.)

Answer:

Equation of BC is x – 2y – 10 = 0

Suppose slope of AB is’m’. AB passes through A (-10, 4)

Equation of AB is y – 4 = m(x + 10) = mx + 10m

⇒ mx – y + (10m + 4) = 0 ………………… (1)

⇒ (m^{2} + 1) = (m + 2)^{2} = m^{2} + 4m + 4

⇒ 4m + 3 = 0 ⇒ m = -3/4

Case (i): Coefficient of m^{2} = 0

⇒ one of the roots is ∞.

Hence AC is a vertical line

∴ Equation of AC is x + 10 = 0;

(∵ y – 4 = \(\frac{1}{0}\) (x + 10))

Case (ii) : m = \(\frac{-3}{4}\)

From (1) equation of AB is

\(\frac{-3}{4}\) – y + (latex]\frac{-30}{4}[/latex] + 4) = 0

⇒ \(\frac{-3 x-4 y-14}{4}\) = 0

⇒ 3x + 4y + 14 = 0

Question 2.

Find the equations of the straight lines passing through the point (1,2) and making an angle of 60° with the line √3 x + y + 2 = 0. (Board Model Paper) (E.Q.)

Answer:

PQ, PR are two lines passing through P(1, 2) and makes an angle of 60° with QR.

Let slope of PQ is m

Equation of PQ is y – 2 = m (x – 1) = mx – m

∴ mx – y + (2 – m) = 0 …………….. ( 1 )

⇒ m^{2} + 1 = (√3m – 1)^{2} = 3m^{2} + 1 – 2√3 m

⇒ 2m^{2} – 2 √3 m = 0

⇒ 2m (m – √3) = 0 ⇒ m = 0 or √3

Case (i): m = 0

Equation of PQ is – y + 2 = 0 or y – 2 = 0

Case (ii): m = √3

Equation of PQ is √3x – y + (2 – √3) = 0

Question 3.

The base of an equilateral triangle is x + y – 2 = 0 and the opposite vertex is (2, -1). Find the equations of the remaining sides. (E.Q.)

Answer:

Equation of BC is x + y – 2 = 0

AB passes through A (2, -1)

Suppose slope of AB is’m’

Equation of AB is y + 1 = m (x – 2) = mx – 2m

⇒ mx – y – (2m + 1) = 0 …………………. (1)

cos 60° = \(\frac{\left|a_1 a_2+b_1 b_2\right|}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}\)

⇒ \(\frac{1}{2}=\frac{|m-1|}{\sqrt{1+1} \sqrt{m^2+1}}\)

⇒ (m^{2} + 1) (2) = 4(m – 1)^{2}

⇒ 2m^{2} + 2 = 4(m^{2} – 2m + 1)

⇒ 2m^{2} – 8m + 2 = 0

⇒ m^{2} – 4m + 1 = 0

⇒ m = \(\frac{4 \pm \sqrt{16-4}}{2}\) = \(\frac{4 \pm \sqrt{12}}{2}\) = \(\frac{4 \pm 2 \sqrt{3}}{2}\) = 2 ± √3

Substituting in (1)

Equation of AB is y + 1 = (2 + √3) (x – 2)

Equation of AC is y + 1 = (2 – √3) (x – 2)

Question 4.

Find the orthocentre of the triangle with the following vertices (E.Q.)

(i) (- 2, – 1), (6, – 1) and (2, 5) (March ’07)

(ii) ( 5, – 2 ), ( – 1, 2 ) and ( 1, 4 ) (March ’12)

Answer:

(i) Let A( – 2, – 1), B( 6, – 1), C( 2, 5 ) are the vertices of ∆ABC

Slope of BC = \(\frac{5+1}{2-6}=\frac{6}{-4}=\frac{-3}{2}\)

AD is perpendicular to BC

Slope of AD = \(\frac{2}{3}\)

Equation of AD is y + 1 = \(\frac{2}{3}\) (x + 2)

⇒ 2x – 3y + 1 = 0 ………. (1)

Slope of AC = \(\frac{5+1}{2+2}=\frac{6}{4}=\frac{3}{2}\) = 2

BE is perpendicular to AC

Slope of BE = \(\frac{-2}{3}\)

Equation of BE is

y + 1 = \(\frac{-2}{3}\) (x – 6)

⇒ 2x + 3y – 9 = 0 ……………… (2)

Solving (1) and (2)

Coordinates of the orthocentre O are \(\left(2, \frac{5}{3}\right)\)

(ii) (5, – 2), (- 1, 2) and (1, 4)

Answer:

Let A(5, -2), B (-1, 2) and C(1, 4) be the vertices of ∆ABC.

Slope of BC =\(\frac{2-4}{-1-1}=\frac{-2}{-2}\) = 1

Slope of AD = – 1

Equation of AD is y + 2 = – 1 (x – 5)

⇒ x + y – 3 = 0 …………… (1)

Slope of AC = \(\frac{-2-4}{5-1}=\frac{-6}{4}=\frac{-3}{2}\)

∴ Slope of BE = \(\frac{2}{3}\)

∴ Equation of BE is y – 2 = \(\frac{2}{3}\) (x + 1)

⇒ 3y – 6 = 2x + 2

⇒ 2x – 3y + 8 = 0 ………………. (2)

Solving (1) and (2) we get the coordinates of the orthocentre ‘O’

∴ Coordinates of orthocentre O are \(\left(\frac{1}{5}, \frac{14}{5}\right)\)

Question 5.

Find the circumcentre of the triangle whose vertices are given below.

(i) (-2, 3), (2, -1) and (4, 0) (March 2011)

(ii) (1, 3), (0, – 2) and (- 3, 1) (May 2006) (E.Q.)

Answer:

(i) (-2, 3), (2, -1) and (4, 0)

Let A(-2, 3), B(2, -1) and C(4, 0) be the vertices of ∆ ABC.

Mid point of side BC; D = \(\left(\frac{4+2}{2}, \frac{-1+0}{2}\right)\)

= \(\left(3,-\frac{1}{2}\right)\)

Slope of BC = \(\frac{-1-0}{2-4}=\frac{1}{2}\)

∴ Slope of SD is – 2.

Equation of SD is y + \(\frac{1}{2}\) = – 2(x – 3)

⇒ 2y + 1 = – 4 (x – 3)

⇒ 4x + 2y – 11 = 0 ……. (2)

E is the mid point of AC Coordinates of E are

⇒ x = \(\frac{12}{8}\) = \(\frac{3}{2}\) and 4\(\left(\frac{3}{2}\right)\) – 2y – 1 = 0

⇒ – 2y + 5 = 0 ⇒ y = \(\frac{5}{2}\)

∴ Coordinates of circumcentre S = \(\left(\frac{3}{2}, \frac{5}{2}\right)\)

(ii) (1, 3), (0, – 2) and (- 3, 1)

Let A(1, 3), B(0, -2) and C(-3, 1) be the vertices of ∆ABC. D is the mid point of BC.

SE is perpendicular to CA

Slope of SE is -2

Equation of SE is y – 2 = – 2 (x + 1)

⇒ 2x + y = 0 (2)

Solving (1) and (2)

Question 6.

Let \(\overline{\text { PS }}\) be the median of the triangle with vertices P( 2, 2 ), Q ( 6, – 1 ) and R ( 7, 3 ). Find the equation of the straight line passing through ( 1, – 1) and parallel to the median PS. (E.Q.)

Answer:

Let P (2, 2), Q (6, -1) and R (7, 3) be the vertices of ∆ABC.

S is the mid point of QR.

AB is parallel to PS and passes through A(1, – 1)

Equation of AB is y + 1 = \(\frac{-2}{9}\) (x – 1)

⇒ 9y + 9 = – 2x + 2

⇒ 2x + 9y + 7 = 0

III.

Question 1.

Find the orthocentre of the triangle formed by the lines x + 2y = 0, 4x + 3y – 5 = 0 and 3x + y = 0 (E.Q.)

Answer:

Equation of AB is x + 2y = 0 …………… (1)

Equation of BC is 4x + 3y – 5 = 0 …………… (2)

Equation of AC is 3x + y = 0 …………………. (3)

Solving (1) and (3) coordinates of A are (0,0)

Solving (1) and (2) we get the coordinates of C

Coordinates of C = (-1. 3)

Slope of BC = \(\frac{-1-3}{2+1}\) = \(\frac{-4}{3}\)

∴ Slope of AD = \(\frac{3}{4}\)

∴ Equation of AD is y – 0 = \(\frac{3}{4}\) (x – 0)

⇒ 3x – 4y = 0 ……… (4)

Slope of AC = \(\frac{0-3}{0+1}\) = – 3

∴ Slope of BE = \(\frac{1}{3}\)

∴ Equation of BE is y + 1 = \(\frac{1}{3}\) (x – 2)

⇒ 3y + 3 = x- 2

⇒ x – 3y – 5 = 0 ……………….. (5)

Solving (4) and (5) we get the coordinates of orthocentre.

⇒ x = -4, y = -3

∴ Coordinates of orthocentre of ∆ABC is 0(- 4, -3)

Question 2.

Find the circumcentre of the triangle whose sides are given by x + y + 2 = 0; 5x – y – 2 = 0 and x – 2y + 5 = 0 (E.Q.)

Answer:

Given lines are x + y + 2 = 0 ………………….. ( 1 )

5x – y – 2 = 0 ………………….. (2)

x – 2y + 5 = 0 ………………….. (3)

Point of intersection of lines (1) and (2) is A = (0, -2)

Point of intersection of lines (2) and (3) is B = (1, 3)

Point of intersection of lines (1) and (3) is C(-3, 1)

D is the mid point of BC;

∴ Coordinates of D = \(\left(\frac{1-3}{2}, \frac{3+1}{2}\right)\)

= (-1, 2)

Slope of BC = \(\frac{3-1}{1+3}=\frac{2}{4}=\frac{1}{2}\)

Slope of SD = -2

Equation of SD is y – 2 = -2 (x + 1)

⇒ 2x + y = 0 …

E is the mid point of AC

(0-3 -2 + 0

∴ Coordinates of E = \(\left(\frac{0-3}{2}, \frac{-2+1}{2}\right)\)

= \(\left(\frac{-3}{2}, \frac{-1}{2}\right)\)

Slope of AC is \(\frac{-2-1}{0+3}\) = – 1

∴ Slope of SE = 1

∴ Equation of SE is y + \(\frac{1}{2}\) = 1 \(\left(x+\frac{3}{2}\right)\)

⇒ 2y + 1 = 2x + 3

⇒ 2x – 2y + 2 = 0

⇒ x – y + 1 = 0 ……………… (5)

Solving (4) and (5) we get the coordinates of circumcentre ‘S’.

Question 3.

Find the equation of the straight lines passing through (1, 1) and which are at a distance of 3 units from (- 2, 3). (E.Q.)

Answer:

Let the line passing through A(l, 1) has slope’m’ then equation of the line is y – 1 = m(x – 1)

⇒ mx – y + (1 – m) = 0 …………… (1)

Distance from (-2, 3) to the line = 3

∴ \(\frac{|m(-2)-3+(1-m)|}{\sqrt{m^2+1}}\) = 3

⇒ (3m + 2)^{2} = 9(m^{2} + 1)

9m^{2} + 12m + 4 = 9m^{2} + 9

⇒ 12m = 5 ⇒ m = \(\frac{5}{12}\)

Coefficient of m^{2} = 0 ⇒ m = ∞

(i) m = ∞

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

⇒ x = 1

(ii) m = \(\frac{5}{12}\)

Substitute in (1)

Equation of the line is y – 1 = \(\frac{5}{12}\) (x – 1)

⇒ 12y – 12 = 5x – 5

⇒ 5x – 12y + 7 = 0

Question 4.

If p and q are the lengths of the perpendiculars from the origin to the straight lines x sec α + y cosec α = a and x cos α – y sin α = a cos 2α. Prove that 4p^{2} + q^{2} = a^{2} (E.Q.)

Answer:

Equation of the line is x sec α + y cosec α = α

⇒ \(\frac{x}{\cos \alpha}+\frac{y}{\sin \alpha}\) = a

⇒ x sin α + y cos α = a sin α cos α

⇒ x sin α + y cos α – a sin α, cos α = 0 …….. (1)

p = length of the perpendicular from 0 on the line (1)

= \(\left|\frac{0+0-a \sin \alpha \cos \alpha}{\sqrt{\sin ^2 \alpha+\cos ^2 \alpha}}\right|\)

⇒ a sin α + cos α = p

⇒ 2p = a sin 2 α

Equation of the other given line is

x cos α – y sin α = a cos 2α ………………. ( 2 )

q = length of the perpendicular from 0 on the line (2)

∴ q = \(\left|\frac{a \cos 2 \alpha}{\sqrt{\cos ^2 \alpha+\sin ^2 \alpha}}\right|\)

⇒ a^{2} cos^{2} 2α = q^{2}

∴ 4p^{2} + q^{2} = a^{2} sin^{2} 2α + a^{2} cos^{2} 2α

= a^{2} (sin^{2} 2α + cos^{2} 2α) = a^{2} (1) = a^{2}

Question 5.

Two adjacent sides of a Parallelogram are given by 4x + 5y = 0 and 7x + 2y = 0 and one diagonal is 1 lx + 7y = 9. Find the equation of the remaining sides and the other diagonal. (E.Q.)

Answer:

Let 4x + 5y = 0 ……………….. (1) and

7x + 2y = 0 …………….. (2) respectively

denote the sides \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\) of the parallelogram OABC

Equation of the diagonal AB is 1 lx + 7y -9 =0 .

Solving (1) and (2) we get O = (0, 0)

Solving (1) and (3) we get

⇒ – 15y + 35 = 12x + 8

⇒ 12x + 15y – 27 = 0

⇒ 4x + 5y – 9 = 0

Question 6.

Find the in – centre of the triangle formed by the following straight lines. (E.Q.)

(i) x + 1= 0, 3x – 4y = 5 and 5x + 12y = 27

(ii) x + y – 7 = 0, x – y + 1 = 0 and x – 3y + 5 = 0

Answer:

(i)x + 1 = 0, 3x – 4y = 5 and 5x + 12y = 27

x + 1 = 0 …………….. (1)

3x – 4y = 5 …………………. (2)

5x + 12y = 27 ………………. (3) are the given equations of lines.

Point of intersection of (1) and (2) is A = (-1,-2)

Point of intersection of (2) and (3) is

(ii) x + y – 7 = 0, x – y + 1 = 0 and x – 3y + 5 = 0

Answer:

x + y – 7 = 0 …………….. (1)

x – y + 1 = 0 …………….. (2)

x – 3y + 5 = 0 …………….. (3)

Solving (1) and (2) we get

∴ In-center of the triangle formed by the sides x + y – 7 = 0, x – y + 1 = 0 and x – 3y + 5 = 0 is (3√5 + 1)

Question 7.

A triangle is formed by the lines ax + by + c = 0, lx + my + n = 0 and px + qy + r = 0. Given that the triangle is not right angled, show that the straight line \(\frac{a x+b y+c}{a p+b q}\) = \(\frac{l \mathbf{x}+\mathbf{m y}+\mathbf{n}}{l \mathbf{p}+\mathbf{m q}}\) passes through the orthocentre of the triangle. (E.Q.)

Answer:

Given equations of lines are

ax + by + c = 0 ……………… (1)

lx + my + n = 0 ……………. (2)

px + qy + r = 0 ……………… (3)

Equation of the line passing through the point of intersection of (1) and (2) is

(ax + by + c) + k (/x + my + n) = 0 (4)

⇒ (a + kl) x + (b + km) y + (c + nk) = 0

It is perpendicular to (3) then

p(a + kl) + q (b + km) = 0

⇒ k = – \(\left(\frac{\mathrm{ap}+\mathrm{bq}}{l \mathrm{p}+\mathrm{mq}}\right)\)

Substitute in (4) we get

(ax + by + c) – \(\left(\frac{\mathrm{ap}+\mathrm{bq}}{l \mathrm{p}+\mathrm{mq}}\right)\) (lx + my + n) = 0

⇒ \(\frac{\mathrm{ax}+\mathrm{by}+\mathrm{c}}{\mathrm{ap}+\mathrm{bq}}=\frac{l \mathrm{x}+\mathrm{my}+\mathrm{n}}{l \mathrm{p}+\mathrm{mq}}\)

is the required equation of the line line passing through the orthocentre of triangle which represents the altitude through A.

Question 8.

The Cartesian equations of the sides

\(\overleftrightarrow{\mathbf{B C}}, \overleftrightarrow{\mathbf{C A}}\) and \(\overleftrightarrow{\mathbf{A B}}\) 0f a triangle are respectively u_{r} = a_{r}x + b_{r}y + c_{r} = 0, r = 1, 2, 3. Show that the equation of the straight line passing through A and bisecting the side \(\overline{\mathbf{B C}}\) is

\(\frac{u_3}{a_3 b_1-a_1 b_3}=\frac{u_2}{a_1 b_2-a_2 b_1}\) (E.Q.)

Answer:

A is a point of intersection of lines u_{2} = 0 and u_{3} = 0

Equation of the line passing through A is u_{2} + λu_{3} = 0

⇒ (a_{2}x + b_{2}y + c_{2}) + λ(a_{3}x + b_{3}y + c_{3}) = 0 …….(1)

⇒ (a_{2} + λa_{3})x + (b_{2} + λb_{3})y + (c_{2} + λc_{3}) = 0

If this is parallel to a_{1}x + b_{1}y + c_{1} = 0 we get