11th NCERT/CBSE Straight Lines Exercise 10.4(supplementary) Questions 4

Hi

Question (1)

Find the equation of line through the intersection of line of lines 3x+4y =7 and x-y+2=0 and whose slope is 5.Solution

lThe equation of required line passing through the points of intersectiopn of l

l

∴ (3x + 4y - 7) + λ(x - y + 2) = 0 ----(1)

(3 + &lamb)x + ( 4 - λ)y - 7 +2λ = 0

Slope of line $\overleftrightarrow {l}= \frac{{ - a}}{b} = \frac{{ - \left( {3 + \lambda } \right)}}{{4 - \lambda }}$

Slope of line $\overleftrightarrow {l}$ =

$\frac{{ - \left( {3 + \lambda } \right)}}{{4 - \lambda }} = 5$

-3-λ = 20 - 5λ

4λ = 23

$\lambda = \frac{{23}}{4}$

Replace value of λ in equation (1) we get

$3x + 4y - 7 + \frac{{23}}{4}\left( {x - y + 2} \right) = 0$

12x + 16y -28 +23x - 23y + 46 = 0

35x - 7y +18 = 0

Question (2)

Find the equation of the line through the intersection of lines x+2y-3 = 0 and 4x-y+7 = 0 and which is parallel to 5x+4y-20 =0Solution

lThe equation of required line passing through the points of intersectiopn of l

l

x+2y-3 + λ(4x - y +7) = 0 ----(1)

(1 +4λ)x + (2-λ)y -3 +7 λ = 0

Slope of line $\overleftrightarrow {l}$ =

$ = \frac{{ - a}}{b} = \frac{{ - \left( {1 + 4\lambda } \right)}}{{2 - \lambda }}$

l

Slpe of l

$\overleftrightarrow {l}$||$\overleftrightarrow {l_3}$

∴ slope of l = slope of l

$\frac{{ - \left( {1 + 4\lambda } \right)}}{{2 - \lambda }} = \frac{{ - 5}}{4}$

4 +16λ = 10 - 5λ

21λ = 6

λ = 6/21 = 2/7

Replacing the value of λ in (1) we get

$x + 2y - 3 + \frac{2}{7}\left( {4x - y + 7} \right) = 0$

7x + 14 y - 21 + 8x - 2y +14 = 0

15x +12y -7 = 0

Question (3)

Find the equation of line through the intersection of the line 2x+3y-4=0 and x-5y=7 that has x-intercept equal to -4Solution

lThe equation of required line which passes through the point of intersection of l

l

2x + 3y -4 + λ(x-5y-7) = 0 ----(1)

(2+λ)x + (3-5λ)y - 4 - 7λ = 0

x intercept of line $ = \frac{{ - c}}{a} = \frac{{ - \left( { - 4 - 7\lambda } \right)}}{{2 + \lambda }}$

x intercept of line = - 4(given)

$\therefore \frac{{ - \left( { - 4 - 7\lambda } \right)}}{{2 + \lambda }} = \frac{{ - 4}}{1}$

-4-7λ = 8 + 4λ

-12 = 11 λ

$\lambda = \frac{{ - 12}}{{11}}$

Replace value of λ in equation (1) we get

$\left( {2x + 3y - 4} \right) - \frac{{12}}{{11}}\left( {x - 5y - 7} \right) = 0$

22x + 33y - 44 -12x + 60y + 84=0

10x + 93y + 40=0

Question (4)

Find the equation of line through the intersection of 5x-3y = 1 and 2x+3y-23=0 and perpendicular to the line 5x-3y-1 = 0Solution

lThe equation of required line passing through the pont of intersection of l

l

5x -3y - 1 + λ(2x +3 y -23) = 0 ----(1)

(5+2λ)x + (-3+3Λ)y - 1 - 23λ = 0

slope of line l = $\frac{{ - a}}{b} = \frac{{ - \left( {5 + 2\lambda } \right)}}{{\left( { - 3 + 3\lambda } \right)}}$

l

slope of l

l ⊥ l

∴ slope of l × slope of l

$\frac{{\left( {5 + 2\lambda } \right)}}{{\left( { - 3 + 3\lambda } \right)}} \times \frac{5}{2} = 1$

5(5+2λ) = 3 (-3+3λ)

25+10λ = -9 + 9λ

λ = -34

Replce the value of λ in equation(3)

(5x-3y-1) - 34(2x+3y-23) = 0

5x-37-1-68x-102y+782 = 0

-63x-105y+781 = 0

63x+105y-781 = 0