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

l1: 3x+4y-7 = 0 and l2: x - y + 2 = 0
The equation of required line passing through the points of intersectiopn of l1 and l2 is given by
l1 + λl2 = 0
∴ (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 =0

Solution

l1: x+2y-3 = 0, l2: 4x-y+7 = 0, l3:5x+4y-20 =0
The equation of required line passing through the points of intersectiopn of l1 and l2 is given by
l1 + λl2 = 0
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 }}$
l3: 5x + 4y -20 = 0
Slpe of l3 = -5/4
$\overleftrightarrow {l}$||$\overleftrightarrow {l_3}$
∴ slope of l = slope of l3
$\frac{{ - \left( {1 + 4\lambda } \right)}}{{2 - \lambda }} = \frac{{ - 5}}{4}$
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 -4

Solution

l1: 2x + 3y - 4 = 0, l2: x -5y-7=0
The equation of required line which passes through the point of intersection of l1 and l2 is given by
l1 + λl2 = 0
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 = 0

Solution

l1: 5x - 3y -1 = 0, l2: 2x+3y-23 = 0, l3: 5x - 3y - 1 = 0
The equation of required line passing through the pont of intersection of l1 and l2 is given by
l1 + λl2 = 0
5x -3y - 1 + λ(2x +3 y -23) = 0 ----(1)
(5+2λ)x + (-3+3Λ)y - 1 - 23λ = 0
slope of line l = $\frac{{ - a}}{b} = \frac{{ - \left( {5 + 2\lambda } \right)}}{{\left( { - 3 + 3\lambda } \right)}}$
l3: 5x - 3y - 1 = 0
slope of l3 $ = \frac{{ - 5}}{{ - 3}} = \frac{5}{3}$
l ⊥ l3
∴ slope of l × slope of l3 = -1
$\frac{{\left( {5 + 2\lambda } \right)}}{{\left( { - 3 + 3\lambda } \right)}} \times \frac{5}{2} = 1$
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
Exercise 10.3 ⇐
⇒ Exercise 10.5