Question

Find the equation of the line passing through the point of intersection of the lines 5x-8y+23=0 and 7x+6y-71=0 and perpendicular to the line 4x-2y=3

Answer 1

The equation of any plane through the line of intersection of planes 5x-8y+23=0 and 7x+6y-71=0.
(5x-8y+23) + k(7x+6y-71)=0
5x-8y+23+7kx+6ky-71k=0
5x+7kx-8y+6ky-71k+23=0
(5+7k)x+(-8+6k)y-71k+23=0

Therefore,this plane is perpendicular to the plane 4x-2y=3.

(5+7k)4+(-8+6k)-2=0
20+28k+16-12k=0
16k+36=0
16k=-36
k= -36/16
k=-9/4

(5x-8y+23)-9/4(7x+6y-71)=0
5x-8y+23-63/4x-54/4y+639/4=0
20x-63x/4 -32y-54y/4 +92+639/4 =0
-43x-86y+731=0
43x+86y-731=0
This is the required equation.

Answer 2

7x + 6y = 71 28x + 24 = 284 ..(1) 5x - 8y = -23 15x – 24y = -69

(2)

Adding (1) and (2), we get,

43x = 215

x = 5

From (2), 8y = 5x + 23 = 25 + 23 = 48 = y = =

= 6

Thus, the required line passes through the point (5, 6).

4x - 2y = 1 =

2y = 4x - 1

y = 2x –1/2

Slope of this line = 2

Slope of the required line = 1/2

The required equation of the line is

y - y1 = m (x1,x2)

y – 6 = (x – 5) ×— 1/2

2y - 12 = -x + 5

x+2y-17=0

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