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

Step-by-step explanation:

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

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