Question 20400
To find the equation of line passing through (3,-2)
and perpendicular to y=2x+4

We know that the perpendicular line for ax+by+c = 0 is -bx+ay+k=0

So the perpendicular line to 2x-y+4=0 is x+2y+k =0 ----->(1)

We have to find the value of k.

Since the line passing through the point(3,-2),it satisfies the equation.

plug x=3 and y = -2 in (1)

3+2(-2)+k =0.

-1 + k = 0.

k=1.

Plug the value of k in (1) we get the required line.

Therefore the required line is, x+2y+1=0.