Answer to Question #138607 in Algebra for Yolanda Hopa

Question

The equation of a line perpendicular to 3y+6x=18 through the point (4;-1)

Answer 1

The gradient of the line 3y+6x=18

Is given by the equation y=mx+c

Where m is the gradient

Rewriting the equation in the form y=mx+c

y=-2x+6

m=-2

For a perpendicular line m1xm2=-1

Therefore the gradient of the perpendicular line is 1/2

Given the point (4,-1) we assume a general point (x,y)

Therefore the gradient is (y+1)/(x-4)=1/2

2y+2=x-4

2y-x=-6