write an equation for a perpendicular line with point (-1,2) for the given line 3x+5y-2=7

Question

I don't know what to do after finding the reciprocal for the slope

Answer 1

Given line 3x+5y-2 = 7

Add 2 to each side.

3x+5y-2+2 = 7+2

Subtract 3x from each side.

3x+5y-3x = 9-3x

5y = -3x+9

Divide to each side by 5.

5y/5 = -3x/5+9/5

y = (-3/5)x+9/5

Given line slope is m1 = -3/5

Perpendicular line slope say m2 is be 5/3.

And the point (-1,2) = (x1,y1)

Perpendicular line equation is y-2 = 5/3(x+1)

3(y-2) = 5(x+1)

3y-6 = 5x+5

Add 6 t0 each side.

3y+6-6 = 5x+5+6

3y = 5x+11

Divide to each side by 3.

3y/3 = (5/3)x+11/3

Required line equation is y = (5/3)x+11/3