whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

Question

need a written equation for line perpendicular to the given line that passes through the given points

Answer 1

-2+6y = 13x+2

Add 2 to each side.

-2+2+6y = 13x+2+2

6y = 13x+4

Divide to each side by 6

6y/6 = (13/6)x+4/6

y = (13/6)x+2/3

Slope of the given line m1 = 13/6

Slope of the perpendicular line say m2 = -6/13.

Point (x1,y1) = (-2,-4)

Equation is y+4 = -6/13(x+2)

y+4 = (-6/13)x-12/13

Subtract 4 to each side.

y+4-4 = (-6/13)x-12/13-4

Required line equation is y = (-6/13)x-52/13

Comments

Typo mistake in last line of above solution.

Perpendicular line is y = (-6/13)x - (64/13)