What is the equation of the line that is perpendicular to the given line and passes through the point (5,3) a) 4x-5y=5 b)5x+4y=37 c)4x+5y=5 d)5x-4y=8

perpendicular lines have a slope that is a negative reciprocal

A) 4x-5y=5

subtract 4x

solve for y

-5y = -4x+5

divide by -5

y = 4/5 x+5   slope is 4/5  perpendicular slope is -5/4

y -y1 =m(x-x1)  point slope form of a line

y-3 = -5/4 (x-5)

B)  5x+4y = 37

subtract 5x

4y =-5x +37

divide by 4

y =-5/4 x +37/4  slope is -5/4 perpendicular slope is 4/5

y -y1 =m(x-x1)  point slope form of a line

y-3 = 4/5 (x-5)

C)4x+5y=5

subtract 4x

5y = -4x +5

divide by 5

y = -4/5 x +1

y =-4/5 x +1  slope is -4/5 perpendicular slope is 5/4

y -y1 =m(x-x1)  point slope form of a line

y-3 = 5/4 (x-5)

D)5x-4y=8

subtract 5x

-4y = -5x+8

divide by -4

y = 5/4 x-2

y =5/4 x +-2 slope is 5/4 perpendicular slope is -4/5

y -y1 =m(x-x1)  point slope form of a line

y-3 = -4/5 (x-5)

what are you trying to solve here

step-by-step explanation: