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)
