Write the equation of a line that is perpendicular to the given line and passes through the given point.
y-3=8/3(x+20;(-2,3)
+128475 Good answer, Solveit.......you are exactly correct....!!!!
A perpendicular line will have the negative reciprocal slope to 8/3 = -3/8
So.....we have
y - 3 = -(3/8)(x - [-2] )
y = (-3/8)x -6/8 + 3
y = (-3/8)x + 18/8
y = (-3/8)x + 9/4
Here is the graph of both lines......https://www.desmos.com/calculator/btb92kqc5y
+2498 y-3=8/3(x+20)
y=8/3x+160/3+3 y=mx+b
our b is 169/3
mx is 8/3x slope is 8/3
formula of perpendicular is -1/m
-1/8/3=-3/8 is the slope of perpendicular line
we have:
m=-3/8
x=-2
y=3
let s put it into y=mx+b
3=-3/8(-2)+b 3=3/4+b b is 9/4
equation:y=-3/8x+9/4
+128475 Good answer, Solveit.......you are exactly correct....!!!!
A perpendicular line will have the negative reciprocal slope to 8/3 = -3/8
So.....we have
y - 3 = -(3/8)(x - [-2] )
y = (-3/8)x -6/8 + 3
y = (-3/8)x + 18/8
y = (-3/8)x + 9/4
Here is the graph of both lines......https://www.desmos.com/calculator/btb92kqc5y
