Find x such that the line through (6, −3) and (7, 6) is perpendicular to the line through (−2, 5) and (x, −1).

Question

Find x such that the line through (6, −3) and (7, 6) is perpendicular to the line through (−2, 5) and (x, −1).

Answer 1

You can just work with the slopes here, finding equations is unnecessary.

Slope of line between (6, -3) and (7,6) = (y2-y1)/ (x2-x1) = (6--3) / (7 -6) = 9/1 = 9

Slope of line between (-2, 5) and (x, -1) = ( -1 - 5) / (x - - 2) = (-6 / (x+2))

These lines are perpendicular so product of slopes is -1

So 9 ( -6 / (x+2)) = -1

-54 / (x+2) = -1

54 = x + 2

x = 52

Answer 2

x+9y = 43 is a line perpendicular to the line through (6,-3) and (7,6)

the line through those two points is y=9x -57. the perpendicular line is y=-x/9 + 43/9. multiply by 9 to get 9y=-x+43 or x+9y=43. plug in the point (x,-1) to solve for x

x+9(-1) = 43

x = 43+9 = 52

x = 52