Perpendicular to line BC containing A

Question

A (14, -2)  B (-3, 15) C (10,10)

Perpendicular to line BC containing A?

Answer 1

first we calculate the slope of line BC:

(10 - 15) / 10 - (-3)   ---> -5/13

we need the slope of the line perpendicular to BC which is:   -1/(-5/13)  ---> 13/5

so, we have the slope of the line perpendicular to line BC. Since we need the equation of the line perpendicular to BC and containing point A, we use the point-slope form to figure out the equation.

y - (-2) = (13/5)*(x - 14)

y + 2 = (13/5)x - 182/5

y = (13/5)x - 182/5 - 10/5

y = (13/5)x - 192/5