A (14, -2) B (-3, 15) C (10,10)
Perpendicular to line BC containing A?
A (14, -2) B (-3, 15) C (10,10)
Perpendicular to line BC containing A?
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