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?

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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**

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