Equations of Altitudes in Triangles

Question

Quick question about how I would solve this question:

Overview:

ABC has the vertices A(3, 4) B(-5, 2) and C(1, -4).

a) Determine an equation for the altitude from C to AB.

I know how to do this part of the question, but not the next.

b) Verify that (-3, 12) lies on the line containing the altitude from C to AB.

Just found part B to be a little confusing and was hoping for some clarification on how to do it.

Thank you.

Answer 1

The slope of AB is (4-2)/(3--5) = 2/8 = 1/4; the slope perpendicular to this is the negative reciprocal, or -1/(1/4) = -4

Thus, the equation of the line, using y = mx + b, and containing the point (1, -4), can be solve as -4 = -4(1) + b

b = 0

y = -4x

b) (3, -12) can easily be checked to be on the line y = -4x; -4(3) = -12

By the way, a fun question c would be to find where the altitude intersects AB

Solve y = -4x; and, for (3,4); (-5,2), as the slope equals 1/4, 4 = 1/4(3) + b; b = 3 1/4

Thus, we have y = -4x and y = 1/4x + 3 1/4; -4x = 1/4x + 3 1/4; - 17/4 x = 13/4;

x = -13/17;y = -4x = 52/17 (-13/17, 52/17)