Question

2. Find equations of altitudes of the
triangle whose vertices are A(2,5), B(6-1)
and C(-4,-3).​

Answer 1

Step-by-step explanation:

Let AD be the altitude from the vertex A of ∆ABC.

Then AD is perpendicular to BC..

Now slope of BC m1 = (y2-y1)/(x2-x1)

= (-3+1)/(-4–60) = -2/-64 = 1/32.

So slope of AD, m = -1/(1/32) = -32 ( since AD is perpendicular to BC)

So equation of AD = y - y1= m(x-x1)

=> y - 5 = -32(x-2) => 32x + y - 5 - 64 = 0

ie 32x + y - 69 = 0