The orthocenter is the intersection of the 3 altitudes.
Draw the triangle with the given coordinates.
Let AH be the 1st altitude and CH', the 2nd
Methodology: we have to find the equation of AH and BC. Once calculated we will equalize them.
a) Slope of BC , m =(-2-2)/(2+2) =-1, AH is perpendicular to BC,
so its slope = 1
and the equation of AH = 1(x) + b. Since we know the coordinates of A, we can find b; -2= 1(- 3) + b and b= 1 and Y(AH) = x+1
b) Slope of AB: m =(-2-2) / (-3+2) = 4, CH' perpendicular to AB,
so its slope = -1/4
and the equation of CH' = -1/4(x) + b. Since we know the coordinates of C, we can find b; -2= -1/4(2) + b and b= -3/2 and Y(CH') = (-1/4)x-3/2
c) Y(AH) = Y(CH')
x+1 = (-1/4).x -3/2, → x = -2, plug in any equation: Y(AH) = x+1;
Y(AH) = - 2 + 1 ; → Y = -1
So the coordinate of the orthocenter are (-2,-1)
