The point of intersection of two perpendicular drawn from any two vertices of the triangle is known as orthocenter.
Let us consider the perpendicular drawn from A is AD and perpendicular drawn from the vertex B is BE.
Slope of AC = [(y2 - y1)/(x2 - x1)]
A (3, 1) and C (-3, 1)
here x1 = 3, x2 = -3, y1 = 1 and y2 = 1
= (1 - 1) / (-3 - 3)
= 0 / (-6)
= 0
Slope of the altitude BE = -1/ slope of AC
= -1/0
Equation of the altitude BE :
(y - y1) = m (x -x1)
Here B (0, 4) and m = -1/0
(y - 4) = -1/0 (x - 0)
10 (y - 4) = -1(x)
x + 10y - 40 = 0 --------(1)
Slope of BC = (y2 - y1)(x2 - x1)]
B (0, 4) and C (-3, 1)
Here x1 = 0, x2 = -3, y1 = 4 and y2 = 1.
= (1 - 4) / (-3 - 0)
= (-3)/(-3)
= 1
Slope of the altitude AD = -1/ slope of AC
= -1/1
= -1
Equation of the altitude AD :
(y - y1) = m (x - x1)
Here A(3, 1) m = -1
(y - 1) = -1 (x - 3)
(y - 1) = -x + 3
x + y - 1 - 3 = 0
x + y - 4 = 0
x = -y + 4--------(2)
Substituting (2) into (1), we get
-y + 4 + 10y - 40 = 0
9y - 36 = 0
y = 36/9
y = 4
By applying y = 4 in (1), we get
x = -4 + 4
x = 0