Question

Find the coordinates of the orthocenter of the triangle whose vertices are A(3, 1), B(0, 4) and C(-3, 1).

Answer 1

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

 

Therefore the orthocenter is (0, 4)

Therefore the orthocenter is (0, 4).