Question

Find the coordinates of the orthocenter of the triangle whose vertices are A(2, −2), B(1, 1), and C(−1, 0).

Answer 1

Let AM and BN be the altitudes of the ΔABC.

Now, slope of BC = `(0 - 1)/(-1 - 1) = 1/2`

Altitude AM a perpendicular to side BC.

∴ slope of altitude AM = –2 and it is passing through A(2, –2).

∴ equation of the altitude AM is y – (–2) = –2(x – 2)

∴ y + 2 = –2x + 4

∴ 2x + y = 2 ....(1)

Slope of side AC = `(0 - (- 2))/(-1 - 2) = -2/3` 

Altitude BN is perpendicular to side AC.

∴ slope of altitude BN = `3/2` and it is passing through B (1, 1).

∴ equation of the altitude BN is

y – 1 = `3/2(x - 1)`

∴ 2y – 2 = 3x – 3

∴ 3x – 2y = 1 ....(2)

The orthocentre H is the point of intersection of the altitudes AM and BN. Hence, we solve equations (1) and (2).

Multiply equation (1) by 2, we get,

4x + 2y = 4 ...(3)

Adding (2) and (3), we get,

   3x – 2y = 1
+ 4x + 2y = 4
    7x = 5

∴  x = `5/7`

∴  from (1), `2(5/7) + y` = 2

∴ y = `2 - 10/7`

= `4/7`

Hence, coordinates of orthocentre H are `(5/7, 4/7)`.