Answer the following question:

The vertices of a triangle are A(1, 4), B(2, 3) and C(1, 6) Find equations of altitudes of ∆ABC

#### Solution

Let AX, BY and CZ be the altitudes through the vertices A, B and C respectively of ∆ABC.

Slope of BC = – 3

∴ Slope of AX = `1/3` ...[∵ AX ⊥ BC]

Since altitude AX passes through (1, 4) and has slope `1/3`,

equation of altitude AX is

y – 4 = `1/3(x - 1)`

∴ 3y – 12 = x – 1

∴ x – 3y + 11 = 0

Since both the points A and C have same x co-ordinates i.e. 1,

the points A and C lie on the line x = 1.

AC is parallel to Y-axis and therefore, altitude

BY is parallel to X-axis

Since the altitude BY passes through B(2, 3),

the equation of altitude BY is y = 3.

Also, slope of AB = – 1

∴ Slope of CZ = 1

Since altitude CZ passes through (1, 6) and has slope 1,

equation of altitude CZ is

y – 6 = 1(x – 1)

∴ x – y + 5 = 0