Question

Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1) and C(– 4, – 3).

Answer 1

A(2, 5), B(6, – 1), C(– 4, – 3) are the vertices of ΔABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ΔABC.

Slope of BC = `(-3 - (- 1))/(4 - 6) = (-2)/(-10) = 1/5`

∴ slope of AD = – 5      ....[∵ AD ⊥ BC]
Since, altitude AD passes through (2, 5) and has slope – 5.
∴ the equation of the altitude AD is
y – 5 = – 5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x + y – 15 = 0

Now, slope of AC = `(-3 - 5)/(-4 - 2) = (-8)/(-6) = 4/3`

∴ Slope of BE = `(-3)/4`    ....[∵ BE ⊥ AC]

Since, altitude BE passes through (6, – 1) and has slope `(-3)/4`.

∴ the equation of the altitude BE is

y – (– 1) = `(-3)/4 (x - 6)`

∴ 4(y + 1) = – 3 (x – 6)
∴ 3x + 4y – 14 = 0 

Also, slope of AB = `(-1 - 5)/(6 - 2) = (-6)/4 = (-3)/2`

∴ Slope of BE = `2/3`    ....[∵ CF ⊥ AB]

Since, altitude CF passes through (– 4, – 3) and has slope `2/3`.

∴ the equation of the altitude CF is

y – (– 3) = `2/3[x - (- 4)]`

∴ 3 (y + 3) = 2 (x + 4)
∴ 2x – 3y – 1 = 0.