Question

Find equations of 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) and 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 the point (2, 5) and has slope –5,

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 a slope `(-3)/4`,

equation of the altitude BE is

y – (– 1) = `(-3)/4(x - 6)`

∴ 4(y + 1) = –3(x – 6)

∴ 4y + 4 = –3x + 18

∴ 3x + 4y – 14 = 0

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

= `(-6)/4`

= `(-3)/2`

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

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

equation of altitude CF is

y – (– 3) = `2/3[x - (- 4)]`

∴ 3(y + 3) = 2(x + 4)

∴ 3y + 9 = 2x + 8

∴ 2x – 3y – 1 = 0