Find equations of altitudes of the triangle whose vertices are A(2,5), B(6,–1) and C(–4,–3).
Solution
Let AD be the altitude through A.
∴ AD is perpendicular to BC and
slope of BC = `(-3 -( - 1))/(-4 - 6)`
= `(-3 + 1)/(-10)`
= `1/5`
∴ slope of AD = – 5
Since altitude AD passes through the point A (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
Let BE be the altitude through B.
∴ BE is perpendicular to AC and
slope of AC = `(-3 - 5)/(-4 - 2)`
= `(-8)/(-6)`
= `4/3`
∴ slope of BE = `-3/4`
Since altitude BE passes through the point B (6, – 1) and has slope `-3/4`, equation of the altitude BE is
y – (– 1) = `-3/4(x - 6)`
∴ 4y + 4 = – 3x + 18
∴ 3x + 4y – 14 = 0
Let CF be the altitude through C.
∴ CF is perpendicular to AB and
slope of AB = `(5 - ( - 1))/(2 - 6)`
= `6/(-4)`
= `-3/2`
∴ slope of CF = `2/3`
Since altitude CF passes through C(– 4, – 3) and has slope `2/3`, the equation of altitude CF is
y – (– 3) = `2/3[x - (- 4)]`
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0
Hence, equations of the altitudes of the ΔABC are
5x + y – 15 = 0, 3x + 4y – 14 = 0 and 2x – 3y – 1 = 0.
