Question

Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(−1, 8), Q(4, −2), and R(−5, −3)

Answer 1

Let A, B, and C be the midpoints of sides PQ, QR, and PR respectively of ΔPQR.

A is the midpoint of side PQ.

∴ A ≡ `((-1 + 4)/2, (8 - 2)/2) = (3/2, 3)`

Slope of side PQ = `(-2 - 8)/(4 - (-1))`

= `(-10)/5`

= – 2

∴ Slope of perpendicular bisector of PQ is `1/2` and it passes through `(3/2, 3)`.

∴ Equation of the perpendicular bisector of side PQ is

y – 3 = `1/2(x - 3/2)`

∴ y –  3 = `1/2((2x - 3)/2)`

∴ 4(y –  3) = 2x –  3

∴ 4y –  12 = 2x – 3

∴ 2x – 4y + 9 = 0

B is the midpoint of side QR

∴ B ≡ `((4 - 5)/2, (-2 - 3)/2) = ((-1)/2, (-5)/2)`

Slope of side QR = `(-3 - (- 2))/(-5 - 4)`

= `(-1)/(-9)`

= `1/9`

∴ Slope of perpendicular bisector of QR is – 9 and it passes through `(-1/2, -5/2)`.

∴ Equation of the perpendicular bisector of side QR is

`y - (-5/2) = -9[x - (-1/2)]`

∴ `(2y + 5)/2 = -9((2x + 1)/2)`

∴ 2y + 5 = –18x – 9

∴ 18x + 2y + 14 = 0

∴ 9x + y + 7 = 0

C is the midpoint of side PR.

∴ C ≡ `((-1 - 5)/2, (8 - 3)/2) = (-3, 5/2)` 

Slope of side PR = `(-3 - 8)/(-5 - (-1)) = (-11)/(-4) = 11/4`

∴ Slope of perpendicular bisector of PR is `-4/11` and it passes through `(-3, 5/2)`.

∴ Equation of the perpendicular bisector of side PR is

`y - 5/2 = -4/11(x + 3)`

∴ `11((2y - 5)/2)` = – 4(x + 3)

∴ 11(2y – 5) = – 8(x + 3)

∴ 22y – 55 = – 8x – 24

∴ 8x + 22y – 31 = 0