The vertices of a triangle are A (1, 4), B (2, 3) and C (1, 6). Find equations of Perpendicular bisectors of sides

#### Solution

Vertices of ΔABC are A(1, 4), B(2, 3) and C(1, 6)

∴ Slope of the perpendicular bisector of BC is `1/3` and the lines pass through `(3/2, 9/2)`.

∴ Equation of the perpendicular bisector of side BC is

`(y - 9/2) = 1/3(x - 3/2)`

∴ `(2y - 9)/2 = 1/3((2x - 3)/2)`

∴ 3(2y – 9) = (2x –3)

∴ 2x – 6y + 24 = 0

∴ x – 3y + 12 = 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, the perpendicular bisector of side AC is parallel to X-axis.

Since, the perpendicular bisector of side AC passes through E(1, 5).

∴ the equation of the perpendicular bisector of side AC is y = 5.

Slope of side AB = `((3 - 4)/(2 - 1))` = – 1

∴ The slope of the perpendicular bisector of AB is 1 and the line passes through `(3/2, 7/2)`.

∴ Equation of the perpendicular bisector of side AB is

`(y - 7/2) = 1(x - 3/2)`

∴ `(2y - 7)/2 = (2x - 3)/2`

∴ 2y – 7 = 2x – 3

∴ 2x – 2y + 4 = 0

∴ x – y + 2 = 0.