Question

Answer the following question:

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

Answer 1

Slope of side BC = ${\displaystyle \left(\frac{6-3}{1-2}\right)=\left(\frac{3}{-1}\right)}$ = – 3

∴ Slope of perpendicular bisector of BC is ${\displaystyle \frac{1}{3}}$ and the line passes through ${\displaystyle (\frac{3}{2},\frac{9}{2})}$.

∴ Equation of the perpendicular bisector of side BC is

${\displaystyle (y-\frac{9}{2})=\frac{1}{3}(x-\frac{3}{2})}$

∴ 3(2y – 9) = (2x – 3)

∴ 6y – 27 = 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, 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 perpendicular bisector of side AC is y = 5.

Slope of side AB = ${\displaystyle \left(\frac{3-4}{2-1}\right)}$ = – 1

∴ Slope of perpendicular bisector of AB is 1 and the line passes through ${\displaystyle (\frac{3}{2},\frac{7}{2})}$.

∴ Equation of the perpendicular bisector of side AB is

${\displaystyle (y-\frac{7}{2})=1(x-\frac{3}{2})}$

∴  2y – 7 = 2x – 3

∴ 2x – 2y + 4 = 0

∴ x – y + 2 = 0