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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
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Solution
Slope of side BC = `((6 - 3)/(1 - 2)) = (3/-1)` = – 3
∴ Slope of perpendicular bisector of BC is `1/3` and the line passes through `(3/2, 9/2)`.
∴ Equation of the perpendicular bisector of side BC is
`(y - 9/2) = 1/3(x - 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 = `((3 - 4)/(2 - 1))` = – 1
∴ Slope of 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 = 2x – 3
∴ 2x – 2y + 4 = 0
∴ x – y + 2 = 0
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