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

Sum

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

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.

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#### APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 11th Standard Maharashtra State Board
Chapter 5 Locus and Straight Line
Miscellaneous Exercise 5 | Q 15. (c) | Page 80
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