TS ICET 23 May 2019 Shift 1 Paper

Question : 117 of 200

Marks: +1, -0

Equation of perpendicular bisector of the line segment joining (13 , - 2) and (- 5 , 10) is

Solution:

The mid-point of A (13, –2) B (–5, 10) is (4, 4)
Slope of AB

={10(-2)}/{-5-13}=12/{-18}=-2/3

∴ Slope of line perpendicular to AB

=3/2

∴ required equation is

y-4=\;{3}/{2}(x-4)

⇒3x-2y = 4.

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