Question 31 The equation of the perpendicular bisector of line segment joining points A (4, 5) and B (−2, 3) is (a) 2x – y + 7 = 0 (b) 3x + 2 y – 7 = 0 (c) 3x – y – 7 =0 (d) 3x + y – 7 = 0
Let Line l be the perpendicular bisector of AB
And let it intersect AB at point P
Let point P be any point on Line l
By symmetry
AP = BP
√(( 𝑥 −4 )2+(𝑦−5)2) = √(( 𝑥 −(−2))2+(𝑦−3)2)
√(( 𝑥 −4 )2+(𝑦−5)2) = √(( 𝑥+2)2+(𝑦−3)2)
Squaring both sides
( 𝒙 −𝟒)𝟐+(𝒚−𝟓)𝟐 = ( 𝒙+𝟐)𝟐+(𝒚−𝟑)𝟐
𝑥^2+16−8𝑥+𝑦^2+25−10𝑦=𝑥^2+4+4𝑥+𝑦^2+9−6𝑦
16−8𝑥+25−10𝑦=4+4𝑥+9−6𝑦
−8𝑥−10𝑦+41=4𝑥−6𝑦+13
0 =4𝑥−6𝑦+13+8𝑥+10𝑦−41
0 =4𝑥+8𝑥−6𝑦+10𝑦+13−41
0 =12𝑥+4𝑦−28
12𝑥+4𝑦−28=0
Dividing by 4 both sudes
𝟑𝒙+𝒚−𝟕=𝟎
So, the correct answer is (d)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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