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

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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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.