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.
Hi, it looks like you're using AdBlock :(
Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.
Please login to view more pages. It's free :)
Teachoo gives you a better experience when you're logged in. Please login :)
Solve all your doubts with Teachoo Black!
Teachoo answers all your questions if you are a Black user!