Straight Line Question 296

Question: The equation of the line bisecting the line segment joining the points (a, b) and $ ({a}’,\ {b}’) $ at right angle, is

Options:

A) $ 2(a-{a}’)x+2(b-{b}’)y=a^{2}+b^{2}-{{{a}’}^{2}}-{{{b}’}^{2}} $

B) $ (a-{a}’)x+(b-{b}’)y=a^{2}+b^{2}-{{{a}’}^{2}}-{{{b}’}^{2}} $

C) $ 2(a-{a}’)x+2(b-{b}’)y={{{a}’}^{2}}+b{{’}^{2}}-a^{2}-b^{2} $

D)None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ m=\frac{-1}{\frac{{b}’-b}{{a}’-a}}=\frac{{a}’-a}{b-{b}’} $ . Midpoint is $ ( \frac{a+{a}’}{2},\frac{b+{b}’}{2} ) $ Therefore equation of line is $ y-( \frac{b+{b}’}{2} )=\frac{{a}’-a}{b-{b}’}( x-\frac{a+{a}’}{2} ) $

$ \Rightarrow 2(b-{b}’)y+2(a-{a}’)x-b^{2}+{{{b}’}^{2}}-a^{2}+{{{a}’}^{2}}=0 $ .