Solveeit Logo
Login

Question

Question: The equation of the line bisecting the line segment joining the points (a, b) and \(\left( a ^ { \pr...

The equation of the line bisecting the line segment joining the points (a, b) and (a′,b′)\left( a ^ { \prime } , b ^ { \prime } \right)at right angle, is.

A

2(a−a′)x+2(b−b′)y=a2+b2−a′2−b′22 \left( a - a ^ { \prime } \right) x + 2 \left( b - b ^ { \prime } \right) y = a ^ { 2 } + b ^ { 2 } - a ^ { \prime 2 } - b ^ { \prime 2 }

B

(a−a′)x+(b−b′)y=a2+b2−a′2−b′2\left( a - a ^ { \prime } \right) x + \left( b - b ^ { \prime } \right) y = a ^ { 2 } + b ^ { 2 } - a ^ { \prime 2 } - b ^ { \prime 2 }

C

2(a−a′)x+2(b−b′)y=a′2+b′2−a2−b22 \left( a - a ^ { \prime } \right) x + 2 \left( b - b ^ { \prime } \right) y = a ^ { \prime 2 } + b ^ { \prime 2 } - a ^ { 2 } - b ^ { 2 }

D

None of these

Answer

2(a−a′)x+2(b−b′)y=a2+b2−a′2−b′22 \left( a - a ^ { \prime } \right) x + 2 \left( b - b ^ { \prime } \right) y = a ^ { 2 } + b ^ { 2 } - a ^ { \prime 2 } - b ^ { \prime 2 }

Explanation

Solution

m=−1b′−ba′−a=a′−ab−b′m = \frac { - 1 } { \frac { b ^ { \prime } - b } { a ^ { \prime } - a } } = \frac { a ^ { \prime } - a } { b - b ^ { \prime } } . Mid point is (a+a′2,b+b′2)\left( \frac { a + a ^ { \prime } } { 2 } , \frac { b + b ^ { \prime } } { 2 } \right)

Therefore equation of line is y−(b+b′2)=a′−ab−b′(x−a+a′2)y - \left( \frac { b + b ^ { \prime } } { 2 } \right) = \frac { a ^ { \prime } - a } { b - b ^ { \prime } } \left( x - \frac { a + a ^ { \prime } } { 2 } \right)

⇒2(b−b′)y+2(a−a′)x−b2+b′2−a2+a′2=0\Rightarrow 2 \left( b - b ^ { \prime } \right) y + 2 \left( a - a ^ { \prime } \right) x - b ^ { 2 } + b ^ { \prime 2 } - a ^ { 2 } + a ^ { \prime 2 } = 0.