question_answer

Suppose A, B are two points on \[2x-y+3=0\] and P (1, 2) is such that \[PA=PB.\] Then the mid-point of AB is

A) \[\left( -\frac{1}{5},\frac{13}{5} \right)\]

B) \[\left( \frac{-7}{5},\frac{9}{5} \right)\]

C) \[\left( \frac{7}{5},\frac{-9}{5} \right)\]

D) \[\left( \frac{-7}{5},\frac{-9}{5} \right)\]

Correct Answer: A

Solution :

[a] Equation of \[AB=2x-y+3=0\] \[\Rightarrow \Delta PAD\cong \Delta PBD\] \[\Rightarrow D\]is foot of perpendicular \[\Rightarrow \] from P to AB \[\frac{\alpha -1}{2}=\frac{\beta -2}{-1}=\frac{-(2\times 1-1\times 2+3)}{4+1}\] \[\frac{\alpha -1}{2}=\frac{\beta -2}{-1}=\frac{-3}{5}\Rightarrow \alpha =\frac{-1}{5},\beta =\frac{13}{5}\]