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

Options

A. `((-1)/(5), (13)/(5))`
B. `((-7)/(5), (9)/(5))`
C. `((7)/(5), (-9)/(5))`
D. `((-7)/(5), (-9)/(5))`

Answer

A

Explanation

From the given information, it is clear that midpoint of AB is the foot of the perpendicular from P(1,2) on the line 2x-y+3=0, which is given by
`(x-1)/(2) = (y-2)/(-1) = -(2(1)-2+3)/((2^(2))+(-1)^(2))`
`"or "(x,y)-=((-1)/(5),(13)/(5))`