If (-2,6) is the image of the point (4,2) with respect to line L=0, then L is:

Answer

3x-2y+5 = 0

Explanation

The midpoint of Q (-2,6) and P (4,2) is
`((-2+4)/(2), (6+2)/(2)), i.e., (1,4)`
and the gradiant of line PQ is
`(2-6)/(4+2)= (-2)/(3)`
Therefore, the slope of L is 3/2.
Hence, the equation of the line which passes through point (1,4) is
`y-4 = (3)/(2) (x-1)`
or 3x-2y+5=0