The point `A (2, 1)` is shifted by `3sqrt2` unit distance parallel to the line `x + y = 1` in the direction of increasing ordinate to reach a point B. Find the image of B by the line `x+y= 1`.

Options

A. (5,-2)
B. (-3,2)
C. (5,4)
D. (-1,4)

Explanation

The equation of a line passing through (2,1) and parallel to x + y = 1 is
`(x - 2)/(cos 3pi //4) = (y-1)/(sin pi//4) or , (x -1)/(-1 //sqrt2) = (y-1)/(1//sqrt2)`
The coordinates of points on this line which are at a distance `3sqrt2` from (2,1) are given by
`(x-2)/(-1//sqrt2) = (y-1)/(1//sqrt2) = pm 3 //sqrt2 implies x = 2 pm 3 , y = 1 pm 3`
But the ordinate of Q is more than that of P (2,1) .
So , the coordinates of Q are (-1,4).
The image of Q(-1,4) in the line x + y = 1 is given by
`(x +1)/(1) = (y-4)/(1) = -2((-1 + 4 -1)/(1^(2) + 1^(2)))`
`implies (x +1)/(1) = (y-4)/(1) = -2 implies x = -3 , y =2`
Thus , the coordinates of the required point are (-3,2)