The middle point of the line segment joining (3, -1)and (1, 1) is shifted by two units (in the sense of increasing y) Perpendicular to the line segment. Then, the coordinates of the point in the new position are
Let P be the middle point of the line segment joining A (3, -1) and B (1, 1).
Then,P=(3+12,−1+12)=(2,0)
Let P be shifted to Q, where PQ = 2 and y – coordinate of Q is greater than that of P (from question).
Now,′m′ of AB=1−(−1)1−3=22=1∴′m′ of PQ=−11=−1
Coordinates of Q by distance formula are (2±2cosθ,0,±2sinθ),where tanθ=1
=(2±2.1√2,0±2.1√2)=(2±√2,±√2)
As y−coordinate of Q,is greater than that of P.
∴Q=(2+√2,√2).This is the required point.
Hence,(c) is the correct answer.