The equation of the given line is x + 3y = 7 … (1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.
Since line (1) is perpendicular to AB,
The mid-point of line segment AB will also satisfy line (1).
Hence, from equation (1), we have
On solving equations (2) and (3), we obtain a = –1 and b = –4.
Thus, the image of the given point with respect to the given line is (–1, –4).
