Find the image of the point (3, 8) with respect to the line x+ 3y = 7 assuming the line to be a plane mirror. 

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.

Slope of AB = (b − 8)/(a − 3), while the slope of line (1) = −1/3.

Since line (1) is perpendicular to AB,

(b − 8)/(a − 3) × (−1/3) = −1

⇒ (b − 8)/(3a − 9) = 1

⇒ b − 8 = 3a − 9

⇒ 3a − b = 1     ...(2)

Midpoint of AB = ((a + 3)/2, (b + 8)/2)

The midpoint of line segment AB will also satisfy line (1).

Hence, from equation (1), we have:

(a + 3)/2 + 3(b + 8)/2 = 7

⇒ a + 3 + 3b + 24 = 14

⇒ a + 3b = −13     ...(3)

On solving equations (2) and (3), we get:

a = −1 and b = −4

Thus, the image of the given point with respect to the given line is (−1, −4).