Find the image of the point (3, 8) with respect to the line x + 3y = 7

1.	Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Answer» Let the image of point P(3, 8) be Q (a, b) with respect to line AB which is given by x + 3y – 7 = 0 …(1) The mid-point of P and Q is \(\big(\frac{3 + a}{2} , \frac{8 + b}{2}\big)\), which lies on line (1). ∴ (1) ⇒ \(\frac{3 + a}{2} +3. \frac{8 + b}{2}\) − 7 = 0 ⇒ 3 + a + 24 + 3b – 14 = 0 ⇒ a + 3b + 13 = 0 …(2) Now, slope of AB = – \(\frac{1}{3}\) and slope of PQ = \(\frac{b−8}{a−3}\) ∴ − \(\frac{1}{3}\) × \(\frac{b−8}{a−3}\) = −1 ⇒ 3a – b – 1 = 0 Solving (2) and (3), we arrive at a = – 1, b = – 4 ∴ Image of (3, 8) is (– 1, – 4).

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

Answer»

Let the image of point P(3, 8) be Q (a, b) with respect to line AB which is given by

x + 3y – 7 = 0 …(1)

The mid-point of P and Q is \(\big(\frac{3 + a}{2} , \frac{8 + b}{2}\big)\), which lies on line (1).

∴ (1) ⇒ \(\frac{3 + a}{2} +3. \frac{8 + b}{2}\) − 7 = 0

⇒ 3 + a + 24 + 3b – 14 = 0

⇒ a + 3b + 13 = 0 …(2)

Now, slope of AB = – \(\frac{1}{3}\)

and slope of PQ = \(\frac{b−8}{a−3}\)

∴ − \(\frac{1}{3}\) × \(\frac{b−8}{a−3}\) = −1

⇒ 3a – b – 1 = 0

Solving (2) and (3), we arrive at a = – 1, b = – 4

∴ Image of (3, 8) is (– 1, – 4).