Question

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 1

Let the equation of line AB be x + 3y = 7 and the coordinates of point P are (3, 8).

y = `- 1/3 "x" + 7/3`

The image of point P will be Q if PQ ⊥ AB, PQ and AB intersect at the point M such that

PM = QM

Slope of line AB = `-1/3`

And slope of PQ = 3

∴ Equation of line PQ,

y – 8 = 3(x – 3)

= 3x – 9

or 3x – y = 1 ….........(i)

Equation of AB x + 3y = 7 ….........(ii)

Multiplying equation (i) by 3 and adding it to equation (ii),

10x = 10 or x = 1

From equation (i) y = 3x – 1

= 3 – 1

= 2

∴ The coordinates of point M are (1, 2).

Let the coordinates of Q be (x₁, y₁)

Point M is the midpoint of line segment PQ

∴ While P(3, 8) is.

∴ `("x"_1 + 3)/2 = 1` or x₁ = −1

`("y"_1 + 8)/2 = 2` or y₁ = −4

∴ The image of P is (−1, – 4).