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

Solution:

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 the line (1) is - 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)

Mid-point of AB = [(a + 3)/2, (b + 8)/2

The mid-point of the line segments 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 obtain

a = - 1 and b = - 4.

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

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 18

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

Summary:

The image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror is (-1, -4)