NCERT Solutions
Class 11 Maths
Straight Lines

Ex.Misc. Q18
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.
Given equation of line is
x + 3y = 7 ......... (1)
Let the point (a, b) is the image of point (3, 8)
Slope of line AB = (b - 8) ÷ (a - 3)
Since the given line x + 3y = 7 is perpendicular to AB
So {(b - 8) ÷ (a - 3)} × (- ) = -1
(Since product of two slopes = -1)
=> (b - 8) ÷ (a - 3) = 3
=> (b - 8) = 3(a - 3)
=> b - 8 = 3a - 9
=> 3a - b = 9 - 8
=> 3a - b = 1 ......... (2)
Now mid-point of AB = {(a + 3) ÷ 3, (b + 8) ÷ 2}
This point also satisfies equation (1)
So, {(a + 3) ÷ 2} + {3(b + 8) ÷ 2} = 7
=> a + 3 + 3b + 24 = 2 × 7
=> a + 3b + 27 = 14
=> a + 3b = 14-27
=> a + 3b = -13 ...... (3)
After solving equation (2) and (3), we get
a = -1, b = -4
Hence, the image of the point (3, 8) with respect to x + 3y = 8 is (-1, -4).