1. Given line equation: x – 2y – 7 = 0 … (i)

Draw a perpendicular from point P (1, 2) on the line

Let P’ be the image of P and let its co-ordinates be (x, y)

The slope of the given line is given as,

2y = x – 7

y = (1/2) x – 7

Slope (m₁) = ½

Let the slope of line segment PP’ be m₂

As PP’ is perpendicular to the given line, product of slopes: m₁ x m₂ = -1

So, ½ x m₂ = -1

m₂ = -2

So, the equation of the line perpendicular to the given line and passing through P (1, 2) is

y – 2 = (-2) (x – 1)

y – 2 = -2x + 2

2x + y – 4 = 0 … (ii)

Let the intersection point of lines (i) and (ii) be taken as M.

Solving both the line equations, we have

Multiplying (ii) by 2 and adding with (i)

x – 2y – 7 = 0

4x + 2y – 8 = 0

———————

5x – 15 = 0

x = 15/5 = 3

Putting value of x in (i), we get

3 – 2y – 7 = 0

2y = -4

y = -4/2 = -2

So, the co-ordinates of M are (3, -2)

Hence, its seen that M should be the mid-point of the line segment PP’

(3, -2) = ((x + 1)/2, (y + 2)/2)

(x + 1)/2 = 3

x + 1 = 6

x = 6 – 1 = 5

And,

(y + 2)/2 = -2

y + 2 = -4

y = -4 – 2 = -6

Therefore, the co-ordinates of P’ are (5, -6).