Let the equation of line AB be x – 4y + 7 = 0

and point C be (-2, 3)

CD is perpendicular to the line AB, and we need to find:

1) Equation of Perpendicular drawn from point C

2) Coordinates of D

Let the coordinates of point D be (a, b)

Also, point D(a, b) lies on the line AB, i.e. point (a, b) satisfy the equation of line AB

Putting x = a and y = b, in equation, we get

a – 4b + 7 = 0 …(i)

Also, the CD is perpendicular to the line AB

and we know that, if two lines are perpendicular then the product of their slope is equal to -1

∴ Slope of AB × Slope of CD = -1

Slope of CD = - 4

Now, Equation of line CD formed by joining the points C(-2, 3) and D(a, b) and having the slope – 4 is

y₂ – y₁ = m(x₂ – x₁)

⇒ b – 3 = (-4)[a – (-2)]

⇒ b – 3 = -4(a + 2)

⇒ b – 3 = -4a – 8

⇒ 4a + b + 5 = 0 …(ii)

Now, our equations are

a – 4b + 7 = 0 …(i)

and 4a + b + 5 = 0 …(ii)

Multiply the eq. (ii) by 4, we get

16a + 4b + 20 = 0 …(iii)

Adding eq. (i) and (iii), we get

a – 4b + 7 + 16a + 4b + 20 = 0

⇒ 17a + 27 = 0

⇒ 17a = -27

Putting the value of a in eq. (i), we get

⇒ 92 – 68b = 0

⇒ -68b = -92