Let the equation of line AB be 4x – 3y + 5 = 0

and point C be (0, 0)

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

4a – 3b + 5 = 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

Now, Equation of line CD formed by joining the points C(0, 0) and D(a, b) and having the slope is

y₂ – y₁ = m(x₂ – x₁)

⇒ 4b = -3a

⇒ 3a + 4b = 0 …(ii)

Now, our equations are

4a – 3b + 5 = 0 …(i)

and 3a + 4b = 0 …(ii)

Multiply the eq. (i) by 4 and (ii) by 3, we get

16a – 12b + 20 = 0 …(iii)

9a + 12b = 0 …(iv)

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

16a – 12b + 20 + 9a + 12b = 0

⇒ 25a + 20 = 0

⇒ 25a = -20

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

⇒ -12 + 20b = 0

⇒ 20b = 12