If p is the length of perpendicular from the origin to the line whose intercepts on the x-axis are a and b, then show that 1/p² = 1/a² + 1/b²

Solution:

It is known that the equation of a line whose intercepts on the axis a and b is

x/a + y/b = 1

bx + ay = ab

bx + ay - ab = 0 ....(1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x\(_1\), y\(_1\)) is given by

d = |Ax\(_1\) + By\(_1\) + C|/√A² + B²

On comparing equation (1) to the general equation of line Ax + By + C = 0 , we obtain

A = b, B = a and C = - ab

Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), We obtain

p = |A(0) + B(0) - ab|/√a² + b²

= |- ab|/√a² + b²

On squaring both sides, we obtain

⇒ p² = (- ab)²/(a² + b²)

⇒ p² (a² + b² ) = a²b²

⇒ (a² + b²)/a²b² = 1/p²

⇒ 1/p² = 1/a² + 1/b²

Hence, we showed 1/p² = 1/a² + 1/b²

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.3 Question 18