Question

If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that `1/p^2 = 1/a^2 + 1/b^2`.

Answer 1

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

`x/a + y/b = 1`

or bx + ay = ab

or bx + ay - ab = 0       ......(1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d = `|Ax_1 + By_1 + C|/sqrt(A^2 + B^2)`.

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|/sqrt(b^2 + a^2)`

= `p = |-ab|/sqrt(a^2 + b^2)`

On sqauring both sides, we obtain

`p^2 = (-ab)^2/(a^2 + b^2)`

= p² (a² + b²) = a²b²

= `(a^2 + b^2)/(a^2b^2) = 1/(p^2)`

= `1/p^2 = 1/a^2 + 1/b^2`.