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/p²=1/a²+1/b²​

Answer 1

Given: P is the length of perpendicular from the origin to the line whose intercepts on the axis are a and b respectively.

To prove: 1/p² = 1/a² + 1/b²

Solution:

We know that equation of a line whose x and y intercept are a and b respectively is,

$\sf \implies \dfrac{x}{a} + \dfrac{y}{b} = 1$

$\sf \implies \dfrac{x}{a} + \dfrac{y}{b} - 1 = 0$

Now, we have a formula to find perpendicular distance of a line from origin whose equation is given:

$\sf \implies p \: = \left| \dfrac{c}{ \sqrt{ {a}^{2} + {b}^{2} } } \right|$

Here p is the perpendicular distance, c is constant of the equation and a & b are x and y intercepts respectively.

So by substituting the known values, we get:

$\sf \implies p \: = \left| \dfrac{ - 1}{ \sqrt{ \dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2} } } } \right|$

By squaring both sides, we get the following results:

$\sf \implies {p}^{2} \: = \dfrac{ ( - 1)^{2} }{\dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2}} }$

$\sf \implies {p}^{2} \: = \dfrac{1}{\dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2}} }$

$\boxed{ \sf \implies \dfrac{1}{ {p}^{2} } = \dfrac{1}{ {a}^{2} } + \dfrac{1}{ {b}^{2}}}$

Therefore, the required result is proved.