Question

if p is the length of perpendicular from the origin to the line whose intercepts on the axis are a&b then show that 1/p2 = 1/a2 +1/b2​

Answer 1

ANSWER

concept : use intercept form of line : x/a + y/b = 1 and then find perpendicular distance of point from line .

equation of line in intercept form is

x/a + y/b = 1

x/a + y/b - 1 = 0 .

now, use formula

distance of point (x1, y1) from line : ax + by + c =0 is |ax1 + by1 + c|/√(a²+b²)

so, it's distance from origin is

P = | 0 + 0 -1|/√(1/a² + 1/b²)

P = 1/√(1/a² + 1/b²)

1/P = √(1/a² + 1/b²)

take square both sides,

1/P² = 1/a² + 1/b²

hence proved.

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Answer 2

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.