Question

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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.

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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Perpendicular distance of a line has been used. We see that the we are given the length of the perpendicular line which is from origin to a line whose intercepts are on the axes a and b. So first we can get the equation of the line with intercepts. Then we can take relation between terms and prove the given thing.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{d\;=\;\bf{\bigg|\dfrac{Ax_{1}\;+\;By_{1}\;+\;C}{\sqrt{A^{2}\;+\;B^{2}}}\bigg|}}}}$

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★ Solution :-

Given,

» Distance of the perpendicular from origin (d) = length of the perpendicular = p

» Coordinates of the origin (x₁, y₁) = (0, 0)

» Given intercepts :: a and b

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~ For the equation of line with intercepts a and b ::

We get the equation of line as,

$\\\;\bf{\mapsto\;\;\orange{\dfrac{x}{a}\;+\;\dfrac{y}{b}\;=\;1}}$

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~ For the value of A, B and C ::

We know that,

$\\\;\bf{\mapsto\;\;\green{Equation\;of\;perpendicular\;line\;=\;Ax\;+\;By\;+\;C\;=\;0}}$

Also, we have,

$\\\;\bf{\mapsto\;\;\bigg(\dfrac{1}{a}\bigg)(x)\;+\;\bigg(\dfrac{1}{b}\bigg)\;=\;1}$

Now comparing this with Equation of Perpendicular Line,

$\\\;\bf{\mapsto\;\;\blue{Ax\;+\;By\;+\;C\;=\;\bigg(\dfrac{1}{a}\bigg)x\;+\;\bigg(\dfrac{1}{b}\bigg)y\;-\;1}}$

From this we get,

→ A = 1/a , B = 1/b , C = -1

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~ For proving the given thing :-

$\\\;\sf{:\rightarrow\;\;d\;=\;\bf{\bigg|\dfrac{Ax_{1}\;+\;By_{1}\;+\;C}{\sqrt{A^{2}\;+\;B^{2}}}\bigg|}}$

$\\\;\sf{:\rightarrow\;\;d\;=\;\bf{\dfrac{\bigg|Ax_{1}\;+\;By_{1}\;+\;C\bigg|}{\sqrt{A^{2}\;+\;B^{2}}}}}$

By applying values, we get

$\\\;\sf{:\Longrightarrow\;\;p\;=\;\bf{\dfrac{\bigg|\bigg(\dfrac{1}{a}\bigg)(0)\;+\;\bigg(\dfrac{1}{b}\bigg)(0)\;-\;1\bigg|}{\sqrt{\bigg(\dfrac{1}{a}\bigg)^{2}\;+\;\bigg(\dfrac{1}{b}\bigg)^{2}}}}}$

$\\\;\sf{:\Longrightarrow\;\;p\;=\;\bf{\dfrac{\bigg|\bigg(\dfrac{0}{a}\bigg)\;+\;\bigg(\dfrac{0}{b}\bigg)\;-\;1\bigg|}{\sqrt{\bigg(\dfrac{1^{2}}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1^{2}}{b^{2}}\bigg)}}}}$

$\\\;\sf{:\Longrightarrow\;\;p\;=\;\bf{\dfrac{\bigg|0\;+\;0\;-\;1\bigg|}{\sqrt{\bigg(\dfrac{1}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1}{b^{2}}\bigg)}}}}$

$\\\;\sf{:\Longrightarrow\;\;p\;=\;\bf{\dfrac{1}{\sqrt{\bigg(\dfrac{1}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1}{b^{2}}\bigg)}}}}$

Now by cross multiplying, we get

$\\\;\sf{:\Longrightarrow\;\;\sqrt{\bigg(\dfrac{1}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1}{b^{2}}\bigg)}\;=\;\bf{\dfrac{1}{p}}}$

Now squaring both sides, we get

$\\\;\sf{:\Longrightarrow\;\;\bigg[\sqrt{\bigg(\dfrac{1}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1}{b^{2}}\bigg)}\;\bigg]^{2}\;=\;\bf{\bigg(\dfrac{1}{p}\bigg)^{2}}}$

$\\\;\sf{:\Longrightarrow\;\;\bigg(\dfrac{1}{a^{2}}\bigg)\;+\;\bigg(\dfrac{1}{b^{2}}\bigg)\;=\;\bf{\bigg(\dfrac{1^{2}}{p^{2}}\bigg)}}$

$\\\;\sf{\red{:\Longrightarrow\;\;\dfrac{1}{a^{2}}\;+\;\dfrac{1}{b^{2}}\;=\;\bf{\dfrac{1}{p^{2}}}}}$

$\\\;\sf{\blue{:\Longrightarrow\;\;\dfrac{1}{p^{2}}\;=\;\dfrac{1}{a^{2}}\;+\;\dfrac{1}{b^{2}}}}$

So we prove the equation.

$\\\;\qquad\qquad\boxed{\boxed{\rm{\purple{Hence,\;\;Proved}}}}$

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★ More formulas to know :-

• Distance between parallel lines is given as ::

$\\\;\sf{\leadsto\;\;d\;=\;\dfrac{\bigg| C_{1}\;-\;C_{2}\bigg|}{\sqrt{A^{2}\;+\;B^{2}}}}$

• Slope of line ::

$\\\;\sf{\leadsto\;\;\tan\theta\;=\;\dfrac{y_{2}\:-\:y_{1}}{x_{2}\:-\:x_{1}}}$

• Equation of line that makes y - intercept c with slope m ::

$\\\;\sf{\leadsto\;\;y\;=\;mx\;+\;c}$

• Equation of line that makes x - intercepts d with slope m ::

$\\\;\sf{\leadsto\;\;y\;=\;m(x\;-\;d)}$

Answer 2

$\large\bold{\underline{\underline{Question:-}}}$

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.

$\large\bold{\underline{\underline{Solution:-}}}$

Given that,

the intercepts are a and b.

Therefore, equation of the line is

$\displaystyle{\sf{ \dfrac{x}{a} + \frac{x}{b} = 1.}}$

And the perpendicular distance (d) of the line$\displaystyle{\sf{ax + by + c = 0}}$ from a point $\displaystyle{\sf{(x_1 , y_1)}}$ is given by

$\displaystyle{\boxed{\sf{\orange{ d\:=\: \frac{|Ax_1 + By_1 + C|}{ \sqrt{A^{2} + B^{2}}}}}}}$

After that,

$\displaystyle{\sf{ \dfrac{x}{a} + \dfrac{x}{b} = 1.}}$

Compare it with $\displaystyle{\sf{ax + by + c = 0}}$

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

Therefore, a = $\displaystyle{\sf{ \frac{1}{a}}}$, b = $\displaystyle{\sf{ \frac{1}{b}}}$, c = $\displaystyle{\sf{ -1}}$

Again, the distance from origin (0, 0) to the line $\displaystyle{\sf{ \dfrac{x}{a} + \dfrac{x}{b} = 1.}}$ is p.

So, distance (d) = p &$\displaystyle{\sf{x_1 = 0\:,\: y_1 =0}}$

Substitute the values in $\displaystyle{\sf{ d\:=\: \dfrac{|Ax_1 + By_1 + C|}{ \sqrt{A^{2} + B^{2}}}}}$ and simplify the equation.

:$\implies{\sf{p\:=\: \dfrac{ \bigg| (0) \dfrac{1}{a} + (0) \dfrac{1}{b} - 1 \bigg|}{ \sqrt{ \dfrac{1}{a}^{2} + \dfrac{1}{b}^{2}}}}}$

:$\implies{\sf{p\:=\: \dfrac{ | 0+ 0 -1|}{ \sqrt{ (\dfrac{1}{a})^{2} + (\dfrac{1}{b})^{2}}}}}$

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

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

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

Squaring both the sides.

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

:$\large\implies{\boxed{\bf{\green{ \dfrac{1}{p^{2}} = \dfrac{1}{a^{2}} + \dfrac{1}{b^{2}}}}}}$

Hence, proved!

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