Question

#Maths Aryabhattas only
Open challenge. [Co-ordinate geometry]
The length of perpendicular from origin to line lx + my + n = 0 is ?
[Ans. $\sf{\frac{ |n| }{ \sqrt{ {l}^{2} + {m}^{2}}}}$
​

Answer 1

Suppose we're asked to find the perpendicular distance of a point $\sf{P(x_1,\ y_1)}$ from the line $\sf{Ax+By+C=0}$ having positive intercepts as shown in the figure.

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Let a line parallel to x axis be drawn from P to our line and meet it at the point Q.

Since P and Q lie on a line parallel to x axis, y coordinates of P and Q are same.

Let the x coordinate of P be h and so the coordinates of Q is $\sf{(h,\ y_1).}$

Since Q is a point on our line, we have,

$\sf{\longrightarrow Ah+By_1+C=0}$

$\sf{\longrightarrow h=-\dfrac{By_1+C}{A}}$

Since $\sf{h>x_1,}$ length of PQ is, by distance formula,

$\sf{\longrightarrow PQ=h-x_1}$

Putting value of h,

$\sf{\longrightarrow PQ=-\dfrac{By_1+C}{A}-x_1}$

$\sf{\longrightarrow PQ=-\dfrac{Ax_1+By_1+C}{A}}$

Let a line parallel to y axis be drawn from P to our line and meet it at the point R.

Since P and R lie on a line parallel to y axis, x coordinates of P and R are same.

Let the y coordinate of P be k and so the coordinates of R is $\sf{(x_1,\ k).}$

Since R is a point on our line, we have,

$\sf{\longrightarrow Ax_1+Bk+C=0}$

$\sf{\longrightarrow k=-\dfrac{Ax_1+C}{B}}$

Since $\sf{k>y_1,}$ length of PR is, by distance formula,

$\sf{\longrightarrow PR=k-y_1}$

Putting value of k,

$\sf{\longrightarrow PR=-\dfrac{Ax_1+C}{B}-y_1}$

$\sf{\longrightarrow PR=-\dfrac{Ax_1+By_1+C}{B}}$

Now consider ΔPQR.

By Pythagoras' Theorem, length of QR,

$\sf{\longrightarrow QR=\sqrt{(PQ)^2+(PR)^2}}$

$\sf{\longrightarrow QR=\sqrt{\left(-\dfrac{Ax_1+By_1+C}{A}\right)^2+\left(-\dfrac{Ax_1+By_1+C}{B}\right)^2}}$

$\sf{\longrightarrow QR=\sqrt{(Ax_1+By_1+C)^2\left(\dfrac{1}{A^2}+\dfrac{1}{B^2}\right)}}$

$\sf{\longrightarrow QR=\sqrt{(Ax_1+By_1+C)^2\left(\dfrac{A^2+B^2}{A^2B^2}\right)}}$

$\sf{\longrightarrow QR=\dfrac{|Ax_1+By_1+C|}{|AB|}\sqrt{A^2+B^2}}$

But, the slope of this line, $\sf{m=-\dfrac{A}{B}}$ is negative as it makes obtuse angle with positive x axis.

$\sf{\longrightarrow -\dfrac{A}{B}<0}$

$\sf{\longrightarrow \dfrac{A}{B}>0}$

This implies A and B have same sign, and so,

$\sf{\longrightarrow AB>0}$

Therefore,

$\sf{\longrightarrow |AB|=AB}$

Thus,

$\sf{\longrightarrow QR=\dfrac{|Ax_1+By_1+C|}{AB}\sqrt{A^2+B^2}}$

In ΔPQR, PS is the altitude drawn from P to QR at S. It's length is $\sf{d}$ which is given by,

$\sf{\longrightarrow d=\dfrac{PQ\cdot PR}{QR}}$

$\sf{\longrightarrow d=\dfrac{-\dfrac{Ax_1+By_1+C}{A}\cdot-\dfrac{Ax_1+By_1+C}{B}}{\dfrac{|Ax_1+By_1+C|}{AB}\sqrt{A^2+B^2}}}$

$\sf{\longrightarrow d=\dfrac{AB(Ax_1+By_1+C)^2}{AB|Ax_1+By_1+C|\sqrt{A^2+B^2}}}$

$\sf{\longrightarrow\underline{\underline{d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}}}}$

We get the same final result in the case of line having,

• positive x intercept and negative y intercept.
• negative x intercept and positive y intercept.
• negative x and y intercepts.

though there are some differences.

In the question,

• $\sf{(x_1,\ y_1)=(0,\ 0)}$

• $\sf{Ax+By+C=lx+my+n=0}$

Then, length of perpendicular from origin to this line is,

$\sf{\longrightarrow d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}}$

$\sf{\longrightarrow d=\dfrac{|l(0)+m(0)+n|}{\sqrt{l^2+m^2}}}$

$\sf{\longrightarrow\underline{\underline{d=\dfrac{|n|}{\sqrt{l^2+m^2}}}}}$

Answer 2

Given ,

The equation of line is

• lx + my + n = 0

On comparing with general equation of line ax + by + c = 0 we , get

• a = l
• b = m
• c = n

We know that , the perpendicular distance from the origin to line ax + by + c = 0 is given by

$\boxed{ \tt{P = \frac{ |c| }{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}$

Thus ,

$\tt \implies p = \frac{ |n| }{ \sqrt{ {(l)}^{2} + {(m)}^{2} } }$

Learn More :

The perpendicular distance of a point (x1 , y1) from a line is given by -

$\boxed{ \tt{D = \frac{ |a x_{1} + by_{1} + c|}{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}$

The distance between two parrallel lines is given by -

$\boxed{ \tt{D = \frac{ | c_{2} - c_{1} | }{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}$

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