#Maths Aryabhattas only
Open challenge. [Co-ordinate geometry]
The length of perpendicular from origin to line lx + my + n = 0 is ?
[Ans. [tex]\sf{\frac{ |n| }{ \sqrt{ {l}^{2} + {m}^{2}}}}[/tex]
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#Maths Aryabhattas only
Open challenge. [Co-ordinate geometry]
The length of perpendicular from origin to line lx + my + n = 0 is ?
[Ans. [tex]\sf{\frac{ |n| }{ \sqrt{ {l}^{2} + {m}^{2}}}}[/tex]
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Verified answer
Suppose we're asked to find the perpendicular distance of a point [tex]\sf{P(x_1,\ y_1)}[/tex] from the line [tex]\sf{Ax+By+C=0}[/tex] having positive intercepts as shown in the figure.
[tex]\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\vector(1,0){50}}\put(0,0){\vector(0,1){50}}\put(15,15){\vector(-1,1){25}}\put(15,15){\vector(1,-1){25}}\put(8,8){\circle*{1}}\put(8,8){\line(1,1){7}}\multiput(8,8)(4,0){4}{\line(1,0){2}}\multiput(8,8)(0,4){4}{\line(0,1){2}}\scriptsize\put(2,4.5){$\sf{P(x_1,\ y_1)}$}\put(23,9){$\sf{Q(h,y_1)}$}\put(9,23){$\sf{R(x_1,\ k)}$}\qbezier(14,14)(14.5,13.5)(15,13)\qbezier(15,13)(15.5,13.5)(16,14)\put(13,-6){$\sf{Ax+By+C=0}$}\put(51,-3){\sf{X}}\put(-3,51){\sf{Y}}\put(-3,-3){\sf{O}}\multiput(8,22)(14,-14){2}{\circle*{1}}\put(16,15){\sf{S}}\put(10,12){\sf{d}}\end{picture}[/tex]
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 [tex]\sf{(h,\ y_1).}[/tex]
Since Q is a point on our line, we have,
[tex]\sf{\longrightarrow Ah+By_1+C=0}[/tex]
[tex]\sf{\longrightarrow h=-\dfrac{By_1+C}{A}}[/tex]
Since [tex]\sf{h>x_1,}[/tex] length of PQ is, by distance formula,
[tex]\sf{\longrightarrow PQ=h-x_1}[/tex]
Putting value of h,
[tex]\sf{\longrightarrow PQ=-\dfrac{By_1+C}{A}-x_1}[/tex]
[tex]\sf{\longrightarrow PQ=-\dfrac{Ax_1+By_1+C}{A}}[/tex]
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 [tex]\sf{(x_1,\ k).}[/tex]
Since R is a point on our line, we have,
[tex]\sf{\longrightarrow Ax_1+Bk+C=0}[/tex]
[tex]\sf{\longrightarrow k=-\dfrac{Ax_1+C}{B}}[/tex]
Since [tex]\sf{k>y_1,}[/tex] length of PR is, by distance formula,
[tex]\sf{\longrightarrow PR=k-y_1}[/tex]
Putting value of k,
[tex]\sf{\longrightarrow PR=-\dfrac{Ax_1+C}{B}-y_1}[/tex]
[tex]\sf{\longrightarrow PR=-\dfrac{Ax_1+By_1+C}{B}}[/tex]
Now consider ΔPQR.
By Pythagoras' Theorem, length of QR,
[tex]\sf{\longrightarrow QR=\sqrt{(PQ)^2+(PR)^2}}[/tex]
[tex]\sf{\longrightarrow QR=\sqrt{\left(-\dfrac{Ax_1+By_1+C}{A}\right)^2+\left(-\dfrac{Ax_1+By_1+C}{B}\right)^2}}[/tex]
[tex]\sf{\longrightarrow QR=\sqrt{(Ax_1+By_1+C)^2\left(\dfrac{1}{A^2}+\dfrac{1}{B^2}\right)}}[/tex]
[tex]\sf{\longrightarrow QR=\sqrt{(Ax_1+By_1+C)^2\left(\dfrac{A^2+B^2}{A^2B^2}\right)}}[/tex]
[tex]\sf{\longrightarrow QR=\dfrac{|Ax_1+By_1+C|}{|AB|}\sqrt{A^2+B^2}}[/tex]
But, the slope of this line, [tex]\sf{m=-\dfrac{A}{B}}[/tex] is negative as it makes obtuse angle with positive x axis.
[tex]\sf{\longrightarrow -\dfrac{A}{B}<0}[/tex]
[tex]\sf{\longrightarrow \dfrac{A}{B}>0}[/tex]
This implies A and B have same sign, and so,
[tex]\sf{\longrightarrow AB>0}[/tex]
Therefore,
[tex]\sf{\longrightarrow |AB|=AB}[/tex]
Thus,
[tex]\sf{\longrightarrow QR=\dfrac{|Ax_1+By_1+C|}{AB}\sqrt{A^2+B^2}}[/tex]
In ΔPQR, PS is the altitude drawn from P to QR at S. It's length is [tex]\sf{d}[/tex] which is given by,
[tex]\sf{\longrightarrow d=\dfrac{PQ\cdot PR}{QR}}[/tex]
[tex]\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}}}[/tex]
[tex]\sf{\longrightarrow d=\dfrac{AB(Ax_1+By_1+C)^2}{AB|Ax_1+By_1+C|\sqrt{A^2+B^2}}}[/tex]
[tex]\sf{\longrightarrow\underline{\underline{d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}}}}[/tex]
We get the same final result in the case of line having,
though there are some differences.
In the question,
Then, length of perpendicular from origin to this line is,
[tex]\sf{\longrightarrow d=\dfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}}[/tex]
[tex]\sf{\longrightarrow d=\dfrac{|l(0)+m(0)+n|}{\sqrt{l^2+m^2}}}[/tex]
[tex]\sf{\longrightarrow\underline{\underline{d=\dfrac{|n|}{\sqrt{l^2+m^2}}}}}[/tex]
Given ,
The equation of line is
On comparing with general equation of line ax + by + c = 0 we , get
We know that , the perpendicular distance from the origin to line ax + by + c = 0 is given by
[tex] \boxed{ \tt{P = \frac{ |c| }{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}[/tex]
Thus ,
[tex] \tt \implies p = \frac{ |n| }{ \sqrt{ {(l)}^{2} + {(m)}^{2} } } [/tex]
Learn More :
The perpendicular distance of a point (x1 , y1) from a line is given by -
[tex] \boxed{ \tt{D = \frac{ |a x_{1} + by_{1} + c|}{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}[/tex]
The distance between two parrallel lines is given by -
[tex] \boxed{ \tt{D = \frac{ | c_{2} - c_{1} | }{ \sqrt{ {(a)}^{2} + {(b)}^{2} } } }}[/tex]
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