Straight Lines

# Perpendicular Distance from a Point on a Line xcosα+ysinα=p

Let $AB:x\cos\alpha+y\sin\alpha=p$ [Normal Form] be the given line and $P(x_1,y_1)$ be the given point. Also, let $PM\perp AB$ and $ON\perp AB$ such that the perpendicular distance from the origin on the line is $p$ i.e. $ON=p$. Then, we have to find $PM$. For this, through $P$, draw $A’B’//AB$ and produce $ON$ upto $ON’$ so that $ON’=p’$.

$PM=N’N=ON’-ON=p’-p$ Equation of $A’B’$ is, $x\cos\alpha+y\sin\alpha=p’$ Since this line passes through $P(x_1,y_1)$, so, $x_1\cos\alpha+y_1\sin\alpha=p’$ $\therefore PM=\pm(p’-p)$ $PM=\pm(x_1\cos\alpha+y_1\sin\alpha-p)$

Now, we find the perpendicular distance drawn from the point $P(x_1,y_1)$ on the line whose equation is $Ax+By+C=0$. For that, we reduce the equation $Ax+By+C=0$ into normal form.

If $C>0$, $p=\frac{C}{\sqrt{A^2+B^2}},\text{ }\cos\alpha=\frac{-A}{\sqrt{A^2+B^2}}$$\text{ and, }\sin\alpha=\frac{-B}{\sqrt{A^2+B^2}}$ If $C<0$, $p=\frac{-C}{\sqrt{A^2+B^2}},\text{ }\cos\alpha=\frac{A}{\sqrt{A^2-B^2}}$$\text{ and, }\sin\alpha=\frac{B}{\sqrt{A^2+B^2}}$ [From: Reduction of Linear Equation into Normal Form (Linear Equation $Ax+By+C=0$)]

For both the cases, the perpendicular distance from the point $(x_1,y_1)$ on the line $x\cos\alpha+y\sin\alpha=p$ is $PM=\pm(x_1\cos\alpha+y_1\sin\alpha-p)$ $=\pm\left(x_1\frac{A}{\sqrt{A^2+B^2}}+y_1\frac{B}{\sqrt{A^2+B^2}}+\frac{C}{\sqrt{A^2+B^2}}\right)$ $=\pm\left(\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\right)$ $=\left|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\right|$

#### Alternative Method

Reduce the given line $Ax+By+C=0$ into the double intercept form, $\frac{x}{\left(-\frac{C}{A}\right)}+\frac{y}{\left(-\frac{C}{B}\right)}=1$ [From: Reduction of Linear Equation into the Double Intercept Form (Linear Equation $Ax+By+C=0$)]

Let the given line meet the axes in $L$ and $M$ so that the coordinates of $L$ and $M$ are $\left(-\frac{C}{A},0\right)$ and $\left(0,-\frac{C}{B}\right)$. Let $PQ$ be the perpendicular distance from $P(x_1,y_1)$ on $LM$. Join $LP$ and $MP$. $\therefore\text{Area of }\Delta LMP$ $=\frac{1}{2}\left|x_1\left(0+\frac{C}{B}\right)+\frac{-C}{B}\left(-\frac{C}{B}-y_1\right)+0(y_1-0)\right|$ $=\frac{1}{2}\left|\frac{Cx_1}{B}+\frac{C^2}{AB}+\frac{Cy_1}{A}\right|$ $=\frac{1}{2}\left|\frac{C}{AB}(Ax_1+By_1+C)\right|$ $=\frac{1}{2}\left|\frac{C}{AB}\right|\left|Ax_1+By_1+C\right|$

Also, $\text{Area of }\Delta LMP=\frac{1}{2}LM.PQ$ $=\frac{1}{2}\sqrt{\left(0+\frac{C}{A}\right)^2+\left(-\frac{C}{B}-0\right)^2}.PQ$ $=\frac{1}{2}\sqrt{\frac{C^2(A^2+B^2)}{A^2B^2}}.PQ$ $=\frac{1}{2}\left|\frac{C}{AB}\right|\sqrt{A^2+B^2}.PQ$

$\therefore\Delta LMP=\frac{1}{2}\left|\frac{C}{AB}\right|\sqrt{A^2+B^2}.PQ$ $=\frac{1}{2}\left|\frac{C}{AB}\right| \left|Ax_1+By_1+C\right|$ $\text{or, }PQ=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}=\left|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\right|$ Thus, the perpendicular distance from $(x_1,y_1)$ on the line $Ax+By+C=0$ is $\left|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}\right|$

### Find the length of perpendicular drawn from $(2,3)$ to the line $8x+15y+24=0$.

Given line is, $8x+15y+24=0$ Perpendicular distance of the given line from $(2,3)$ is, $\pm\left(\frac{8×2+15×3+24}{\sqrt{8^2+15^2}}\right)=\pm\frac{85}{17}=5\text{ units}$

### If $p$ is the length of the perpendicular dropped from the origin on the line $\frac{x}{a}+\frac{y}{b}=1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$.

Given line is, $\frac{x}{a}+\frac{y}{b}=1$ $bx+ay-ab=0$ Perpendicular distance of the given line from origin is, $p=\pm\left(\frac{b×0+a×0-ab}{\sqrt{b^2+a^2}}\right)=\pm\frac{ab}{\sqrt{a^2+b^2}}$ $p^2=\frac{a^2b^2}{a^2+b^2}$ $\frac{1}{p^2}=\frac{a^2+b^2}{a^2b^2}$ $\therefore\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$

### Find the distance between the parallel lines $2x-5y-6=0$ and $6x-15y+11=0$.

Given lines are, $2x-5y-6=0$ $6x-15y+11=0$ Put $x=3$ in $2x-5y-6=0$, then, $2×3-5y-6=0$ $\therefore y=0$ Hence, $(3,0)$ is a point on the line $2x-5y-6=0$. Thus, perpendicular distance from $(3,0)$ on the line $6x-15y+11=0$ is, $\pm\left(\frac{6×3-15×0+11}{\sqrt{6^2+15^2}}\right)=\pm\frac{29}{\sqrt{261}}=\frac{29}{\sqrt{261}}\text{ units}$

### What are the points on the axis of x whose perpendicular distance from the straight line $\frac{x}{a}+\frac{y}{b}=1$ is a?

Let $P(h,0)$ be the point on x-axis. Given line is, $\frac{x}{a}+\frac{y}{b}=1$ $bx+ay-ab=0$ Perpendicular distance of the given line from $P$ is, $a=\pm\left(\frac{bh+a×0-ab}{\sqrt{b^2+a^2}}\right)$ $a=\pm\frac{bh-ab}{\sqrt{a^2+b^2}}$ $\frac{a}{b}\sqrt{a^2+b^2}=\pm(h-a)$ $\therefore\frac{a}{b}\sqrt{a^2+b^2}=h-a\text{ and }\frac{a}{b}\sqrt{a^2+b^2}=-h+a$ $h=\frac{a}{b}\sqrt{a^2+b^2}+a\text{ or }h=a-\frac{a}{b}\sqrt{a^2+b^2}$ Thus, the points on the x-axis are, $\left(\frac{a}{b}\sqrt{a^2+b^2}+a,0\right)\text{ and }\left(a-\frac{a}{b}\sqrt{a^2+b^2},0\right)$

### Find the equation of the line which is at right angles to $3x+4y=12$, such that its perpendicular distance from the origin is equal to the length of the perpendicular from $(3,2)$ on the given line.

Given line is, $3x+4y-12=0$ Its perpendicular distance from $(3,2)$ is, $\pm\left(\frac{3×3+4×2-12}{\sqrt{9+16}}\right)=\pm\frac{5}{5}=1$ and, $\text{Slope of the given line}=-\frac{3}{4}$ Since the required line is perpendicular to the given line, so, $\text{Slope of the required line }(m)=\frac{4}{3}$ Equation of the required line is given by, $y=mx+c$ $y=\frac{4}{3}x+c$ $4x-3y+3c=0$ Its perpendicular distance from the origin is, $1=\pm\left(\frac{4×0-3×0+3c}{\sqrt{16+9}}\right)$ $1=\pm\frac{3c}{5}$ $\therefore c=\pm\frac{5}{3}$ Thus, required equation of line is, $4x-3y\pm 5=0$