Warning: preg_match(): Compilation failed: quantifier does not follow a repeatable item at offset 1 in /home/customer/www/mathsdiscussion.com/public_html/wp-includes/class-wp.php on line 231
Distance of a point from line. - Distance between lines

Distance of a point from line.

Perpendicular distance of a point from the line.

The length of perpendicular distance of point $P(x_1,y_1)$ from line $ax+by+c=0$.

Let foot of perpendicular from point $P$ on line $L\,:\,ax+by+c=0$ be $Q(h,k)$

$$\therefore ah+bk+c=0$$ $$m_L.m_{PQ}=-1$$

$$\Rightarrow -(\frac{a}{b})(\frac{k-y_1}{h-x_1})=-1$$

$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{a(h-x_1)+b(k-y_1)}{a^2+b^2}$$

$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{(ah+bk+c)-(ax_1+by_1+c)}{a^2+b^2}$$

$$\Rightarrow \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}$$

$$\Rightarrow (h-x_1)=-a( \frac{(ax_1+by_1+c)}{a^2+b^2})$$

$$And \;\; (k-y_1)=-b( \frac{(ax_1+by_1+c)}{a^2+b^2})$$

$$\Rightarrow PQ=\sqrt{(h-x_1)^2+(k-y_1)^2}=\cfrac{\vert ax_1+by_1+c \vert}{\sqrt{a^2+b^2}}$$

$\therefore$ Distance of point $P(x_1,y_1)$ from the line $ax+by+c=0$ is $PQ= \cfrac{\vert ax_1+by_1+c \vert}{\sqrt{a^2+b^2}}$

NOTE : – (i) Distance of origin $O(0,0)$ from the line $L:ax+by+c=0$ is $OQ=\cfrac{\vert c \vert}{\sqrt{a^2+b^2}}$

(ii) Distance between two parallel lines .

(a) If origin is on side with respect to the two lines.

Distance between line $L_1$ and $L_2$ is $AB=OB-OA$

$$=\cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}}- \cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}}$$

$$= \frac{\vert c_2-c_1 \vert}{\sqrt{a^2+b^2}}$$

Since $c_1.c_2\gt 0$ origin lie on same side with respect to the lines.

(b) If origin lie between the two lines.

If origin lies between the lines then $c_1.c_2 \lt 0$.

And Distance between lines $L_1\;,\,L_2$ is $AB=OA+OB$.

$$=\cfrac{\vert c_1 \vert}{\sqrt{a^2+b^2}}+ \cfrac{\vert c_2 \vert}{\sqrt{a^2+b^2}}$$ $$= \cfrac{\vert c_1-c_2 \vert}{\sqrt{a^2+b^2}} \;,Since \;\;c_1.c_2 \lt 0$$

Foot of Perpendicular.

Foot of perpendicular from the given point to fixed line.

Let $Q(h,k)$ be the foot of perpendicular from $P(x_1,y_1)$ on the line $L\;:\;ax+by+c=0$

$$\Rightarrow ah+bk+c=0 \;—–(1)$$ $$Since\;\;\;m_L.m_{PQ}=-1$$

$$\therefore -\Bigl(\frac{a}{b}\Bigr)\Bigl( \frac{k-y_1}{h-x_1} \Bigr)=-1$$

$$\Rightarrow \cfrac{h-x_1}{a}=\cfrac{k-y_1}{b}=(\frac{(ah+bk+c)-(ax_1+by_1+c)}{a^2+b^2}$$

Using (i) we get $\cfrac{h-x_1}{a}=\cfrac{k-y_1}{b}=-\Bigl(\cfrac{ax_1+by_1+c}{a^2+b^2}\Bigr)$

Where (h,k) is foot of perpendicular from $(x_1,y_1)$ on the line $L\;: \;ax+by+c=0$

NOTE :- let $R(\alpha,\beta)$ be image of point $P(x_1,y_1)$ with respect to line $L\;:\;ax+by+c=0$

$\therefore Q(h,k)=Q\Bigl(\cfrac{\alpha +x_1}{2},\cfrac{\beta + y_1}{2}\Bigr)$ Lies on line $L=0$

$$\therefore a(\frac{\alpha+x_1}{2})+b(\frac{\beta+y_1}{2})+c=0$$

$$\therefore a\alpha +b\beta+c+(ax_1+by_1+c)=0$$

$$\Rightarrow \;a\alpha+b\beta+c=-(ax_1+by_1+c)\,——(2)$$

$$m_{PR}.m_L=-1\;\;\Rightarrow (\cfrac{\beta-y_1}{\alpha-x_1})(-\cfrac{a}{b})=-1$$

$$\Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{a(\alpha-x_1)+b(\beta-y_1)}{a^2+b^2}$$

$$\Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{(a\alpha+b\beta+c)-(ax_1+by_1+c)}{a^2+b^2}$$

Using (2) we get $\Rightarrow \cfrac{\alpha – x_1}{a}=\cfrac{\beta-y_1}{b}=\cfrac{-2(ax_1+by_1+c)}{a^2+b^2}$

Where $(\alpha,\beta)$ is image of $(x_1,y_1)$ with respect to line $ax+by+c=0$