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Distance of a point from line. - Distance between lines

Distance of a point from line.

distance between paralen lines also

Perpendicular distance of a point from the line.

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

Perpendicular distance of point P from line L : 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 parallel line with origin on same side with respect to 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.

Distance between parallel lines with origin between the 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 $L\;:\;ax+by+c=0$

Foot of perpendicular from $P(x_1,y_1)$ on line L is $Q(h,k)$

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

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

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

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 $ax+by+c=0$

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