Question

Derive the formula for image $B(l,m)$ of a point $P(p,q)$ w.r.t a line $ax+by+c=0$.

• A. $\frac{l+p}{a}=\frac{m+q}{b}=\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$
• B. $\frac{l-p}{a}=\frac{m-q}{b}=-\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$
• C. $\frac{l+p}{a}=\frac{m+q}{b}=-\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$
• D. $\frac{l-p}{a}=\frac{m-q}{b}=\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$

Answer 1

Answer: (B)

$\frac{l-p}{a}=\frac{m-q}{b}=-\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$

Let $ax+by+c=0$ be the line and $P(p,q)$ be the point.

The image of the point $P$ in the line mirror $ax+by+c=0$ is the point $B(l,m)$ such that $PB$ is perpendicular to $ax+by+c=0$ and the midpoint L of $PB$ lies on the line $ax+by+c=0$

The slope of the line $ax+by+c=0$ is $-\frac{a}{b}$

Thus the slope of $PB$ is $\frac{b}{a}$.

If $PB$ makes an angle $\theta$ with $x-$axis. Then $tan\theta =\frac{b}{a}\Rightarrow sin\theta =\frac{b}{\sqrt{a^2+b^2}},cos\theta =\frac{a}{\sqrt{a^2+b^2}}$

Equation of $PB$ in distance form is $\frac{x-p}{cos\theta }=\frac{y-q}{sin\theta }$

Let $PL=r$. Then the coordinates of $L$ are given by

$\frac{x-p}{cos\theta }=\frac{y-q}{sin\theta }=r$

So, the coordinates of $L$ are $(p+rcos\theta ,q+rsin\theta )$

Since $L$ lies on $ax+by+c=0$,

We get $a(p+rcos\theta )+b(q+rsin\theta )+c=0$

$\Rightarrow r=-\frac{ap+bq+c}{acos\theta +bsin\theta }$

$\Rightarrow r=-\frac{ap+bq+c}{\sqrt{a^2+b^2}}[\because sin\theta =\frac{b}{\sqrt{a^2+b^2}},cos\theta =\frac{a}{\sqrt{a^2+b^2}}]$

The coordinates of $B$ are given by $\frac{x-p}{cos\theta }=\frac{y-q}{sin\theta }=2r$ or $\frac{x-p}{a}=\frac{y-q}{b}=-\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$

Thus, the image $B(l,m)$ of the point $P(p,q)$ on the line $ax+by+c=0$ is $\frac{l-p}{a}=\frac{m-q}{b}=-\frac{2(ap+bq+c)}{\sqrt{a^2+b^2}}$