Question

What is the reflection of the point $(5,2)$ in the line $x=-3$
A. $(-11,2)$
B. $(-11,-2)$
C. $(11,-2)$
D. $(11,2)$

Answer 1

Hint: in this question first draw the figure. Use the property of reflection as image and object distance are at equal distance from the mirror. Also take distance as perpendicular distance, then find the equation of line which is perpendicular to the line of mirror and passes through point (5,2). Then find the point of intersection of mirror line and line perpendicular to it and passes through. This gives a coordinate which is the mid-point of point P and image point. Then apply the external division formula to find the image of the given point.

Complete step-by-step answer:
In this question the straight line $x=-3$ play the role of mirror with respect to which we have to find the reflection of point (5, -3)
suppose AB is the line of mirror, so equation of mirror is
$x=-3$ ------------------------------(1)
Let $P(5,2)$ be the point whose reflection is to be found out.
Let $Q(\alpha ,\beta )$ be the image of point P with respect to the line mirror (1). Now we draw the figure.

So, we can say that
$PQ\bot AM\text{ and PL=LQ}$ as we know that image distance is equal to object distance
Here L is the foot of perpendicular from point P to Line AB.
Now since PL is perpendicular to the line (1) and passes through (5,2), so here we have to find the equation of PL.
Let slope of line PL is $m$
Slope of line (1) is $\infty$ as it is parallel to x axis
So, we can the property that product of slope of two perpendicular lines is -1, hence we can write
$\begin{aligned} & m(\infty )=-1 \\\ & \Rightarrow m=0 \\\ \end{aligned}$
Hence PL is parallel to y axis. So, we can write equation of PL as
$x=k$
As line PL is passes through (5,2), so equation of PL is
$y=2$ -------------------------------(2)
Hence from (1) and (2) we get the coordinate of L, so
$L\equiv (-3,2)$
Since Q is the image the image of P, therefor $PL=LQ\text{ or }\dfrac{PL}{QL}=\dfrac{2}{1}$
Thus, Q divides PL externally in the ratio $2:1$
Now here we use external division formula which says that if any point $Q\equiv (\alpha ,\beta )$ divides a line LP externally in the ratio $m:n$ then
$\therefore \alpha =\dfrac{m({{x}_{2}})-n({{x}_{1}})}{m-n}\text{and }\beta \text{=}\dfrac{m({{y}_{2}})-n({{y}_{1}})}{m-n}$
$\begin{aligned} & L({{x}_{1}},{{y}_{1}})\equiv (-3,2)\text{ and} \\\ & P({{x}_{2}},{{y}_{2}})\equiv (5,2)\text{ } \\\ \end{aligned}$
also
$m:n\equiv 1:2$
Putting the above values, we get
$\therefore \alpha =\dfrac{1(5)-2.(-3)}{1-2}=-11\text{ and }\beta \text{=}\dfrac{1(2)-2(2)}{1-2}=2$
Hence
$\alpha =-11,\text{ and }\beta =2$
So reflection of $(5,2)$ with respect to line $x=-3$ is $(-11,2)$
Hence option A is correct.

Note: The given question can be solved by using the formula below directly
The image of a point $({{x}_{1}},{{y}_{1}})$ with respect to the line mirror $ax+by+c=0$ be $Q(x,y)$ is given by
$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=-\dfrac{2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}}$
In multiple choice questions we use the above formula in order to get the result fast.