Question

The image of a point $\left( {2,1} \right)$ with respect to the line $x + 1 = 0$ is:
(A) $\left( {2,5} \right)$
(B) $\left( {0,5} \right)$
(C) $\left( { - 4,1} \right)$
(D) $\left( { - 2, - 3} \right)$

Answer 1

Hint: The distance of a point from a line is the same as the distance of its mirror image from the line. And the line formed by the point and its mirror image is always perpendicular to the mirror line.

Complete step-by-step answer:

The line given in the question is $x + 1 = 0$. And we have to determine the image of point $\left( {2,1} \right)$. Let the point $\left( {2,1} \right)$ is $P$.

Let the image of the point $P$ is point $Q\left( {a,b} \right)$. So points $P\left( {2,1} \right)$ and $Q\left( {a,b} \right)$are mirror images of each other with respect to line $x + 1 = 0$. Thus, the line $PQ$ will be perpendicular to the line $x + 1 = 0$.

Let the equation of line $PQ$ is $y = c$.

And since it is passing through point $P\left( {2,1} \right)$, we will satisfy this point with the equation of line.

Thus, the equation of line $PQ$ is $y = 1$.

This line $y = 1$ is also passing through $Q\left( {a,b} \right)$, so we have $b = 1$.

Thus the coordinates of Q are $\left( {a,1} \right)$.

Further, points $P\left( {2,1} \right)$ and $Q\left( {a,1} \right)$ are equidistant from line $x + 1 = 0$ because they are mirror images with respect to this line.

We know that the distance of a point$\left( {h,k} \right)$ from a line $lx + my + n = 0$ is given as:

$ \Rightarrow D = \left| {\dfrac{{lh + mk + n}}{{\sqrt {{l^2} + {m^2}} }}} \right|$.

Applying this formula for the distances of $P\left( {2,1} \right)$ and $Q\left( {a,1} \right)$ from line $x + 1 = 0$, we’ll get:

$
   \Rightarrow \left| {\dfrac{{2 + 1}}{{\sqrt {{1^2} + 0} }}} \right| = \left| {\dfrac{{a + 1}}{{\sqrt {{1^2} + 0} }}} \right|, \\

   \Rightarrow \left| {a + 1} \right| = \left| 3 \right|, \\

   \Rightarrow a + 1 = 3{\text{ or }}a + 1 = - 3, \\

   \Rightarrow a = 2{\text{ or }}a = - 4 \\

$

We can’t take $a = 2$ because in that case points P and Q will coincide. Hence $a = - 4$ is valid.

Thus the image of point $\left( {2,1} \right)$ with respect to the line $x + 1 = 0$ is $\left( { - 4,1} \right)$. (C) is the correct option.

Note: In the formula of the distance between a point and a line,
$ \Rightarrow D = \left| {\dfrac{{lh + mk + n}}{{\sqrt {{l^2} + {m^2}} }}} \right|$,
We always take modulus signs just to make sure that the distance cannot be negative.