Question

$P\left( {2,1} \right)$ is the image of point $Q\left( {4,3} \right)$ about the line
A. $x + y = 3$
B. $x - y = 1$
C. $3x + y = 5$
D. $ - x + 2y = 0$

Answer 1

Hint: For this question, imagine that you are standing in front of a mirror. You will observe that your image seems to be at a distance equal to the distance between you and the mirror. You will also find that you and the image are a straight line which is perpendicular to the plane of the mirror. This question is in 2-D though and so the line connecting these points will be perpendicular to the line acting as a mirror.

Formulas used:
Equation of line:
$\dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Midpoint \[M\left( {x,y} \right)\] of points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$:
$x = \dfrac{{{x_1} + {x_2}}}{2},y = \dfrac{{{y_1} + {y_2}}}{2}$
Equation of line perpendicular to $y = mx + c$:
$y = - \dfrac{1}{m}x + c'$

Complete Step by Step Solution:
Here in this question, the first step is to find the line perpendicular to the line joining $P\left( {2,1} \right)$ and $Q\left( {4,3} \right)$. For this, first we need to find the line P and Q.
So the line will be
$\
  \dfrac{{y - {y_1}}}{{x - {x_1}}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \\
  \dfrac{{y - 1}}{{x - 2}} = \dfrac{{3 - 1}}{{4 - 2}} \\
  \dfrac{{y - 1}}{{x - 2}} = \dfrac{2}{2} \\
  \dfrac{{y - 1}}{{x - 2}} = 1 \\
  y - 1 = x - 2 \\
  y = x - 1 \\
\ $
Now the line perpendicular to it will be written by the formula mentioned above.
For line $y = mx + c$, perpendicular line is $y = - \dfrac{1}{m}x + c'$. So for line $y = x - 1$ perpendicular line will be
$\
  y = - \dfrac{1}{1}x + c' \\
  y = - x + c' \\
\ $
Now to find the exact line, we need to find $c'$ and to do so, we must know a point which is on that line. We know that the midpoint of the object and the image which is $P\left( {2,1} \right)$ and $Q\left( {4,3} \right)$ respectively, lies on this line.
So we find that midpoint.
Midpoint \[M\left( {x,y} \right)\] of points $P\left( {2,1} \right)$ and $Q\left( {4,3} \right)$:
$\
  x = \dfrac{{{x_1} + {x_2}}}{2},y = \dfrac{{{y_1} + {y_2}}}{2} \\
  x = \dfrac{{2 + 4}}{2},y = \dfrac{{1 + 3}}{2} \\
  x = \dfrac{6}{2},y = \dfrac{4}{2} \\
  x = 3,y = 2 \\
\ $
Putting this in the line $y = - x + c'$, we find the value of $c'$
$\
  y = - x + c' \\
  2 = - 3 + c' \\
  c' = 2 + 3 \\
  c' = 5 \\
\ $
So the equation of line becomes
$\
  y = - x + 5 \\
  x + y = 5 \\
\ $

Note:
Whenever working with reflections, the distance between the object and the image is twice that of the distance between the object and the foot of the perpendicular and so this approach could also have been taken to solve this question.