Question

The image of a point $A\left( {3,8} \right)$ in the line $x + 3y - 7 = 0$ is
A. $( - 1, - 4)$
B. $( - 3, - 8)$
C. $(1, - 4)$
D. $(3,8)$

Answer 1

Hint: Firstly, we will suppose the coordinates of the image of the point given. Then we determine the midpoint of the point and its image. Since the midpoint lies on the given line, we substitute the coordinates of the midpoint in the given equation and hence obtain an equation. Since the line and the given line are perpendicular, by using the relationship between the slopes of two perpendicular lines we obtain another equation. On solving the two equations obtained we can determine the value of a and b.

Complete step by step solution:
Let us consider the image of $A\left( {3,8} \right)$ to be $B\left( {a,b} \right)$.

let $M{\text{ }}$be the midpoint of $AB$.
Then the coordinates of ${\text{M }}$ are $\left( {\dfrac{{3 + a}}{2},\dfrac{{8 + b}}{2}} \right)$
The point ${\text{M }}$lies on the line $x + 3y = 7$.
$\therefore \,\,\dfrac{{3 + a}}{2} + 3\left( {\dfrac{{8 + b}}{2}} \right) = 7$
$ \Rightarrow 3 + a + 24 + 3b = 14$
$ \Rightarrow a + 3b + 13 = 0...(1)$
Since the line AB and CD are perpendicular, Slope of AB X Slope of CD = -1
Slope of AB= $\dfrac{{b - 8}}{{a - 3}}$
Slope of CD= $ - \dfrac{1}{3}$
$ \Rightarrow \dfrac{{b - 8}}{{a - 3}} \times \left( { - \dfrac{1}{3}} \right) = - 1$
$ \Rightarrow \,\,b - 8 = 3a - 9$
$ \Rightarrow \,\,3a - b - 1 = 0...(2)$
On solving (1) and (2) we get,
$a = - 1,b = - 4$
Therefore, the image of the point $\left( {3,{\text{ }}8} \right)$ with respect to the line $x{\text{ }} + {\text{ }}3y{\text{ }} = {\text{ }}7$ is
(− 1, − 4).

Option ‘A’ is correct

Note: In order to solve the given question, one must know the midpoint formula. The midpoint formula for the points $({x_1},{y_1})$ and $({x_2},{y_2})$ is $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$. One must also know to find the slope of a line when two points are given and when the equation of line is given. Two lines are said to be perpendicular if the product of their slopes is −1.