The image of $\left( 2,-3 \right)$ in the y-axis is
a) $\left( 2,3 \right)$
b) $\left( -2,3 \right)$
c) $\left( -2,-3 \right)$
d) $\left( 2,-3 \right)$
VerifiedHint: Let us assume that the point is $A(2,-3)$ and the image of the point about y-axis is $A'(x,y)$. Let B is midpoint of line segment AA’ and lies perpendicularly on y-axis, i.e. $x=0$. So, point B is $(0,3)$ as it satisfies both y-axis and line segment AA’. Since point B divides line segment AA’ in ratio $1:1$ internally, so by using section formula $B(x,y)=\left( \dfrac{m\times {{x}_{2}}+n\times {{x}_{1}}}{m+n},\dfrac{m\times {{y}_{2}}+n\times {{y}_{1}}}{m+n} \right)$, find point $A'(x,y)$
Complete step-by-step solution:
As we have assumed that, the point is $A(2,-3)$, and the image of the point about the y-axis is $A'(x,y)$. Also, point B is given as $(0,3)$. So, we have the following diagram:
Since point B divides line segment AA’ in ratio $1:1$ internally, so by using section formula $B(x,y)=\left( \dfrac{m\times {{x}_{2}}+n\times {{x}_{1}}}{m+n},\dfrac{m\times {{y}_{2}}+n\times {{y}_{1}}}{m+n} \right)$
We get:
$B(0,-3)=\left( \dfrac{1\times x+1\times 2}{1+1},\dfrac{1\times y+1\times -3}{1+1} \right)......(1)$
Now, further solving equation (1), we get:
$B(0,-3)=\left( \dfrac{2+x}{2},\dfrac{-3+y}{2} \right)......(2)$
From equation (2), we get two equations as:
$\dfrac{2+x}{2}=0......(3)$
$\dfrac{-3+y}{2}=-3......(4)$
Now, solve equation (3) and equation (4) to get the value of x and y.
We get:
x=-2......(5)
y=-3......(6)
Hence, we have $A'(-2,-3)$ as the image of $A(2,-3)$ about y-axis.
Therefore, option (c) is the correct answer.
Note: The image of the point is at the same distance from the line as the point itself is from the line. Therefore, we can say that the point that satisfies both, the line is given and the line segment made by point and its image divides the line segment in ratio $1:1$ internally. While applying the section formula, some might interchange the values of ratios given, i.e. m in place of n and vice versa. So, take care of writing the given ratio and co-ordinates at the right place or you might end up with a wrong answer. Also include the sign or the co-ordinates alongside as it might change the whole value.
