Question

Assuming that the straight line works as the plane mirror for a point, find the image of the point $\left( {1,2} \right)$ in the line $x - 3y + 4 = 0$.

Answer 1

To solve this question first we assume a random point that is the image of the given value. Then we apply the image theorem for the general equation. We are able to find all the unknown variables by comparing the general equation and the given equation. from there we are able to find the coordinate of an image of the given point on the given line.

Complete step-by-step solution:
Let, the image of the point $\left( {h,k} \right)$
Let the given point is $\left( {{x_1},{y_1}} \right) = \left( {1,2} \right)$
General equation of the line is $ax + by + c = 0$
The given equation of the line is $x - 3y + 4 = 0$.
By image theorem of the general equation.
$\dfrac{{h - {x_1}}}{a} = \dfrac{{k - {y_1}}}{b} = \dfrac{{ - 2\left( {a{x_1} + b{y_1} + c} \right)}}{{{a^2} + {b^2}}}$ ……(i)
On comparing the given equation and the general equation of the line. We are able to find the value of the $a,b,c$.
On comparing the values are.
$a = 1$,
$b = - 3$ and
$c = 4$.
On putting all these values in equation (i).
$\dfrac{{h - 1}}{1} = \dfrac{{k - 2}}{{ - 3}} = \dfrac{{ - 2\left( {1 + - 6 + 4} \right)}}{{10}}$
On further solving
$\dfrac{{h - 1}}{1} = \dfrac{{k - 2}}{{ - 3}} = \dfrac{2}{{10}}$
On simplifying
$\dfrac{{h - 1}}{1} = \dfrac{{k - 2}}{{ - 3}} = \dfrac{1}{5}$
From here first we equate the first and third term in order to get the value of $h$. Then we compare the second and third terms in order to get the value of $k$.
On equating first and third term.
$\dfrac{{h - 1}}{1} = \dfrac{1}{5}$
On rearranging
$h - 1 = \dfrac{1}{5}$
On further solving
$h = \dfrac{1}{5} + 1$
On taking the LCM in denominator.
$h = \dfrac{{1 + 5}}{5}$
$h = \dfrac{6}{5}$
The image of the x coordinate along the given line is
$\Rightarrow h = \dfrac{6}{5}$
Now on equating second and third term
$\dfrac{{k - 2}}{{ - 3}} = \dfrac{1}{5}$
On rearranging this equation
$k - 2 = \dfrac{{ - 3}}{5}$
On further solving.
$k = \dfrac{{ - 3}}{5} + 2$
On taking LCM in the denominator
$k = \dfrac{{ - 3 + 10}}{5}$
$k = \dfrac{7}{5}$
The image of the y coordinate of the line is
$\Rightarrow k = \dfrac{7}{5}$
Final answer:
The image of the point $\left( {1,2} \right)$ on the line $x - 3y + 4 = 0$ is
$\Rightarrow \left( {\dfrac{6}{5},\dfrac{7}{5}} \right)$

Note: Students are making mistakes in the formula only they make changes if the formula by their earlier remembrance they directly put the square root in denominator of the third term or they forgot to multiply with the factor $- 2$. If we have to find the image on a plane then there is an extra term with z also and we have to equate three equations to find all the coordinates of the image.