Question

Find the image of the point $(2,1)$ with respect to the line mirror $x + y - 5 = 0$

Answer 1

Hint: Firstly, we will suppose the image of the given number. Then, calculate the slope of the image. Further we will find the midpoint of the given line to get the result.


Complete step by step solution:
Let the image of $A(1,2)\,\,be\,\,B(a,b)$.Let $C$ be the midpoint of $AB$.
The midpoint formula $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Then coordinates of $M$ are $\left[ {\dfrac{{2 + a}}{2},\dfrac{{1 + b}}{2}} \right]$

Point $C$ lies on the line $x + y - 5 = 0$ $....(1)$
Now, we put the values of $x$ and $y$ in equation (1), we have,
$\dfrac{{2 + a}}{2} + \dfrac{{1 + b}}{2} - 5 = 0$
$ \Rightarrow \dfrac{{2 + a + 1 + b - 10}}{2} = 0$
\[ \Rightarrow 3 + a + b - 10 = 0 \times 2\]
$ \Rightarrow a + b - 7 = 0$
$ \Rightarrow a + b = 0 + 7$
$ \Rightarrow a + b = 7$ ……(ii)
Then, the lines $x + y - 5 = 0$ and $AB$ are perpendicular then, the product of their slope is$ - 1$.
The points are $A(2,1)\,and\,\,B(a,b)$, then
The slope of $AB = \dfrac{{b - 1}}{{a - 2}}$
And the line $x + y - 5 = 0$
$x + y = 0 + 5$
$y = 5 - x$
$y = x + 5$
Thus, slope of line$({m_2}) = - 1$
${m_1} \times {m_2} = - 1$
Putting the values of slopes, we will get,
$\dfrac{{b - 1}}{{a - 2}} \times - 1$
$ = - 1$
$ - \dfrac{{(b - 1)}}{{a - 2}} = \dfrac{{ - 1}}{1}$
Now, by cross multiplying the values , we have
$b - 1 = a - 2$
$ \Rightarrow - 1 + 2 = a - b$
$ \Rightarrow 1 = a - b$
$ \Rightarrow a - b = 1$ …..(3)
Adding equation (2) and (3), we will get
$a + b + a = 7 + 1$
$2a = 8$
$a = \dfrac{8}{4}$
$a = 4$
Now, from equation (3)
$a - b = 1$
$4 - b = 1$
$ \Rightarrow 4 - 1 = b$
$ \Rightarrow 3 = b$
$ \Rightarrow b = 3$
Hence, the image of point $(2,1)$ with respect to the line mirror $x + y - 5 = 0 $is$ (4,3)$.


Note: Students should substitute the exact value of \[{x_1},{x_2},{y_1} and {y_2}\] in the midpoint formula in accordance to the values given in the question