Question

If $(-2,6)$ is the image of the point $(4,2)$ with respect to line $L=0$, then $L=$

• A. $3x-2y+5$
• B. $3x-2y+10$
• C. $2x+3y-5$
• D. $6x-4y-7$

Answer 1

The correct option is A

$3x-2y+5$

The explanation for the correct option:

Step 1: Evaluating the given data.

Let the point $A=(-2,6)$and the point $B=(4,2)$ . and Let the intersection point be marked as $P$.

Since $(-2,6)$ is the image of the point $(4,2)$ with respect to line $L$,

The line $L$ will be a perpendicular bisector of $AB$i.e. $AP=BP$

Step 2: Calculating slopes of the lines.

Let $m_1,m_2$ represent the slopes of line $AB$ and $L$ respectively.

Now, the slope $m_1$ can be calculated as $m_1=\frac{2-6}{4+2}$

$\Rightarrow m_1=\frac{-2}{3}$

It is known that the relation of the slopes of two perpendicular lines is given by ${\mathit{m}}_{\mathbf{1}}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}$

Therefore, on substituting the value of $m_1$ we get the slope $m_2$ as

$\Rightarrow \frac{-2}{3}{m}_2=-1$

$\Rightarrow m_2=\frac{3}{2}$

Step 3: Finding the coordinates of $P$:

Since is the midpoint of the line $AB$, its coordinates can be calculated as $P=\left(\frac{-2+4}{2},\frac{6+2}{2}\right)$

$\Rightarrow P=(1,4)$

Step 4: Finding the equation of the line $L$:

It is known that the equation of the line is given as $\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathit{m}\mathbf{(}\mathit{x}\mathbf{-}{\mathit{x}}_{\mathbf{1}}\mathbf{)}\mathbf{}\mathbf{}.......\left(ii\right)$

Substituting the values with respect to the line $L$ in equation (ii)

$\Rightarrow$ $y-4=\frac{3}{2}(x-1)$

$\Rightarrow$ $2y-8=3x-3$

$\Rightarrow$$\mathbf{}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0}$

Hence, Option (A) is correct.