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# Find the image of the point $(2,2)$ with respect to the line $x+y+2=0$.

$(a)\;(-4,-4)\qquad(b)\;(-4,-3)\qquad(c)\;(-4,2)\qquad(d)\;(4,4)$

Given line is $x+y+2=0$..........(i)
Let $B(x_1,y_1)$ be the image point of $A(2,2)$ with respect to $x+y+2=0$, and let $C$ be the intersecting point of $AB$ and the given line (i)
Since $B$ is the image of $A$, $C$ is the mid point of $A$ and $B$.
Let the slope of the line $x+y+2=0$ be $m_1$
$\Rightarrow\:m_1=-1$
Let the slope of the line $AB$ be $m_2$
Since $AB$ is $\perp$ to $x+y+2=0$,
$m_1\times m_2=-1$.
$\Rightarrow\:m_2=1$
$\therefore$ The equation of $AB$ is given by
$y-2=(1)(x-2)$
$\Rightarrow\:$ Equation of $AB$ is $y-x=0$......(ii)
Solving (i) and (ii) we get the intersecting point $C$
$i.e., C(-1,-1)$
Step 2
Also we know that $C$ is the mid point of $A$ and $B$.
$\therefore$ From midpoint formula $C$ is given by
$\bigg(\large\frac{x_1+2}{2},\frac{y_1+2}{2}\bigg)$
$\Rightarrow\:\large\frac{x_1+2}{2}$$=-1 and \large\frac{y_1+2}{2}$$=-1$
$\Rightarrow\:x_1=-4$ and $y_1=-4$.
Hence $B(-4,-4)$
Hence (a) is the correct answer.
edited Mar 17, 2014