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

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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.

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