$(a)\;(1,0)\qquad(b)\;(-2,0)\qquad(c)\;(-1,0)\qquad(d)\;(2,0)$

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- If two lines are $\perp$ then the product of their slopes is $-1$

Given line is $x+y+1=0$.....(i)

Let the foot of $\perp$ of the point $(2,3)$ on (i) be $D(x_1,y_1)$.

Slope of the line $x+y+1=0$ is $m_1=-1$

$\Rightarrow\: $ Slope of the $\perp$ is $1$

$i.e.,m_2=1$ (slope of perpendicular)

$\therefore$ Equation of the perpendicular line passing through $(2,3)$ will be

$(y-3)=1(x-2)$

$y-x=1$.......(ii)

Now finding the intersection of the lines $y-x=1$ & $x+y+1=0$

$y=x+1$-------(i)

$x+y+1=0$-------(ii)

Putting (1) & (2)

$2x+2=0$

$x=-1$

So $y=0$

Hence foot of perpendicular will be $(-1,0)$.

Hence (c) is the correct answer.

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