Find the coordinates of foot of the perpendicular drawn from the point (2,3) to the line $x+y+1=0$.

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

Toolbox:
• 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.
edited Mar 17, 2014