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A straight line cuts off an intercept of 2 units on the positive direction of $\mathrm{x}$ -axis and passes through the point $(-3,5)$. What is the foot of the perpendicular drawn from the point $(3,3)$ on this line?
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The correct answer is:
$(1,1)$
The given line passes through $(-3,5)$ and $(2,0)$. Its equation is
$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right)$
$\Rightarrow(y-5)=\left(\frac{0-5}{2+3}\right)(x+3)$
$\Rightarrow y=-x+2$ ...(1)
Slope $=m=-1$
and slope of perpendicular line $=-\frac{1}{m}=1$
Equation of this line passing through $(3,3)$ is:
$(y-3)=1(x-3)$
$\Rightarrow \quad y=x .$
From eq. (1) we get, $x=-x+2$
$\Rightarrow x=1$ and $y=1 .$
$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right)$
$\Rightarrow(y-5)=\left(\frac{0-5}{2+3}\right)(x+3)$
$\Rightarrow y=-x+2$ ...(1)
Slope $=m=-1$
and slope of perpendicular line $=-\frac{1}{m}=1$
Equation of this line passing through $(3,3)$ is:
$(y-3)=1(x-3)$
$\Rightarrow \quad y=x .$
From eq. (1) we get, $x=-x+2$
$\Rightarrow x=1$ and $y=1 .$
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