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What are the co-ordinates of the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0 ?$
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The correct answer is:
$(5,6)$
Let B be the foot of perpendicular $\mathrm{AB}$.
Now, $x+y-11=0$
$\Rightarrow \mathrm{y}=-\mathrm{x}+11$ ...(1)
$\Rightarrow$ Slope $=-1$ ...(2)
Since, $A B$ is perpendicular to $x+y-11=0$ product of their slopes $=-1$ $\Rightarrow-1+$ Slope of $A B=-1$
$\Rightarrow$ Slope of $\mathrm{AB}=1$
Now, equation of $\mathrm{AB}$ is given as $y-3=1(x-2) \quad$ (using slope point form) $\Rightarrow y-x=1$ ...(3)
Now, foot of perpendicular
$=$ Point of intersection of line $A B$ and $x+y-11=0$ So, on solving equation (1) and (2) we get $x=5, y=6$.
Hence, $B=(5,6)$.
Now, $x+y-11=0$
$\Rightarrow \mathrm{y}=-\mathrm{x}+11$ ...(1)
$\Rightarrow$ Slope $=-1$ ...(2)
Since, $A B$ is perpendicular to $x+y-11=0$ product of their slopes $=-1$ $\Rightarrow-1+$ Slope of $A B=-1$
$\Rightarrow$ Slope of $\mathrm{AB}=1$
Now, equation of $\mathrm{AB}$ is given as $y-3=1(x-2) \quad$ (using slope point form) $\Rightarrow y-x=1$ ...(3)
Now, foot of perpendicular
$=$ Point of intersection of line $A B$ and $x+y-11=0$ So, on solving equation (1) and (2) we get $x=5, y=6$.
Hence, $B=(5,6)$.
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