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What is the foot of the perpendicular from the point $(2,3)$ on the line $x+y-11=0 ?$
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Verified Answer
The correct answer is:
$(5,6)$
We know a line perpendicular to a given line $a x-b y+c=0$ is $b x+a y+k=0$
The equation of line perpendicular to given line
$$
x+y-11=0 ...(i)
$$
is $-x+y+\lambda=0$ ...(ii)
Since, this equation passes through $(2,3)$. Therefore $(2,3)$ satisfies the equation of line $\therefore \quad-2+3+\lambda=0$
$\Rightarrow \quad \lambda=-1$
From Eq. (ii), $-x+y-1=0$
$\Rightarrow \quad y=x+1$
And from eq. (i), $x+x+1-11=0$
$\Rightarrow \quad 2 x=10$
$\Rightarrow \quad x=5$
Hence, coordinates of foot of perpendicular from $(2,3)$ to given line is $(5,6)$
The equation of line perpendicular to given line
$$
x+y-11=0 ...(i)
$$
is $-x+y+\lambda=0$ ...(ii)
Since, this equation passes through $(2,3)$. Therefore $(2,3)$ satisfies the equation of line $\therefore \quad-2+3+\lambda=0$
$\Rightarrow \quad \lambda=-1$
From Eq. (ii), $-x+y-1=0$
$\Rightarrow \quad y=x+1$
And from eq. (i), $x+x+1-11=0$
$\Rightarrow \quad 2 x=10$
$\Rightarrow \quad x=5$
Hence, coordinates of foot of perpendicular from $(2,3)$ to given line is $(5,6)$
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