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A variable line passes through a fixed point $P$. The algebraic sum of the perpendiculars drawn from the points $(2,0)$, $(0,2)$ and $(1,1)$ on the line is zero. Find the coordinates of the point $P$.
[Hint: Let the slope of the line be $m$. Then the equation of the line passing through the fixed point $P\left(x_1, y_1\right)$ is $y-y_1$ $=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$. Taking the algebraic sum of perpendicular distances equal to zero, we get $y-1=m(x-1)$. Thus $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\left.(1,1).\right]$
[Hint: Let the slope of the line be $m$. Then the equation of the line passing through the fixed point $P\left(x_1, y_1\right)$ is $y-y_1$ $=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$. Taking the algebraic sum of perpendicular distances equal to zero, we get $y-1=m(x-1)$. Thus $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\left.(1,1).\right]$
Solution:
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Verified Answer
Let $\mathrm{P}$ be $\left(\mathrm{x}_1, \mathrm{y}_1\right)$, line is $\mathrm{y}-\mathrm{y}_1=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$ $\Rightarrow \mathrm{mx}-\mathrm{y}+\left(\mathrm{y}_1-\mathrm{mx}_1\right)=0$ $\Rightarrow \mathrm{y}_1-\mathrm{mx}_1=\mathrm{y}-\mathrm{mx}$
Sum of perpendiculars distances from $(2,0)$, $(0,2),(1,1)$ on $(\mathrm{i})=0$
$$
\begin{aligned}
&\Rightarrow \frac{2 m+y_1-m x_1}{\sqrt{m^2+1}}+\frac{-2+y_1-m x_1}{\sqrt{m^2+1}}+\frac{m-1+y_1-m x_1}{\sqrt{m^2+1}}=0 \\
&\Rightarrow 2 \mathrm{~m}+3\left(\mathrm{y}_1-\mathrm{mx}_1\right)-2+\mathrm{m}-1=0
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow 2 \mathrm{~m}+3(\mathrm{y}-\mathrm{mx})-3+\mathrm{m}=0[\text { Using (ii) }] \\
&\Rightarrow 3(\mathrm{y}-\mathrm{mx})+3 \mathrm{~m}-3=0 \Rightarrow \mathrm{y}-\mathrm{mx}+\mathrm{m}-1=0 \\
&\Rightarrow \mathrm{y}-1=\mathrm{m}(x-1) \text {, c.f. with (i) } \\
&\therefore \mathrm{y}_1=1, \mathrm{x}_1=1 \Rightarrow \mathrm{P}(1,1)
\end{aligned}
$$
Sum of perpendiculars distances from $(2,0)$, $(0,2),(1,1)$ on $(\mathrm{i})=0$
$$
\begin{aligned}
&\Rightarrow \frac{2 m+y_1-m x_1}{\sqrt{m^2+1}}+\frac{-2+y_1-m x_1}{\sqrt{m^2+1}}+\frac{m-1+y_1-m x_1}{\sqrt{m^2+1}}=0 \\
&\Rightarrow 2 \mathrm{~m}+3\left(\mathrm{y}_1-\mathrm{mx}_1\right)-2+\mathrm{m}-1=0
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow 2 \mathrm{~m}+3(\mathrm{y}-\mathrm{mx})-3+\mathrm{m}=0[\text { Using (ii) }] \\
&\Rightarrow 3(\mathrm{y}-\mathrm{mx})+3 \mathrm{~m}-3=0 \Rightarrow \mathrm{y}-\mathrm{mx}+\mathrm{m}-1=0 \\
&\Rightarrow \mathrm{y}-1=\mathrm{m}(x-1) \text {, c.f. with (i) } \\
&\therefore \mathrm{y}_1=1, \mathrm{x}_1=1 \Rightarrow \mathrm{P}(1,1)
\end{aligned}
$$
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