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A straight line \(L_1\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle O P Q\) is sq. units
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Verified Answer
The correct answer is:
\(\frac{50}{3}\)
\(A=(3,1)\)
Let slope of line be \(m\)
\(\begin{array}{ll}
\therefore \quad & y-y_1=m\left(x-x_1\right) \quad \text { [be the required line] } \\
& y-1=m(x-3)
\end{array}\)
\(\begin{aligned} & y-1=m x-3 m \\ & m x-y+(1-3 m)=0 \quad \ldots (i) \end{aligned}\)
The greatest distance of line from origin passes through \(A(3,1)\) is perpendicular to the given line.
\(\therefore \quad O A \perp P Q\)
Slope of \(O A \times\) Slope of \(\mathrm{PQ}=-1\)
\(\frac{1}{3} \times m=-1 \Rightarrow m=-3\)
Put, \(m=-3\) in Eq. (i),
\(\begin{aligned}
-3 x-y+(1-3)(-3) & =0 \\
-3 x-y+(10) & =0 \\
3 x+y+10 & =0
\end{aligned}\)
\(\therefore\) Area of \(\triangle P O Q=\frac{1}{2}\left|\frac{c^2}{a b}\right|=\frac{1}{2}\left|\frac{100}{3 \times 1}\right|=\frac{50}{3} \mathrm{sq}\) units.
Hence, option (c) is correct.
Let slope of line be \(m\)
\(\begin{array}{ll}
\therefore \quad & y-y_1=m\left(x-x_1\right) \quad \text { [be the required line] } \\
& y-1=m(x-3)
\end{array}\)
\(\begin{aligned} & y-1=m x-3 m \\ & m x-y+(1-3 m)=0 \quad \ldots (i) \end{aligned}\)
The greatest distance of line from origin passes through \(A(3,1)\) is perpendicular to the given line.
\(\therefore \quad O A \perp P Q\)
Slope of \(O A \times\) Slope of \(\mathrm{PQ}=-1\)
\(\frac{1}{3} \times m=-1 \Rightarrow m=-3\)
Put, \(m=-3\) in Eq. (i),
\(\begin{aligned}
-3 x-y+(1-3)(-3) & =0 \\
-3 x-y+(10) & =0 \\
3 x+y+10 & =0
\end{aligned}\)
\(\therefore\) Area of \(\triangle P O Q=\frac{1}{2}\left|\frac{c^2}{a b}\right|=\frac{1}{2}\left|\frac{100}{3 \times 1}\right|=\frac{50}{3} \mathrm{sq}\) units.
Hence, option (c) is correct.
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