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Lines $x+y=1$ and $3 y=x+3$ intersect the ellipse $x^{2}+9 y^{2}=9$ at the points $P, Q$ and $R$. The area of the $\triangle P Q R$ is
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Verified Answer
The correct answer is:
$\frac{18}{5}$
Given equation of ellipse
$$
x^{2}+9 y^{2}=9
$$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
and equation of lines
$$
x+y=1 \text { and } 3 y=x+3
$$
or $\quad \frac{x}{(-3)}+\frac{y}{1}=1$
$\therefore$ Area of $\Delta P Q R=\left|\begin{array}{c|ccc}0 & 1 & 1 \\ -3 & 0 & 1 \\ 9 / 5 & -4 / 5 & 1\end{array}\right| \mid$
$=\frac{1}{2}\left[\frac{24}{5}+\frac{12}{5}\right]$
$=\frac{1}{2} \times \frac{36}{5}=\frac{18}{5}$
$$
x^{2}+9 y^{2}=9
$$
$\Rightarrow \quad \frac{x^{2}}{9}+\frac{y^{2}}{1}=1$
and equation of lines
$$
x+y=1 \text { and } 3 y=x+3
$$
or $\quad \frac{x}{(-3)}+\frac{y}{1}=1$
$\therefore$ Area of $\Delta P Q R=\left|\begin{array}{c|ccc}0 & 1 & 1 \\ -3 & 0 & 1 \\ 9 / 5 & -4 / 5 & 1\end{array}\right| \mid$
$=\frac{1}{2}\left[\frac{24}{5}+\frac{12}{5}\right]$
$=\frac{1}{2} \times \frac{36}{5}=\frac{18}{5}$
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