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If the area of the triangle formed by the lines $y=x+c$ and $2 x^2+5 x y+3 y^2=0$ is $\frac{1}{20}$ sq. units, then $c=$
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Verified Answer
The correct answer is:
$\pm {1}$
$$
\begin{aligned}
& \text { } 2 x^2+5 x y+3 y^2=0 \\
& (x+y)(2 x+3 y)=0
\end{aligned}
$$
Let $\triangle \mathrm{ABC}$ be the region formed by
$\begin{aligned} & \text { Now, area of } \triangle \mathrm{ABC}=\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & 1 \\ -\frac{c}{2} & \frac{c}{2} & 1 \\ -\frac{3 c}{5} & \frac{2 c}{5} & 1\end{array}\right| \\ & \Rightarrow \frac{1}{20}=0+0+1\left[\left(-\frac{c}{2}\right)\left(\frac{2 c}{5}\right)-\left(\frac{c}{2}\right)\left(-\frac{3 c}{5}\right)\right] \\ & \Rightarrow c^2=1 \Rightarrow c= \pm 1\end{aligned}$
\begin{aligned}
& \text { } 2 x^2+5 x y+3 y^2=0 \\
& (x+y)(2 x+3 y)=0
\end{aligned}
$$
Let $\triangle \mathrm{ABC}$ be the region formed by
$\begin{aligned} & \text { Now, area of } \triangle \mathrm{ABC}=\left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|=\left|\begin{array}{ccc}0 & 0 & 1 \\ -\frac{c}{2} & \frac{c}{2} & 1 \\ -\frac{3 c}{5} & \frac{2 c}{5} & 1\end{array}\right| \\ & \Rightarrow \frac{1}{20}=0+0+1\left[\left(-\frac{c}{2}\right)\left(\frac{2 c}{5}\right)-\left(\frac{c}{2}\right)\left(-\frac{3 c}{5}\right)\right] \\ & \Rightarrow c^2=1 \Rightarrow c= \pm 1\end{aligned}$
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