The area of the triangle formed by the lines represented by $3 x+y+15=0$ and $3 x^2+12 x y-13 y^2=0$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 3 x+y+15=0 ...(i)\ & 3 x^2+12 x y-13 y^2=0...(ii) \end{aligned}$ Point of intersection of (2) is $(0,0)$ From (i) and (ii), $\begin{aligned} & \Rightarrow x=\frac{9 \pm \sqrt{3}}{2}, y=\frac{12 \pm 3 \sqrt{3}}{2} \ & \Rightarrow 	ext { Area }=\frac{1}{2}\left[x_1\left(y_2-y_3ight)+x_2\left(y_3-y_1ight)+x_3\left(y_1-y_2ight)ight] \end{aligned}$ Here $\mathrm{A}(0,0), \mathrm{B}\left(\frac{9+\sqrt{3}}{2}, \frac{12-2 \sqrt{3}}{2}ight), C\left(\frac{9-\sqrt{3}}{2}, \frac{-12-3 \sqrt{3}}{2}ight)$ are the vertices of triangle. $\begin{aligned} & =\frac{1}{2}\left[0+\left(\frac{9+\sqrt{3}}{2}ight)\left(\frac{12-3 \sqrt{3}}{2}ight)+\left(\frac{9-\sqrt{3}}{2}ight)\left(\frac{-12-3 \sqrt{3}}{2}ight)ight] \ & =\frac{15 \sqrt{3}}{2} \end{aligned}$

The area of the triangle formed by the lines represented by $3 x+y+15=0$ and $3 x^2+12 x y-13 y^2=0$ is

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