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Area of the triangle formed by the line $y^2-9 x y+18 x^2=0$ and $y=9$ is
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Verified Answer
The correct answer is:
$\frac{27}{4}$ sq. units
$$
\begin{aligned}
& y^2-9 x y+18 x^2=0 \\
& \therefore(y-3 x)(y-6 x)=0
\end{aligned}
$$
Thus three lines forming triangle are $y=3 x, y=6 x, y=9$
Their point of intersections are $(0,0),(3,9),\left(\frac{3}{2}, 9\right)$
$\therefore$ Area of triangle
$=\frac{1}{2} \times \frac{3}{2} \times 9$
$=\frac{27}{4}$ sq. units
\begin{aligned}
& y^2-9 x y+18 x^2=0 \\
& \therefore(y-3 x)(y-6 x)=0
\end{aligned}
$$
Thus three lines forming triangle are $y=3 x, y=6 x, y=9$
Their point of intersections are $(0,0),(3,9),\left(\frac{3}{2}, 9\right)$
$\therefore$ Area of triangle
$=\frac{1}{2} \times \frac{3}{2} \times 9$
$=\frac{27}{4}$ sq. units
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