Area of the triangle formed by the lines $y^{2} - 9 x y + 18 x^{2} = 0$ and $y = 9$ is :

\frac{27}{4}

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\frac{9}{4}

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27

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Solution

The correct option is A $\frac{27}{4}$
Given pair of straight line $y^{2} - 9 x y + 18 x^{2} = 0$
Let the above line represents lines $y = m_{1} x$ and $y = m_{2} x$
$\Rightarrow (y - m_{1} x) (y - m_{2} x) = y^{2} - 9 x y + 18 x^{2}$
$\Rightarrow y^{2} - (m_{1} + m_{2}) x + m_{1} m_{2} = y^{2} - 9 x y + 18 x^{2}$
On comparing, we get
$m_{1} + m_{2} = 9$ and $m_{1} m_{2} = 18$
$\Rightarrow m_{1} = 3, 6, m_{2} = 6, 3$
So, the equations are $y = 3 x, y = 6 x$
Now, point of intersection of $y = 3 x, y = 6 x, y = 9$ are $(0, 0), (3, 9) (\frac{3}{2}, 9)$
Area $= \frac{1}{2} \times \frac{3}{2} \times 9$ $= \frac{27}{4}$ sq.units

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