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

\frac{27}{4}

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\frac{27}{2}

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\frac{27}{8}

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27

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Solution

The correct option is A $\frac{27}{4}$
$18 x^{2} - 9 x y + y^{2} = 0$
$18 x^{2} - 6 x y - 3 x y + y^{2} = 0$
$6 x (3 x - y) - y (3 x - y) = 0$
$(6 x - y) (3 x - y) = 0$
$y = 6 x, y = 3 x$
$9 = 6 x \Rightarrow x = \frac{9}{6} = \frac{3}{2}$
$9 = 3 x \Rightarrow x = 3$
Area $\Rightarrow$
$\frac{1}{2} ∣ ∣ ∣ \begin{matrix} 0 & 3 & 3 / 2 & 0 0 & 9 & 9 & 0 \end{matrix} ∣ ∣ ∣$
$\frac{1}{2} (27 - \frac{9.3}{2}) = \frac{1}{2} (27 (\frac{1 - 1}{2}))$
$\frac{1}{2} (\frac{27}{2})$
$= \frac{27}{4}$ sq units

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Similar questions

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

y^{2} - 9 x y + 18 x^{2} = 0

y = 9

y^{2} - 9 x y + 18 x^{2} = 0

y = 9

If area of the triangle formed by the lines $y^{2} - 9 x y + 18 x^{2} = 0$ and y=9 is A sq. unit find the value of 4A.

4 x^{2} - 9 x y - 9 y^{2} = 0

x = 2

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