The area of the triangle formed by the lines | vedclass

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(C) The given equation is $y^2 - 9xy + 18x^2 = 0$. Factoring the quadratic equation,we get $(y - 3x)(y - 6x) = 0$. Thus,the two lines represented by t

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$\frac{27}{4}$ sq. units

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(C) The given equation is $y^2 - 9xy + 18x^2 = 0$. Factoring the quadratic equation,we get $(y - 3x)(y - 6x) = 0$. Thus,the two lines represented by the equation are $y = 3x$ and $y = 6x$. The third line is $y = 9$. To find the vertices of the triangle,we find the intersection points: $1$. Intersection of $y = 3x$ and $y = 6x$ is $(0, 0)$. $2$. Intersection of $y = 3x$ and $y = 9$ is $(3, 9)$. $3$. Intersection of $y = 6x$ and $y = 9$ is $(\frac{3}{2}, 9)$. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Area $= \frac{1}{2} |0(9 - 9) + 3(9 - 0) + \frac{3}{2}(0 - 9)|$ Area $= \frac{1}{2} |0 + 27 - \frac{27}{2}| = \frac{1}{2} |\frac{54 - 27}{2}| = \frac{1}{2} 	imes \frac{27}{2} = \frac{27}{4}$ sq. units.

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

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