The area of the triangle formed by the lines | vedclass

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(C) The given pair of lines is $3x^2 - 4xy + y^2 = 0$. Factoring the quadratic expression: $3x^2 - 3xy - xy + y^2 = 0$ $\Rightarrow 3x(x - y) - y(x -

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$36 	ext{ sq units}$

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(C) The given pair of lines is $3x^2 - 4xy + y^2 = 0$. Factoring the quadratic expression: $3x^2 - 3xy - xy + y^2 = 0$ $\Rightarrow 3x(x - y) - y(x - y) = 0$ $\Rightarrow (3x - y)(x - y) = 0$. Thus,the two lines are $L_1: 3x - y = 0$ and $L_2: x - y = 0$. The third line is $L_3: 2x - y = 6$. To find the vertices of the triangle,we find the intersection points: $1$. Intersection of $L_1$ and $L_2$: $(0, 0)$. $2$. Intersection of $L_1$ and $L_3$: $3x - y = 0$ and $2x - y = 6$. Subtracting gives $x = -6$,so $y = -18$. Point: $(-6, -18)$. $3$. Intersection of $L_2$ and $L_3$: $x - y = 0$ and $2x - y = 6$. Subtracting gives $x = 6$,so $y = 6$. Point: $(6, 6)$. The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Area $= \frac{1}{2} |0(-18 - 6) + (-6)(6 - 0) + 6(0 - (-18))| = \frac{1}{2} |0 - 36 + 108| = \frac{1}{2} |72| = 36 	ext{ sq units}$.

The area of the triangle formed by the lines $3x^2 - 4xy + y^2 = 0$ and $2x - y = 6$ is:

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