The area of the triangle formed by the lines | vedclass

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Question

(C) The given equation is $4x^2 - 9xy - 9y^2 = 0$. Factorizing the quadratic expression: $4x^2 - 12xy + 3xy - 9y^2 = 0$ $4x(x - 3y) + 3y(x - 3y) = 0$

Vedclass · Accepted Answer

$10/3$

Vedclass · Answer

(C) The given equation is $4x^2 - 9xy - 9y^2 = 0$. Factorizing the quadratic expression: $4x^2 - 12xy + 3xy - 9y^2 = 0$ $4x(x - 3y) + 3y(x - 3y) = 0$ $(4x + 3y)(x - 3y) = 0$. Thus,the two lines are $L_1: 4x + 3y = 0$ and $L_2: x - 3y = 0$. The third line is $L_3: x = 2$. 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 $L_3$: Substituting $x = 2$ into $4x + 3y = 0$,we get $8 + 3y = 0$,so $y = -8/3$. Vertex is $(2, -8/3)$. $3$. Intersection of $L_2$ and $L_3$: Substituting $x = 2$ into $x - 3y = 0$,we get $2 - 3y = 0$,so $y = 2/3$. Vertex is $(2, 2/3)$. 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 $= \frac{1}{2} |0(2/3 - (-8/3)) + 2(-8/3 - 0) + 2(0 - 2/3)|$ Area $= \frac{1}{2} |0 - 16/3 - 4/3| = \frac{1}{2} |-20/3| = 10/3$.

The area of the triangle formed by the lines represented by $4x^2 - 9xy - 9y^2 = 0$ and the line $x = 2$ is

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