The triangle formed by the lines 2x^2 - 3xy - | vedclass

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(D) Step $1$: Factorize the pair of lines $2x^2 - 3xy - 2y^2 = 0$. We have $2x^2 - 4xy + xy - 2y^2 = 2x(x - 2y) + y(x - 2y) = (2x + y)(x - 2y) = 0$. T

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Right angled isosceles

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(D) Step $1$: Factorize the pair of lines $2x^2 - 3xy - 2y^2 = 0$. We have $2x^2 - 4xy + xy - 2y^2 = 2x(x - 2y) + y(x - 2y) = (2x + y)(x - 2y) = 0$. The lines are $L_1: 2x + y = 0$ and $L_2: x - 2y = 0$.Step $2$: Observe that the product of the slopes $m_1 = -2$ and $m_2 = 1/2$ is $-2 	imes (1/2) = -1$. Thus, the lines $L_1$ and $L_2$ are perpendicular, meeting at the origin $(0, 0)$.Step $3$: The third line is $L_3: 3x - y = 7$. The triangle is formed by the intersection of $L_1, L_2, L_3$. Since $L_1 \perp L_2$, the triangle is right-angled at the origin.Step $4$: Calculate the lengths of the sides from the origin to the line $L_3$. The distance from $(0,0)$ to $L_1$ is $d_1 = |(3(0) - (0) - 7) / \sqrt{3^2 + (-1)^2}| = 7 / \sqrt{10}$. The vertices on $L_3$ are found by solving systems: $L_1 \cap L_3 \implies 2x + y = 0, 3x - y = 7 \implies x = 1, y = -2$. $L_2 \cap L_3 \implies x - 2y = 0, 3x - y = 7 \implies x = 2, y = 1$.Step $5$: The lengths of the sides forming the right angle are $OA = \sqrt{1^2 + (-2)^2} = \sqrt{5}$ and $OB = \sqrt{2^2 + 1^2} = \sqrt{5}$. Since $OA = OB$, the triangle is isosceles. Thus, it is a right-angled isosceles triangle.

The triangle formed by the lines $2x^2 - 3xy - 2y^2 = 0$ and $3x - y = 7$ is...

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