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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

Right angled isosceles

Vedclass · Answer

(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...

Similar Questions

If one of the lines given by the equation $2x^2 + axy + 3y^2 = 0$ coincides with one of those given by the equation $2x^2 + bxy - 3y^2 = 0$,while the other two lines are perpendicular to each other,then the values of $a$ and $b$ are

Suppose a triangle of area $27$ sq. units is formed by $18x^2 - 9xy + y^2 = 0$ and $y = c$. Then the centroid of the triangle is

If $\frac{x^2}{a} + \frac{y^2}{b} + \frac{2xy}{h} = 0$ represents a pair of straight lines and the slope of one line is twice the other,then $ab : h^2$ is

The triangle formed by the lines $2x^2+xy-6y^2=0$ and $x+y-1=0$ is

If the slopes of both the lines given by $x^2 + 2hxy + 6y^2 = 0$ are positive and the angle between these lines is $\operatorname{Tan}^{-1}\left(\frac{1}{7}\right)$,then the product of the perpendiculars from the point $(1, 1)$ to these lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module