The centroid of the triangle formed by the li | vedclass

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

Question

(C) The equation $6x^2+xy-y^2=0$ can be factored as $-(y-3x)(y+2x)=0$,which gives the lines $y=3x$ and $y=-2x$.To find the vertices of the triangle,we

Vedclass · Accepted Answer

$\left(\frac{-1}{3}, \frac{7}{3}\right)$

Vedclass · Answer

(C) The equation $6x^2+xy-y^2=0$ can be factored as $-(y-3x)(y+2x)=0$,which gives the lines $y=3x$ and $y=-2x$.To find the vertices of the triangle,we find the intersection points of these lines with $x+3y=10$:$1$. Intersection of $y=3x$ and $x+3y=10$: $x+3(3x)=10 \implies 10x=10 \implies x=1, y=3$. Vertex is $(1,3)$.$2$. Intersection of $y=-2x$ and $x+3y=10$: $x+3(-2x)=10 \implies -5x=10 \implies x=-2, y=4$. Vertex is $(-2,4)$.$3$. Intersection of $y=3x$ and $y=-2x$ is the origin $(0,0)$.The centroid is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+1-2}{3}, \frac{0+3+4}{3}\right) = \left(-\frac{1}{3}, \frac{7}{3}\right)$.

The centroid of the triangle formed by the lines $x+3y=10$ and $6x^2+xy-y^2=0$ is

Similar Questions

If the slope of one of the lines represented by $a x^2+2 h x y+b y^2=0$ is the square of the other,then $\frac{a+b}{h}+\frac{8 h^2}{a b}$ is

Assertion $(A)$: The lines $2x^2 + 5xy + 2y^2 = 0$ and $x - 2y + 1 = 0$ form a right-angled triangle.
Reason $(R)$: The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of perpendicular lines if $a + b = 0$.
Choose the correct answer.

If $\left(\frac{2}{3}, 0\right)$ is the centroid of the triangle formed by the lines $4x^2-y^2=0$ and $lx+my+n=0$,then $l+m+n=$

If $2x^2 - 3xy + y^2 = 0$ represents two sides of a triangle and $x + y - 1 = 0$ is its third side,then the distance between the orthocenter and the circumcentre of that triangle is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module