The three sides of a triangle are given by (x | vedclass

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

Question

(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,a triangle is form

Vedclass · Accepted Answer

$(\frac{-7}{3}, \frac{2}{3})$

Vedclass · Answer

(C) The equation of the sides is given by $(x^2+7xy+2y^2)(y-1)=0$. This implies the sides are $x^2+7xy+2y^2=0$ and $y-1=0$. However,a triangle is formed by three straight lines. The equation $x^2+7xy+2y^2=0$ represents two lines passing through the origin $(0,0)$. Let these lines be $L_1$ and $L_2$. The third line is $L_3: y=1$. To find the vertices,we find the intersection points: $1$. Intersection of $L_1$ and $L_2$: Both pass through the origin,so one vertex is $V_1 = (0,0)$. $2$. Intersection of $L_1$ and $L_3$: Substitute $y=1$ into $x^2+7xy+2y^2=0$,we get $x^2+7x+2=0$. Let the roots be $x_1, x_2$. The vertices are $V_2 = (x_1, 1)$ and $V_3 = (x_2, 1)$. Using the quadratic formula for $x^2+7x+2=0$,the sum of roots $x_1+x_2 = -7$. The centroid $(G_x, G_y)$ is given by $(\frac{0+x_1+x_2}{3}, \frac{0+1+1}{3}) = (\frac{-7}{3}, \frac{2}{3})$.

The three sides of a triangle are given by $(x^2+7xy+2y^2)(y-1)=0$. Then the centroid of that triangle is

Similar Questions

The lines represented by the equation $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$ and the line $4x - 3y - 2 = 0$:

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

The image of the pair of lines represented by $ax^2 + 2hxy + by^2 = 0$ by the line mirror $y = 0$ is

The difference of the slopes of the lines represented by the equation ${x^2}(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$ is:

If the slopes of the lines represented by $ax^{2} + 2hxy + by^{2} = 0$ are in the ratio $1 : 3$,then $h^{2} : ab =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module