Question

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

• A. $\left( \frac{1}{3}, -\frac{7}{3} \right)$
• B. $\left( \frac{1}{3}, \frac{7}{3} \right)$
• C. $\left( -\frac{1}{3}, \frac{7}{3} \right)$
• D. $\left( \frac{1}{3}, \frac{7}{3} \right)$

Hint:

Remember that a homogeneous quadratic equation in $x$ and $y$ of the form $Ax^2 + Bxy + Cy^2 = 0$ represents a pair of straight lines passing through the origin. To find the centroid, first find the vertices by solving the line equations in pairs.

Answer 1

The Correct Option is A

Step 1: Identify the equations of the lines.One line is given as $L_1: x + 3y = 10$.The equation $6x^2 + xy - y^2 = 0$ represents a pair of straight lines passing through the origin.Factorize the quadratic equation: \[ 6x^2 + xy - y^2 = 0 \] \[ 6x^2 + 3xy - 2xy - y^2 = 0 \] \[ 3x(2x + y) - y(2x + y) = 0 \] \[ (3x - y)(2x + y) = 0 \] This gives two lines: $L_2: 3x - y = 0 \Rightarrow y = 3x$ $L_3: 2x + y = 0 \Rightarrow y = -2x$
Step 2: Find the vertices of the triangle by finding the intersection points of these three lines.Vertex 1 (Intersection of $L_2$ and $L_3$):Since both $L_2$ and $L_3$ pass through the origin, their intersection point is $V_1 = (0, 0)$.Vertex 2 (Intersection of $L_1$ and $L_2$):Substitute $y = 3x$ into $x + 3y = 10$: \[ x + 3(3x) = 10 \] \[ x + 9x = 10 \] \[ 10x = 10 \] \[ x = 1 \] Then $y = 3(1) = 3$. So, $V_2 = (1, 3)$.Vertex 3 (Intersection of $L_1$ and $L_3$):Substitute $y = -2x$ into $x + 3y = 10$: \[ x + 3(-2x) = 10 \] \[ x - 6x = 10 \] \[ -5x = 10 \] \[ x = -2 \] Then $y = -2(-2) = 4$. So, $V_3 = (-2, 4)$.
Step 3: Calculate the centroid of the triangle with vertices $V_1(0,0)$, $V_2(1,3)$, and $V_3(-2,4)$.The centroid $(G_x, G_y)$ of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: \[ G_x = \frac{x_1 + x_2 + x_3}{3} \] \[ G_y = \frac{y_1 + y_2 + y_3}{3} \] \[ G_x = \frac{0 + 1 + (-2)}{3} = \frac{-1}{3} \] \[ G_y = \frac{0 + 3 + 4}{3} = \frac{7}{3} \] Therefore, the centroid is $\left( -\frac{1}{3}, \frac{7}{3} \right)$.