Question

If \(2x^2 - 5xy + 2y^2 = 0\) represents two sides of a triangle whose centroid is \((1,1)\), then the equation of the third side is:

• A. \(x + y + 3 = 0\)
• B. \(x - y - 3 = 0\)
• C. \(x + y - 3 = 0\)
• D. \(x - y + 3 = 0\)

Hint:

To find a third side given two lines and centroid, find the third vertex using centroid formula, then find the line connecting the two known points.

Answer 1

The Correct Option is C

Given: \(2x^2 - 5xy + 2y^2 = 0\) This represents a pair of straight lines through the origin. To find their point of intersection (which is the origin), and to find the third side of the triangle such that the centroid is at \( (1, 1) \), we assume the intersection point is \(A = (0, 0)\), and the other two vertices lie on the lines represented by the given equation. Let the points of intersection of the lines with some arbitrary constants be \(B = (x_1, y_1)\), \(C = (x_2, y_2)\). Then the centroid \(G\) of triangle \(ABC\) is: \[ G = \left( \frac{0 + x_1 + x_2}{3}, \frac{0 + y_1 + y_2}{3} \right) \] Given \(G = (1,1)\), so: \[ \frac{x_1 + x_2}{3} = 1 \Rightarrow x_1 + x_2 = 3
\frac{y_1 + y_2}{3} = 1 \Rightarrow y_1 + y_2 = 3 \] Using section formula or geometry, we find that the third side connects \(B\) and \(C\), whose midpoint is: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3}{2}, \frac{3}{2} \right) \] The line joining \(B\) and \(C\) has midpoint \( \left( \frac{3}{2}, \frac{3}{2} \right) \) and passes through that point. Slope from the earlier line analysis turns out to give the equation: \[ x + y - 3 = 0 \]