Question

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

• A. \((3,12)\)
• B. \((12,3)\)
• C. \((-12,3)\)
• D. \((-3,12)\)

Hint:

For triangle formed by a quadratic and line, solve for vertices, then use centroid formula \((x_1+x_2+x_3)/3, (y_1+y_2+y_3)/3\). Use area to find any unknown parameter.

Answer 1

The Correct Option is A

Step 1: Find vertices.
Triangle formed by intersection of \(18x^2 - 9xy + y^2 = 0\) and \(y = c\). Solve \(18x^2 - 9x c + c^2 = 0\) to find x-coordinates of vertices.

Step 2: Solve quadratic.
\[ 18x^2 - 9cx + c^2 = 0 \implies x = \frac{9c \pm \sqrt{(9c)^2 - 72 c^2}}{36} = \frac{9c \pm 3c}{36} \implies x = \frac{1}{4}c, \frac{1}{2}c \]

Step 3: Vertices coordinates.
Vertices: \((0,0), (\frac{1}{4}c, c), (\frac{1}{2}c, c)\)

Step 4: Centroid formula.
\[ G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0 + c/4 + c/2}{3}, \frac{0 + c + c}{3}\right) = \left(\frac{3c/4}{3}, \frac{2c}{3}\right) = \left(c/4, 2c/3\right) \]

Step 5: Use area condition to find c.
Area \(= 27 = \frac{1}{2} \times base \times height = \frac{1}{2} \times (c/2 - 0) \times c = c^2/4 \implies c = 12\)

Step 6: Compute centroid coordinates.
\[ G = \left(\frac{12}{4}, \frac{2 \cdot 12}{3}\right) = (3,12) \]

Step 7: Final conclusion.
Hence, the centroid is \[ \boxed{(3,12)} \]