Question

Let A(1, 0), B(6, 2) and C \( \left( {{3 \over 2},6} \right) \) be the vertices of a triangle ABC. If P is a Point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then the length of the line segment PQ, where Q is the point \( \left( { - {7 \over 6}, - {1 \over 3}} \right) \), is ________.

Hint:

When the triangles APC, APB and BPC have equal areas, P is the centroid of triangle ABC. Then use the distance formula for PQ.

Answer 1

Correct Answer: 5

Step 1: The three triangles APB, BPC and APC have equal areas (each is one third of the area of triangle ABC), so P is the centroid of triangle ABC.
Step 2: With \(A(1, 0), B(6, 2)\) and \(C\left(\dfrac{3}{2}, 6\right)\): \(P = \left(\dfrac{1 + 6 + 3/2}{3}, \dfrac{0 + 2 + 6}{3}\right) = \left(\dfrac{17}{6}, \dfrac{8}{3}\right)\).
Step 3: For \(Q\left(-\dfrac{7}{6}, -\dfrac{1}{3}\right)\): \(PQ^2 = \left(\dfrac{17}{6} + \dfrac{7}{6}\right)^2 + \left(\dfrac{8}{3} + \dfrac{1}{3}\right)^2 = 4^2 + 3^2 = 25\).
So \(PQ = 5\).
Answer: 5