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Let ` A ( 1, 0 ) , B ( 6, 2 ) ` and ` C (( 3 ) /(2), 6 ) ` 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 ` ( - ( 7 ) /(6), - ( 1 ) /(3)) , ` is _________.

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To solve the problem step by step, we will first find the coordinates of point P, which is the centroid of triangle ABC, and then calculate the distance between points P and Q. ### Step 1: Find the coordinates of the centroid P of triangle ABC. The vertices of triangle ABC are: - A(1, 0) - B(6, 2) - C(3/2, 6) The formula for the centroid (P) of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) is given by: \[ P\left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \] Substituting the coordinates of A, B, and C: \[ P\left(\frac{1 + 6 + \frac{3}{2}}{3}, \frac{0 + 2 + 6}{3}\right) \] Calculating the x-coordinate: \[ x_P = \frac{1 + 6 + \frac{3}{2}}{3} = \frac{7 + \frac{3}{2}}{3} = \frac{\frac{14}{2} + \frac{3}{2}}{3} = \frac{\frac{17}{2}}{3} = \frac{17}{6} \] Calculating the y-coordinate: \[ y_P = \frac{0 + 2 + 6}{3} = \frac{8}{3} \] Thus, the coordinates of point P are: \[ P\left(\frac{17}{6}, \frac{8}{3}\right) \] ### Step 2: Identify the coordinates of point Q. The coordinates of point Q are given as: \[ Q\left(-\frac{7}{6}, -\frac{1}{3}\right) \] ### Step 3: Calculate the distance PQ using the distance formula. The distance \(d\) between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Substituting the coordinates of P and Q: \[ d = \sqrt{\left(-\frac{7}{6} - \frac{17}{6}\right)^2 + \left(-\frac{1}{3} - \frac{8}{3}\right)^2} \] Calculating the x-component: \[ -\frac{7}{6} - \frac{17}{6} = -\frac{24}{6} = -4 \] Calculating the y-component: \[ -\frac{1}{3} - \frac{8}{3} = -\frac{9}{3} = -3 \] Thus, we have: \[ d = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] ### Final Answer: The length of the line segment PQ is \(5\). ---
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