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Step-by-Step Solution
Step 1: Identify the coordinates of the triangle's vertices
The given vertices are
A(1, 0),
B(6, 2), and
C( \frac{3}{2} , 6).
Step 2: Recognize the special point P inside the triangle
P is a point inside triangle ABC such that the areas of triangles APC, APB, and BPC are equal.
In coordinate geometry, the unique point that divides a triangle into three regions of equal area is the centroid.
Step 3: Compute the centroid P
The centroid (P) of a triangle with vertices
(x_1, y_1) ,
(x_2, y_2) ,
(x_3, y_3)
is given by
\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right).
Substituting the given coordinates:
P_x = \frac{1 + 6 + \frac{3}{2}}{3}, \quad
P_y = \frac{0 + 2 + 6}{3}.
Simplify each component:
P_x = \frac{7 + \frac{3}{2}}{3} = \frac{\frac{14}{2} + \frac{3}{2}}{3}
= \frac{\frac{17}{2}}{3}
= \frac{17}{6},
P_y = \frac{8}{3}.
Hence
P\left(\frac{17}{6}, \frac{8}{3}\right).
Step 4: Note the coordinates of point Q
The given point is
Q\left(-\frac{7}{6}, -\frac{1}{3}\right).
Step 5: Apply the distance formula to find PQ
The distance between two points (x_1, y_1) and (x_2, y_2) is
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
Here, substituting
\displaystyle P\left(\frac{17}{6}, \frac{8}{3}\right) and
\displaystyle Q\left(-\frac{7}{6}, -\frac{1}{3}\right) :
PQ = \sqrt{\left(\frac{17}{6} - \left(-\frac{7}{6}\right)\right)^2
+ \left(\frac{8}{3} - \left(-\frac{1}{3}\right)\right)^2}.
Simplify inside each parenthesis:
\frac{17}{6} - \left(-\frac{7}{6}\right) = \frac{17}{6} + \frac{7}{6} = \frac{24}{6} = 4,
\frac{8}{3} - \left(-\frac{1}{3}\right) = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3.
Hence
PQ = \sqrt{(4)^2 + (3)^2}
= \sqrt{16 + 9}
= \sqrt{25}
= 5.
Step 6: Final Answer
Therefore, the length of the line segment PQ is 5.
Reference Image
