If A(1,1),B(6,5)C, `((3)/(2),2)` are vertices of `triangleABC`. A point P is such that area of `trianglePAB, trianglePAC,trianglePBC` are equal also `Q((-7)/(6),(-1)/(3))` then length of PQ is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the length of \( PQ \) given the points \( A(1,1) \), \( B(6,5) \), \( C\left(\frac{3}{2}, 2\right) \), and \( Q\left(-\frac{7}{6}, -\frac{1}{3}\right) \), we will follow these steps:

### Step 1: Find the coordinates of point P
Since the areas of triangles \( PAB \), \( PAC \), and \( PBC \) are equal, point \( P \) must be the centroid of triangle \( ABC \). The centroid \( P \) can be calculated using the formula:

\[ P\left(x_P, y_P\right) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right) \]

Substituting the coordinates of points \( A \), \( B \), and \( C \):

\[ x_P = \frac{1 + 6 + \frac{3}{2}}{3} = \frac{1 + 6 + 1.5}{3} = \frac{8.5}{3} = \frac{17}{6} \]

\[ y_P = \frac{1 + 5 + 2}{3} = \frac{8}{3} \]

Thus, the coordinates of point \( P \) are:

\[ P\left(\frac{17}{6}, \frac{8}{3}\right) \]

### Step 2: Use the distance formula to find \( PQ \)
Now, we will find the distance \( PQ \) using the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Where \( P\left(\frac{17}{6}, \frac{8}{3}\right) \) and \( Q\left(-\frac{7}{6}, -\frac{1}{3}\right) \).

Substituting the coordinates into the distance formula:

\[ PQ = \sqrt{\left(-\frac{7}{6} - \frac{17}{6}\right)^2 + \left(-\frac{1}{3} - \frac{8}{3}\right)^2} \]

Calculating the differences:

\[ x_2 - x_1 = -\frac{7}{6} - \frac{17}{6} = -\frac{24}{6} = -4 \]

\[ y_2 - y_1 = -\frac{1}{3} - \frac{8}{3} = -\frac{9}{3} = -3 \]

Now substituting these values back into the distance formula:

\[ PQ = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Final Answer
The length of \( PQ \) is \( 5 \).

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