If ‘O’ the ortho center of triangle formed by vertices $A (1, - 3), B (7, 2), C (2, 5)$ and P,Q,R are the ortho centres of the triangles $AOC, BOC, AOB$ then the area of $ΔPQR$ is.

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$\frac{43}{2}$

answer is C.

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Detailed Solution

If O is orthocentre of a triangle ABC then the ortho center of the triangle formed by and any other two vertices of triangle ABC is third vertex.
Hence, the vertices P,Q,R are same as the vertices A,B,C
The area of the triangle PQR is same as the area of the triangle ABC
$\begin{array}{l} Area of triangle ABC = \frac{1}{2} ∣ \begin{array}{cc} x_{2} - x_{1} & y_{2} - y_{1} \\ x_{3} - x_{1} & y_{3} - y_{1} \end{array} \\ = \frac{1}{2} |\begin{array}{ll} 1 - 7 & 7 - 2 \\ - 3 - 2 & 2 - 5 \end{array}| \\ = \frac{1}{2} |\begin{array}{cc} - 6 & 5 \\ - 5 & - 3 \end{array}| \\ = \frac{1}{2} | 43 | \end{array}$
Therefore, the area of the triangle PQR is equal to $\frac{43}{2} sq units$