If the area of a triangle with vertices (a, 5 | vedclass

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(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y

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$9$

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(D) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $(a, 5), (6, 7)$ and $(2, 3)$,and the area is $10$.Substituting the values:$10 = \frac{1}{2} |a(7 - 3) + 6(3 - 5) + 2(5 - 7)|$$20 = |a(4) + 6(-2) + 2(-2)|$$20 = |4a - 12 - 4|$$20 = |4a - 16|$This gives two cases:Case $1$: $4a - 16 = 20 \implies 4a = 36 \implies a = 9$Case $2$: $4a - 16 = -20 \implies 4a = -4 \implies a = -1$Since it is given that $a \in N$ (natural number),we discard $a = -1$.Therefore,$a = 9$.

If the area of a triangle with vertices $(a, 5), (6, 7)$ and $(2, 3)$ is $10$ square units,then $a = \dots$ (where $a \in N$).

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