Find the area of the triangle formed by the p | vedclass

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(A) 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(

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$0$

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(A) 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 $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Substituting the given coordinates $P(-1.5, 3)$,$Q(6, -2)$,and $R(-3, 4)$:Area $= \frac{1}{2} |-1.5(-2 - 4) + 6(4 - 3) + (-3)(3 - (-2))|$Area $= \frac{1}{2} |-1.5(-6) + 6(1) - 3(5)|$Area $= \frac{1}{2} |9 + 6 - 15|$Area $= \frac{1}{2} |0| = 0$ square units.Since the area is $0$,the points $P$,$Q$,and $R$ are collinear,meaning they lie on the same straight line and do not form a triangle.

Find the area of the triangle formed by the points $P(-1.5, 3)$,$Q(6, -2)$,and $R(-3, 4)$ (in square units).

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