(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: $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x

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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: $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Substituting the given vertices $(5, 2)$,$(2/3, 2)$,and $(-4, 3)$: $\Delta = \frac{1}{2} |5(2 - 3) + \frac{2}{3}(3 - 2) + (-4)(2 - 2)|$ $= \frac{1}{2} |5(-1) + \frac{2}{3}(1) + (-4)(0)|$ $= \frac{1}{2} |-5 + \frac{2}{3}|$ $= \frac{1}{2} |\frac{-15 + 2}{3}|$ $= \frac{1}{2} |-\frac{13}{3}| = \frac{13}{6}$ square units.

Class 12 Mathematics 3 and 4 .Determinants and Matrices Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If the vertices of a triangle are $(5, 2)$,$(2/3, 2)$,and $(-4, 3)$,then the area of the triangle is

A
$28/6$
B
$5/2$
C
$43$
D
$13/6$

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If the vertices of a triangle are $(5, 2)$,$(2/3, 2)$,and $(-4, 3)$,then the area of the triangle is

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