The area of the triangle with vertices (3, 8) | vedclass

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

Vedclass · Accepted Answer

$122$

Vedclass · Answer

(C) The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ is given by $	ext{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 vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$:$	ext{Area} = \frac{1}{2} |3(2 - 1) + (-4)(1 - 8) + 5(8 - 2)|$$	ext{Area} = \frac{1}{2} |3(1) - 4(-7) + 5(6)|$$	ext{Area} = \frac{1}{2} |3 + 28 + 30| = \frac{61}{2}$.Given that $	ext{Area} = \frac{P}{4}$, we have $\frac{P}{4} = \frac{61}{2}$.Multiplying both sides by $4$, we get $P = \frac{61 	imes 4}{2} = 122$.

The area of the triangle with vertices $(3, 8)$, $(-4, 2)$ and $(5, 1)$ is $\frac{P}{4}$. The value of $P$ is:

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