Three points are A(6, 3),B(-3, 5),and C(4, -2 | 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 $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_

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$\left| \frac{x + y - 2}{7} \right|$

Vedclass · Answer

(A) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area of $\Delta ABC = \frac{1}{2} |6(5 - (-2)) + (-3)(-2 - 3) + 4(3 - 5)| = \frac{1}{2} |6(7) - 3(-5) + 4(-2)| = \frac{1}{2} |42 + 15 - 8| = \frac{1}{2} |49| = 24.5$.Area of $\Delta PBC = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - y) + 4(y - 5)| = \frac{1}{2} |7x + 6 + 3y + 4y - 20| = \frac{1}{2} |7x + 7y - 14| = \frac{7}{2} |x + y - 2|$.Ratio = $\frac{	ext{Area}(\Delta PBC)}{	ext{Area}(\Delta ABC)} = \frac{\frac{7}{2} |x + y - 2|}{\frac{49}{2}} = \left| \frac{x + y - 2}{7} ight|$.

Three points are $A(6, 3)$,$B(-3, 5)$,and $C(4, -2)$. If $P(x, y)$ is a point,then the ratio of the area of $\Delta PBC$ to the area of $\Delta ABC$ is:

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