If A(6, 3),B(-3, 5),C(4, -2),and D(x, 3x) are | 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_

Vedclass · Accepted Answer

$\frac{11}{8}$

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 $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.For $\Delta ABC$ with $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$:$	ext{Area}(\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{49}{2}$.For $\Delta DBC$ with $D(x, 3x)$,$B(-3, 5)$,$C(4, -2)$:$	ext{Area}(\Delta DBC) = \frac{1}{2} |x(5 - (-2)) + (-3)(-2 - 3x) + 4(3x - 5)| = \frac{1}{2} |7x + 6 + 9x + 12x - 20| = \frac{1}{2} |28x - 14| = |14x - 7|$.Given $\frac{	ext{Area}(\Delta DBC)}{	ext{Area}(\Delta ABC)} = \frac{1}{2}$,we have $2 	imes 	ext{Area}(\Delta DBC) = 	ext{Area}(\Delta ABC)$.$2 |14x - 7| = \frac{49}{2} \Rightarrow |14x - 7| = \frac{49}{4}$.Case $1$: $14x - 7 = \frac{49}{4}$ $\Rightarrow 14x = \frac{49}{4} + 7 = \frac{77}{4}$ $\Rightarrow x = \frac{77}{56} = \frac{11}{8}$.Case $2$: $14x - 7 = -\frac{49}{4}$ $\Rightarrow 14x = 7 - \frac{49}{4} = -\frac{21}{4}$ $\Rightarrow x = -\frac{21}{56} = -\frac{3}{8}$.Since $\frac{11}{8}$ is in the options,the correct value is $\frac{11}{8}$.

If $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$,and $D(x, 3x)$ are four points. If the ratio of the area of $\Delta DBC$ and $\Delta ABC$ is $1 : 2$,then the value of $x$ is:

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