Let ABC be a triangle. Let a point P divide A | vedclass

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(B) Let the vertices be $A(\vec{a})$,$B(\vec{0})$,and $C(\vec{c})$.Since $P$ divides $AB$ in ratio $1:2$,$\vec{p} = \frac{1\vec{b} + 2\vec{a}}{3} = \f

Vedclass · Accepted Answer

$\frac{2k}{7}$

Vedclass · Answer

(B) Let the vertices be $A(\vec{a})$,$B(\vec{0})$,and $C(\vec{c})$.Since $P$ divides $AB$ in ratio $1:2$,$\vec{p} = \frac{1\vec{b} + 2\vec{a}}{3} = \frac{2}{3}\vec{a}$.Since $Q$ divides $BC$ in ratio $1:2$,$\vec{q} = \frac{1\vec{c} + 2\vec{b}}{3} = \frac{1}{3}\vec{c}$.The line $AQ$ is given by $\vec{r} = (1-t)\vec{a} + t(\frac{1}{3}\vec{c})$.The line $CP$ is given by $\vec{r} = (1-s)\vec{c} + s(\frac{2}{3}\vec{a})$.Equating the coefficients of $\vec{a}$ and $\vec{c}$ for the intersection point $D$:$1-t = \frac{2}{3}s$ and $\frac{1}{3}t = 1-s$.Solving these,we get $s = \frac{2}{7}$ and $t = \frac{6}{7}$.Thus,$\vec{d} = \frac{1}{7}\vec{a} + \frac{2}{7}\vec{c}$.The area of $	riangle BCD$ is given by $\frac{1}{2} |\vec{b} 	imes \vec{d} + \vec{d} 	imes \vec{c} + \vec{c} 	imes \vec{b}|$. Since $\vec{b} = 0$,this simplifies to $\frac{1}{2} |\vec{d} 	imes \vec{c}| = \frac{1}{2} |(\frac{1}{7}\vec{a} + \frac{2}{7}\vec{c}) 	imes \vec{c}| = \frac{1}{14} |\vec{a} 	imes \vec{c}|$.Since the area of $	riangle ABC$ is $k = \frac{1}{2} |\vec{a} 	imes \vec{c}|$,the area of $	riangle BCD$ is $\frac{1}{7} k$.

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