ABCD is a parallelogram and P is the mid-poin | vedclass

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(A) Let the position vectors of vertices $A, B, D$ be $\vec{0}, \vec{b}, \vec{d}$ respectively. Then the position vector of $C$ is $\vec{b}+\vec{d}$.S

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, D$ be $\vec{0}, \vec{b}, \vec{d}$ respectively. Then the position vector of $C$ is $\vec{b}+\vec{d}$.Since $P$ is the mid-point of $AD$,the position vector of $P$ is $\frac{\vec{d}}{2}$.Let $Q$ divide $AC$ in the ratio $\lambda:1$ and $BP$ in the ratio $\mu:1$.The position vector of $Q$ on $AC$ is $\frac{\lambda(\vec{b}+\vec{d}) + 1(\vec{0})}{\lambda+1} = \frac{\lambda}{\lambda+1}\vec{b} + \frac{\lambda}{\lambda+1}\vec{d}$.The position vector of $Q$ on $BP$ is $\frac{\mu(\frac{\vec{d}}{2}) + 1(\vec{b})}{\mu+1} = \frac{1}{\mu+1}\vec{b} + \frac{\mu}{2(\mu+1)}\vec{d}$.Comparing the coefficients of $\vec{b}$ and $\vec{d}$:$\frac{\lambda}{\lambda+1} = \frac{1}{\mu+1}$ and $\frac{\lambda}{\lambda+1} = \frac{\mu}{2(\mu+1)}$.From these,$\frac{1}{\mu+1} = \frac{\mu}{2(\mu+1)} \Rightarrow \mu = 2$.Substituting $\mu=2$ into the first equation: $\frac{\lambda}{\lambda+1} = \frac{1}{2+1} = \frac{1}{3}$.$3\lambda = \lambda+1 \Rightarrow 2\lambda = 1 \Rightarrow \lambda = \frac{1}{2}$.Thus,the ratio $AQ:QC = \lambda:1 = \frac{1}{2}:1 = 1:2$.

$ABCD$ is a parallelogram and $P$ is the mid-point of the side $AD$. The line $BP$ meets the diagonal $AC$ in $Q$. Then,the ratio of $AQ:QC$ is equal to

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