$P Q R S$ is a quadrilateral and $\mathbf{P Q}=\mathbf{a}, \mathbf{Q R}=\mathbf{b}, \mathbf{S…

$P Q R S$ is a quadrilateral and $\mathbf{P Q}=\mathbf{a}, \mathbf{Q R}=\mathbf{b}, \mathbf{S P}=\mathbf{a}-\mathbf{b}, M$ is the mid-point of $Q R$ and $X$ is a point on $\mathbf{S M}$ such that $\mathbf{S X}=\frac{4}{5} \mathbf{S M}$. If $\mathbf{S M}=m(4 \mathbf{a}-\mathbf{b})$ and $\mathbf{S X}=n(4 \mathbf{a}-\mathbf{b})$, then $m+n=$
  1. $9 / 10$
  2. $10 / 9$
  3. $11 / 9$
  4. $4 / 3$

Solution


$\begin{aligned} & S Q=S P+P Q=(a-b)+a=2 a-b \\ & S M=S Q+Q M=(2 a-b)+\frac{b}{2}=2 a-\frac{b}{2}\end{aligned}$ $\Rightarrow \quad \mathbf{S M}=\frac{1}{2}(4 \mathbf{a}-\mathbf{b})$ $\therefore \quad m=\frac{1}{2}$ $\mathbf{S X}=\frac{4}{5} \mathbf{S M}$ (given) $=\frac{4}{5} \cdot \frac{1}{2}(4 a-b)$ $\Rightarrow \quad S X=\frac{2}{5}(4 a-b)$ $\therefore \quad n=\frac{2}{5}$ Hence, $m+n=\frac{1}{2}+\frac{2}{5}=\frac{9}{10}$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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