### Straight Lines Question 1

#### Question 1 - 24 January - Shift 1

Let $PQR$ be a triangle. The points $A, B$ and $C$ are on the sides QR, RP and PQ respectively such that $\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac{1}{2}$. Then $\frac{Area(\triangle PQR)}{Area(\triangle ABC)}$ is equal to

(1) 4

(2) 3

(3) 2

(4) $\frac{5}{2}$

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Area of Triangle

Let $P$ is $\overrightarrow{{}0}, Q$ is $\vec{q}$ and $R$ is $\vec{r}$

$A$ is $\frac{2 \vec{q}+\vec{r}}{3}, B$ is $\frac{2 \vec{r}}{3}$ and $C$ is $\frac{\vec{q}}{3}$

Area of $\triangle PQR$ is $=\frac{1}{2}|\overrightarrow{{}q} \times \overrightarrow{{}r}|$

Area of $\triangle ABC$ is $\frac{1}{2}|\overrightarrow{{}AB} \times \overrightarrow{{}AC}|$

$\overrightarrow{{}AB}=\frac{\overrightarrow{{}r}-2 \overrightarrow{{}q}}{3}, \overrightarrow{{}AC}=\frac{-\overrightarrow{{}r}-\overrightarrow{{}q}}{3}$

Area of $\triangle ABC=\frac{1}{6}|\overrightarrow{{}q} \times \overrightarrow{{}r}|$

$\frac{Area(\triangle PQR)}{Area(\triangle ABC)}=3$