MCQ
The origin divides the line segment $AB$ joining the points $A (1,-3)$ and $B (-3,9)$ in the ratio :
- A$3: 1$
- ✓$1: 3$
- C$2: 3$
- D$1: 1$
✓
Answer
Correct option: B.
$1: 3$
Let the ratio be $k : 1$
Using section formula,
$(x, y)=\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
$\Rightarrow(0,0)=\left(\frac{k \times-3+1 \times 1}{k+1}, \frac{k \times 9+1 \times-3}{k+1}\right)$
$\Rightarrow(0,0)=\left(\frac{-3 k+1}{k+1}, \frac{9 k-3}{k+1}\right)$
$\therefore 0=\frac{-3 k+1}{k+1} $ and $0=\frac{9 k-3}{k+1}$
$\Rightarrow 0=-3 k+1 $and $ 0=9 k-3$
$\Rightarrow 3 k=1 $ and $ 9 k=3$
$\therefore \frac{k}{1}=\frac{1}{3}$ and $\frac{k}{1}=\frac{3}{9}=\frac{1}{3}$
Therefore, the required ratio is $1: 3$.
Using section formula,
$(x, y)=\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
$\Rightarrow(0,0)=\left(\frac{k \times-3+1 \times 1}{k+1}, \frac{k \times 9+1 \times-3}{k+1}\right)$
$\Rightarrow(0,0)=\left(\frac{-3 k+1}{k+1}, \frac{9 k-3}{k+1}\right)$
$\therefore 0=\frac{-3 k+1}{k+1} $ and $0=\frac{9 k-3}{k+1}$
$\Rightarrow 0=-3 k+1 $and $ 0=9 k-3$
$\Rightarrow 3 k=1 $ and $ 9 k=3$
$\therefore \frac{k}{1}=\frac{1}{3}$ and $\frac{k}{1}=\frac{3}{9}=\frac{1}{3}$
Therefore, the required ratio is $1: 3$.
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