MCQ
The point $P$ which divides the line segment joining the points $A(2,- 5)$ and $B(5, 2)$ in the ratio $2 : 3$ lies in the quadrant :
- A$I$
- B$II$
- C$III$
- ✓$IV$
✓
Answer
Correct option: D.
$IV$
$ A(2,-5) \Rightarrow x_1=2, y_1=-5$
$ B(5,2) \Rightarrow x_2=5, y_2=2 $
$m_1= 2, m_2= 3$
Let the coordinates of $P$ be $(x, y)$
$\therefore$ By section formula,
$\text{x}=\frac{\text{m}_1\text{x}_2+\text{m}_2\text{x}_1}{\text{m}_1+\text{m}_2},\text{y}=\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}$
$\text{x}=\frac{2\times5+3\times2}{2+3},\text{y}=\frac{-5\times3+2\times2}{2+3}$
$\text{x}=\frac{10+6}{5},\text{y}=\frac{-15+4}{5}$
$\text{x}=\frac{16}{5},\text{y}=\frac{-11}{5}$
$\therefore\ \text{P}=\Big(\frac{16}{5},\frac{-11}{5}\Big)$
$P$ lies in the $IV$ quadrant.
$ B(5,2) \Rightarrow x_2=5, y_2=2 $
$m_1= 2, m_2= 3$
Let the coordinates of $P$ be $(x, y)$
$\therefore$ By section formula,
$\text{x}=\frac{\text{m}_1\text{x}_2+\text{m}_2\text{x}_1}{\text{m}_1+\text{m}_2},\text{y}=\frac{\text{m}_2\text{y}_1+\text{m}_1\text{y}_2}{\text{m}_1+\text{m}_2}$
$\text{x}=\frac{2\times5+3\times2}{2+3},\text{y}=\frac{-5\times3+2\times2}{2+3}$
$\text{x}=\frac{10+6}{5},\text{y}=\frac{-15+4}{5}$
$\text{x}=\frac{16}{5},\text{y}=\frac{-11}{5}$
$\therefore\ \text{P}=\Big(\frac{16}{5},\frac{-11}{5}\Big)$
$P$ lies in the $IV$ quadrant.
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