MCQ
The point which divides the line segment joining the points $(7, – 6)$ and $(3, 4)$ in ratio $1 : 2$ internally lies in the :
- A$I$ quadrant.
- B$II$ quadrant.
- C$III$ quadrant.
- ✓$IV$ quadrant.
✓
Answer
Correct option: D.
$IV$ quadrant.
If $P(x, y)$ divides the line segment joining $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_2\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ internally in the ratio $m : n$ then
$\text{x}=\frac{1(3)+2(7)}{1+2},\text{y}=\frac{1(4)+2(-6)}{1+2}\ [$by section formula$]$
$\Rightarrow\text{x}=\frac{3+14}{3},\text{y}=\frac{-4+12}{3}$
$\Rightarrow\text{x}=\frac{17}{3},\text{y}=-\frac{8}{3}$
So, $\big(\text{x},\text{ y})=\Big(\frac{17}{3},-\frac{8}{3}\Big) $ lies in $IV$ quadrant.
$[$Since, in $IV$ quadrant, $x-$ coordinate is positive and $y-$coordinate is negative$]$
$\text{x}=\frac{1(3)+2(7)}{1+2},\text{y}=\frac{1(4)+2(-6)}{1+2}\ [$by section formula$]$
$\Rightarrow\text{x}=\frac{3+14}{3},\text{y}=\frac{-4+12}{3}$
$\Rightarrow\text{x}=\frac{17}{3},\text{y}=-\frac{8}{3}$
So, $\big(\text{x},\text{ y})=\Big(\frac{17}{3},-\frac{8}{3}\Big) $ lies in $IV$ quadrant.
$[$Since, in $IV$ quadrant, $x-$ coordinate is positive and $y-$coordinate is negative$]$
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