If the point P(2, 1) lies on the line segment joining points A(4,…

MCQ

If the point $P(2, 1) $ lies on the line segment joining points $A(4, 2)$ and $B(8, 4),$ then :

A
$\text{AP}=\frac{1}{3}\text{AB}$
B
$\text{AP}=\text{PB}$
C
$\text{PB}=\frac{1}{3}\text{AB}$
✓
$\text{AP}=\frac{1}{2}\text{AB}$

✓

Answer

Correct option: D.

$\text{AP}=\frac{1}{2}\text{AB}$

Given that, the point $P(2, 1)$ lies on the line segment joining the points $A(4, 2)$ and $B(8, 4),$ which shows in the figure below :

Now, distance between $A(4, 2)$ and $(2, 1),$
$\text{AP}=\sqrt{(2-4)^2+(1-2)^2}$
$\Big[\ \therefore$ distance between two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$, $\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\text{AP}=\sqrt{(-2)^2+(-1)^2}$
$\text{AP}=\sqrt{4+1}$
$\text{AP}=\sqrt{5}$
Distance between $A(4, 2)$ and $B(8, 4),$
$\text{AB}=\sqrt{(8-4)^2+(4-2)^2}$
$\text{AB}=\sqrt{(4)^2+(2)^2}$
$\text{AB}=\sqrt{16+4}$
$\text{AB}=\sqrt{20}$
$\text{AB}=2\sqrt{5}$
Distance between $B(8, 4)$ and $P(2, 1),$
$\text{BP}=\sqrt{(8-2)^2+(4-1)^2}$
$\text{BP}=\sqrt{(6)^2+(3)^2}$
$\text{BP}=\sqrt{36+9}$
$\text{BP}=\sqrt{45}$
$\text{BP}=3\sqrt{5}$
$\therefore\text{AB}=2\sqrt{5}=2\text{AP}$
$\text{AP}=\frac{\text{AB}}{2}$
Hence, required condition is $\text{AP}=\frac{\text{AB}}{2}.$