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 $(2, 1),$
$\text{AP}=\sqrt{(2-4)^2+(1-2)^2}$
$\left[\therefore\right.$ distance between two points $\left( x _1, y _1\right)$ and $\left.\left( x _2, y _2\right), d =\sqrt{\left( x _2- x _1\right)^2+\left( y _2- y _1\right)^2}\right]$
$\text{AP}=\sqrt{(-2)^2+(-1)^2}$
$\text{AP}=\sqrt{4+1}$
$\text{AP}=\sqrt{5}$
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 $\text{AP}=\frac{\text{AB}}{2}.$