If the point $ P(2,1) $ lies on the line segment joining points $ A(4,2) $ and $ B(8,4) $, then
(A) $ \mathrm{AP}=\frac{1}{3} \mathrm{AB} $
(B) $ \mathrm{AP}=\mathrm{PB} $
(C) $ \mathrm{PB}=\frac{1}{3} \mathrm{AB} $
(D) $ \mathrm{AP}=\frac{1}{2} \mathrm{AB} $

AcademicMathematicsNCERTClass 10

Given:

The point \( P(2,1) \) lies on the line segment joining points \( A(4,2) \) and \( B(8,4) \).

To do:

We have to choose the correct option.

Solution:

We know that,

The distance between two points $\left(x_{1}, y_{1}\right)$ and $(x_{2}, y_{2})$, d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

The distance between $A(4,2)$ and $P(2,1)$ is,

$A P=\sqrt{(2-4)^{2}+(1-2)^{2}}$

$=\sqrt{(-2)^{2}+(-1)^{2}}$

$=\sqrt{4+1}$

$=\sqrt{5}$

The distance between $A(4,2)$ and $B(8,4)$ is,

$A B =\sqrt{(8-4)^{2}+(4-2)^{2}}$

$=\sqrt{(4)^{2}+(2)^{2}}$

$=\sqrt{16+4}$

$=\sqrt{20}$

$=2 \sqrt{5}$

The distance between $B(8,4)$ and $P(2,1)$ is,

$B P=\sqrt{(8-2)^{2}+(4-1)^{2}}$

$=\sqrt{6^{2}+3^{2}}$

$=\sqrt{36+9}$

$=\sqrt{45}$

$=3 \sqrt{5}$

$A B =2 \sqrt{5}$

$=2 A P$

$A P=\frac{A B}{2}$

Hence, the required condition is \( \mathrm{AP}=\frac{A B}{2} \).

raja
Updated on 10-Oct-2022 13:28:26

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