Find the ratio in which $P( 4,\ m)$ divides the line segment joining the points $A( 2,\ 3)$ and $B( 6,\ –3)$. Hence find $m$.

Academic Mathematics NCERT Class 10

Given: $P( 4,\ m)$ divides the line segment joining the points $A( 2,\ 3)$ and $B( 6,\ –3)$

To do: To find the ratio of division and to find the value of $m$.

Solution:

Let us Suppose the point $P( 4,\ m)$ divides the line segment joining the points $A( 2,\ 3)$ and $B(6, -3)$ in the ratio

$k:1$.

Using section formula, we have $P(x, y)=( \frac{nx_{1}+mx_{2}}{m+n}, \frac{ny_{1}+my_{2}}{m+n})$

Co-ordinates of point $P(4, m)=( \frac{2×1+6×k}{k+1}, \frac{1×3+k×-3}{k+1})$

$\Rightarrow (4, m)=( \frac{6k+2}{k+1}, \frac{3-3k}{k+1})$

$\frac{6k+2}{k+1}=4 .....................( 1)$

$\frac{3-3k}{k+1}=m ......................( 2)$

From $( 1)$,

$\Rightarrow 6k+2=4( k+1)$

$\Rightarrow 6k+2=4k+4$

$\Rightarrow 6k-4k=4-2$

$\Rightarrow 2k=2$

$\Rightarrow k=1$

Putting the above value in $( 2)$,

$m=\frac{3-3k}{k+1}$

$\Rightarrow m=\frac{3-3(1)}{1+1}$

$\Rightarrow m=0$

Therefore,the division ratio $=k:1=1:1\ and\ m=0$.

Simply Easy Learning

Updated on: 10-Oct-2022

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