# The line segment joining the points $A(3,2)$ and $B(5,1)$ is divided at the point $P$ in the ratio $1: 2$ and it lies on the line $3 x-18 y+k=0$. Find the value of $k$.

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Given:

The line segment joining the points $A( 3,\ 2)$ and $B( 5,\ 1)$ isdivided at the point $P$ in the ratio $1: 2$ and it lies on the line $3 x-18 y+k=0$.

To do:

We have to find the value of $k$.

Solution:

The line segment $AB$ is divided at the point $P$ in the ratio $1: 2$

This implies, $AP: PB = 1:2$

Using section formula, we have,

$( x,\ y)=( \frac{mx_{2}+nx_{1}}{m+n}, \frac{my_{2}+ny_{1}}{m+n})$

Then, coordinates of $P$ are,

$P=( \frac{1\times5+2\times3}{1+2}, \frac{1\times(1)+2\times2}{1+2})$

$\Rightarrow P=( \frac{5+6}{3}, \frac{1+4}{3})$

$\Rightarrow P=( \frac{11}{3}, \frac{5}{3})$

The point $P( \frac{11}{3}, \frac{5}{3})$ lies on the line $3x-18y+k=0$.

This implies, point $P( \frac{11}{3}, \frac{5}{3})$ satisfies the above equation.

$\Rightarrow 3(\frac{11}{3})-18(\frac{5}{3})+k=0$

$\Rightarrow 11-30+k=0$

$\Rightarrow k-19=0$

$\Rightarrow k=19$

Therefore, the value of $k$ is $19$.

Updated on 10-Oct-2022 13:28:51