 # The line segment joining the points A(3, 2)

Question:

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

3x – 18y + k = 0. Find the value of k.

Solution:

Given that, the line segment joining the points 4(3,2) and 6(5,1) is divided at the point P in the ratio 1 : 2.

$\therefore \quad$ Coordinate of point $P=\left\{\frac{5(1)+3(2)}{1+2}, \frac{1(1)+2(2)}{1+2}\right\}$

$=\left(\frac{5+6}{3}, \frac{1+4}{3}\right)=\left(\frac{11}{3}, \frac{5}{3}\right)$

$\left[\because\right.$ by section formula for internal ratio $\left.\equiv\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}\right)\right]$

But the point $P\left(\frac{11}{3}, \frac{5}{3}\right)$ lies on the line $3 x-18 y+k=0$.  [given]

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

$\Rightarrow$ $11-30+k=0$

$\Rightarrow$ $k-19=0 \Rightarrow k=19$

Hence, the required value of k is 19.