Question

Question 15
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.

Answer 1

Given that the line segment joining the points A(3,2) and B(5,1) is divided at the point P in the ratio 1:2.
$\therefore \text{Coordinate of point P}\equiv \{\frac{5\left(1\right)+3\left(2\right)}{1+2},\frac{1\left(1\right)+2\left(2\right)}{1+2}\}\equiv (\frac{5+6}{3},\frac{1+4}{3})\equiv (\frac{11}{3},\frac{5}{3})[\because \text{By section formula, for internal ratio}=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})]\text{But the point}P(\frac{11}{3},\frac{5}{3})\text{lies on the line 3x-18y+k=0}.\therefore 3\left(\frac{11}{3}\right)-18\left\{\frac{5}{3}\right\}+k=0\Rightarrow 11-30+k=0\Rightarrow k-19=0\Rightarrow k=19\text{Hence, the required value of k is 19.}$