Question

The line segment joining $A(2,1)$ and $B(5,-8)$ is trisected at the points $P$ and $Q$. If $P$ is closer to point $A$ and lies on the line $2x-y+k=0$, find the value of $k$.

Answer 1

For point $P$ :
$m_1:m_2=AP:PB=1:2(x_1,y_1)=A(2,1)\text{and}(x_2,y_2)=B(5,-8)$
$\therefore \text{Point}P=(\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})=(\frac{1\times 5+2\times 2}{1+2},\frac{1\times -8+2\times 1}{1+2})=(3,-2)$$\because P(3,-2)$ lies on the line $2x-y+k=0$
$\therefore$ On substituting $x=3$ and $y=-2$ in the given equation $2x-y+k=0$; we get : $2\times 3-(-2)+k=0\Rightarrow k=-8$