Point P divides the line segment …

Point P divides the line segment joining the points A(2, 1) and B(5, -8) such that $\frac { A P } { A B } = \frac { 1 } { 3 }$. If P lies on the line 2x - y + k =0,We have to find the value of k.

$\frac { A P } { A B } = \frac { 1 } { 3 } \Rightarrow \frac { A P } { A P + P B } = \frac { 1 } { 3 }$
$\Rightarrow$ 3AP = AP + PB $\Rightarrow$ 2 AP = PB
$\Rightarrow \frac { A P } { P B } = \frac { 1 } { 2 } \Rightarrow A P : P B = 1 : 2$
So, P divides AB in the ratio 1:2
$\therefore$ coordinates of P are
$\left( \frac { 1 \times 5 + 2 \times 2 } { 1 + 2 } , \frac { 1 \times ( - 8 ) + 2 \times 1 } { 1 + 2 } \right) = \left( \frac { 9 } { 3 } , \frac { - 6 } { 3 } \right) = ( 3 , - 2 )$
Since P(3, -2) lies on 2x - y + k =0, we have
2 $\times$ 3 -(-2) + k = 0 $\Rightarrow$ 6 + 2 + k =0 $\Rightarrow$ = -8.
Hence, the required value of k is -8.