Point P divides the line segment …

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

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