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

 

 

Answers (1)

Given, \frac{AP}{AB}=\frac{1}{3}\Rightarrow \frac{AP}{AP+PB}=\frac{1}{3}   

    \frac{AP}{PB}=\frac{1}{2}                   A=(2,1);B(5,-8);P=(m,n)

Applying section formula,

Coordinates of P(m,n)=\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 )=\left ( 3,-2 \right )

Since, P lies on 2x-y+k=0

                \Rightarrow 2\times 3-(-2)+k=0

                \Rightarrow k=-8

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Safeer PP

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