Question

Point P divides the line segment joining the point A(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.

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

We have,
$\frac{AP}{AB}=\frac{1}{3}$$\Rightarrow \frac{AP}{AP+PB}=\frac{1}{3}$$\Rightarrow 3AP=AP+BP$
$\Rightarrow 2AP=BP$
$\Rightarrow \frac{AP}{BP}=\frac{1}{2}$
So, P divides AB in the ratio 1:2.

$\therefore CoordinatesofPare(\frac{1\times 5+2\times 2}{1+2},\frac{1\times -8+2\times 1}{1+2})=(3,2)$

$Since,P(3,2)liesontheline2x-y+k=0.$

$\therefore 2\times 3-2+k=0\Rightarrow k=-4$

TYPE III ON DETERMINATION OF THE TYPE OF A GIVEN QUADRILATERAL