Question

Find the ratio in which the point $P$ whose abscissa is $3$ divides the line joining $A(6,5)$ and $B(-1,4)$ and hence find the coordinates of $P$.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then

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

& for mid points we have, $m:n=1:1$.

Let the ratio in which $P$ divides $AB$ be $r:1$

Given Abscissa of $P$ is $3$

$A(6,5),B(-1,4)$

Co-ordinates of $P$ is $(\frac{-r+6}{r+1},\frac{4r+5}{r+1})$

$⟹\frac{-r+6}{r+1}=3⟹-r+6=3r+3$

$⟹4r=3⟹r=\frac{3}{4}$

So the ratio is $3:4$

Coordinates of $P$ is $(\frac{4\left(6\right)+3(-1)}{4+3},\frac{4\left(5\right)+3\left(4\right)}{4+3})=(3,\frac{32}{7})$