Question

Find the ratio in which the point $(-1,k)$ divides the line joining the points $(-3,10)$ and $(6,-8)$, and find the value of k.

Answer 1

We know that the section formula states that if a point $P(x,y)$ lies on line segment $AB$ joining the points $A(x_1,y_1)$ and $B(x_2,y_2)$and satisfies $AP:PB=m:n$, then we say that $P$ divides internally $AB$ in the ratio $m:n$. The coordinates of the point of division has the coordinates

$P=(\frac{{mx}_2{+nx}_1}{m+n},\frac{{my}_2{+ny}_1}{m+n})$

Let $P(-1,k)$ divides the line segment $AB$ joining the points $A(-3,10)$ and $B(6,-8)$ in the ratio $m:n$, then using section formula we get,

$P=(\frac{{mx}_2{+nx}_1}{m+n},\frac{{my}_2{+ny}_1}{m+n})\Rightarrow (-1,k)=(\frac{(m\times 6)+(n\times -3)}{m+n},\frac{(m\times -8)+(n\times 10)}{m+n})\Rightarrow (-1,k)=(\frac{6m-3n}{m+n},\frac{-8m+10n}{m+n})\Rightarrow -1=\frac{6m-3n}{m+n}$
$\Rightarrow -1(m+n)=6m-3n\Rightarrow -m-n=6m-3n\Rightarrow 6m+m=3n-n\Rightarrow 7m=2n\Rightarrow \frac{m}{n}=\frac{2}{7}\Rightarrow m:n=2:7$

Hence, the point $(-1,k)$ divides the line segment joining the points $(-3,10)$ and $(6,-8)$ in the ratio $2:7$.