Question

In what ratio does the point $\left(-4,6\right)$ divide the line segment joining the points $A\left(-6,10\right)$ and $\mathrm{B}\left(3,-8\right)$.

Answer 1

Compute the required ratio.

In the question, it is given that the point $\left(-4,6\right)$ divides the line segment joining the points $A\left(-6,10\right)$ and $\mathrm{B}\left(3,-8\right)$.

We know that if a point $(x,y)$ divides the line segment joining the points $(x_1,y_1)$ and $({\mathrm{x}}_2,{\mathrm{y}}_2)$ in $m:n$. Then $\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right)$

Let the ratio be $k:1$.

Here $\mathrm{m}=\mathrm{k},\mathrm{n}=1$

$$\begin{gathered} \Rightarrow -4=\frac{3\mathrm{k}-6}{\mathrm{k}+1} \\ \Rightarrow -4\mathrm{k}-4=3\mathrm{k}-6 \\ \Rightarrow \mathrm{k}=\frac{2}{7} \end{gathered}$$

and,$6=\frac{-8\mathrm{k}+10}{\mathrm{k}+1}$

$$\begin{gathered} \Rightarrow 6\mathrm{k}+6=-8\mathrm{k}+10 \\ \Rightarrow 14\mathrm{k}=4 \\ \Rightarrow \mathrm{k}=\frac{2}{7} \end{gathered}$$

Since from both equations values of $\mathrm{k}$ is coming same so $\mathrm{k}=\frac{2}{7}$.

Therefore, the required ratio is $2:7$.

Hence $\left(-4,6\right)$ divides the line segment in ratio $2:7$.