Question

Find the ratio in which the line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$.

Answer 1

Step 1: Defining the problem

Let point $(-3,10)$ be denoted as $A$, point as $(6,-8)$ be denoted as $B$ and point $(-1,6)$ be denoted as $C$.

Section formula gives us the ratio at which a point divides a line segment. If a point $C$ divides a line segment in the ratio $\mathrm{m}:\mathrm{n}$, then,

$C(x,y)=\left(\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}},\frac{{\mathrm{my}}_2+{\mathrm{ny}}_1}{\mathrm{m}+\mathrm{n}}\right)$

where, $x_1$ and $x_2$ are the $x$-coordinates and $y_2$ and $y_1$ are the $y$-coordinates of the vertices of the line segment.

Given, $x=-1$, $y=6$, $x_1=-3$, $y_1=10$, $x_2=6$ and $y_2=-8$

Step 2: Compute ratio of $\mathrm{m}$ and $\mathrm{n}$

$$\begin{gathered} x=\frac{{\mathrm{mx}}_2+{\mathrm{nx}}_1}{\mathrm{m}+\mathrm{n}} \\ \Rightarrow -1=\frac{\mathrm{m}·6+\mathrm{n}·(-3)}{\mathrm{m}+\mathrm{n}} \\ \Rightarrow -\mathrm{n}-\mathrm{m}=6\mathrm{m}-3\mathrm{n} \\ \Rightarrow 7\mathrm{m}=2\mathrm{n} \\ \Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=\frac{2}{7} \\ \Rightarrow \mathrm{m}:\mathrm{n}=2:7 \end{gathered}$$

Therefore, the ratio in which the line segment joining the points $(-3,10)$ and $(6,-8)$ is divided by $(-1,6)$ is $2:7$.