Question

$A(2,5),B(-1,2)andC(5,8)$are the vertices of a triangle $ABC$, ‘$M$’ is a point on $AB$ such that $AM:MB=1:2$. Find the coordinates of ‘$M$’. Hence, find the equation of the line passing through the points $C$ and $M$.

Answer 1

Step 1- Finding the coordinates of $M$:

Given that $M$ is a point which divides the line segment $AB$ internally in the ratio $1:2$

Using section formula , $M(x,y)=\left(\frac{m{x}_B+n{x}_A}{m+n},\frac{m{y}_B+n{y}_A}{m+n}\right)$

Here , $m=2,n=1,x_B=2,x_A=-1,y_B=2and{y}_A=5$

$$\begin{gathered} M(x,y)=\left(\frac{2x_B+x_A}{3},\frac{2y_B+y_A}{3}\right) \\ \Rightarrow M(x,y)=\left(\frac{2\times 2-1}{3},\frac{2\times 5+2}{3}\right) \\ \Rightarrow M(x,y)=\left(\frac{4-1}{3},\frac{10+2}{3}\right) \\ \Rightarrow M(x,y)=\left(\frac{3}{3},\frac{12}{3}\right) \\ \Rightarrow M(x,y)=\left(1,4\right) \end{gathered}$$

Thus , the coordinates of $M$ are $(1,4)$

Step2- Finding the equation of line $CM$:

Now, the equation of line passing through the point $C(5,8)$ and $M(1,4)$ is ,

$\Rightarrow y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$$\begin{gathered} y-8=\frac{4-8}{1-5}(x-5) \\ y-8=\frac{-4}{-4}(x-5) \\ y-8=x-5 \\ x-y+3=0 \end{gathered}$$

Therefore , the equation of line passing the points $C$ and $M$ is $x-y+3=0$.