Question

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

Answer 1

$M$ is a point which divides the line segment $AB$ internally in the ratio $1:2$ because it is on $AB$ and not on the extended portion of $AB$.

To find the coodrinates of a point $(X,Y)$ which divides a line segment joining $(x_A,y_A)$ and $(x_B,y_B)$ internally in the ratio $m:n$, we use the formula:

$X=\frac{m{x}_B+n{x}_A}{m+n}$ and

$Y=\frac{m{y}_B+n{y}_A}{m+n}$

So the point $M$ can be found with the formula:

$x_M=\frac{2x_A+x_B}{3}=\frac{2\times 2-1}{3}=1$ and

$y_M=\frac{2y_A+y_B}{3}=\frac{2\times 5+2}{3}=4$

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

The equation of the line passing through points $C$ and $M$ can be obtained using the two-point form of line equation as:

$\frac{y-y_C}{x-x_C}=\frac{y_M-y_C}{x_M-x_C}$

$\frac{y-8}{x-5}=\frac{4-8}{1-5}=1$

$\Rightarrow x-y+3=0$

Thus the equation of the line passing through the points $C$ and $M$ is $x-y+3=0$