Determine the ratio in which the point P (m, 6)

Solution:

The co-ordinates of a point which divided two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ internally in the ratio $m: n$ is given by the formula,

$(x, y)=\left(\left(\frac{m x_{2}+n x_{1}}{m+n}\right),\left(\frac{m y_{2}+n y_{1}}{m+n}\right)\right)$

Here we are given that the point P(m,6) divides the line joining the points A(−4,3) and B(2,8) in some ratio.

Let us substitute these values in the earlier mentioned formula.

$(m, 6)=\left(\left(\frac{m(2)+n(-4)}{m+n}\right),\left(\frac{m(8)+n(3)}{m+n}\right)\right)$

Equating the individual components we have

$6=\left(\frac{m(8)+n(3)}{m+n}\right)$

$6 m+6 n=8 m+3 n$

$2 m=3 n$

$\frac{m}{n}=\frac{3}{2}$

We see that the ratio in which the given point divides the line segment is.

Let us now use this ratio to find out the value of ‘m’.

$(m, 6)=\left(\left(\frac{m(2)+n(-4)}{m+n}\right),\left(\frac{m(8)+n(3)}{m+n}\right)\right)$

$(m, 6)=\left(\left(\frac{3(2)+2(-4)}{3+2}\right),\left(\frac{3(8)+2(3)}{3+2}\right)\right)$

Equating the individual components we have

$m=\frac{3(2)+2(-4)}{3+2}$

$m=-\frac{2}{5}$

Thus the value of ' $m$ ' is $-\frac{2}{5}$.