Find the ratio in which the point (2, y)

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(2,y) divides the line joining the points A(−2,2) and B(3,7) in some ratio.

Let us substitute these values in the earlier mentioned formula.

$(2, y)=\left(\left(\frac{m(3)+n(-2)}{m+n}\right),\left(\frac{m(7)+n(2)}{m+n}\right)\right)$

Equating the individual components we have

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

$2 m+2 n=3 m-2 n$

$m=4 n$

$\frac{m}{n}=\frac{4}{1}$

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 ‘y’

$(2, y)=\left(\left(\frac{m(3)+n(-2)}{m+n}\right),\left(\frac{m(7)+n(2)}{m+n}\right)\right)$

$(2, y)=\left(\left(\frac{4(3)+1(-2)}{4+1}\right),\left(\frac{4(7)+1(2)}{4+1}\right)\right)$

Equating the individual components we have

$y=\frac{4(7)+1(2)}{4+1}$

$y=6$

Thus the value of ' $y$ ' is 6 .