Question

Find the coordinates of the point which divides the line segment joining the points A(4,-3) and B(9,7) in the ratio 3:2.

Answer 1

We know that by section formula, the co-ordinates of the points which divide internally the line segment joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$ is

$(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

The end points of AB are A(4,-3) and B(9,7).

Therefore,

$(x_1=4,y_1=-3)$ and $(x_2=9,y_2=7)$

Also, $m=3$ and $n=2$

Let the required point be P(x,y).

Using section formula,

$x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n}$

$x=\frac{3\times 9+2\times 4}{5}=7,y=\frac{3\times 7+\times -3}{5}=3$

Therefore, the required point is $P(7,3)$.