Question

Find the abscissa of the point which divides the line segment joining the points $(6,3)$ and $(-4,5)$ in the ratio $3:2$ internally.

Answer 1

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ internally in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$
Substituting $(x_1,y_1)\equiv (6,3)$ and $(x_2,y_2)\equiv (-4,5)$ and $m=3,n=2$ in the section formula, we get the point as,

$(x,y)=(\frac{3(-4)+2\left(6\right)}{3+2},\frac{3\left(5\right)+2\left(3\right)}{3+2})=(\frac{12-12}{3+2},\frac{15+6}{2+3})=(0,\frac{21}{5})$

The required abscissa is $0$.