Question

The coordinates of A and B are $(1,2)$ and $(2,3)$ respectively. Find the coordinates of R on the line segment AB so that $\frac{AR}{RB}=\frac{4}{3}$.

• A. $\left(\begin{array}{c}-\frac{11}{7},-\frac{18}{7}\end{array}\right)$
• B. $\left(\begin{array}{c}\frac{11}{7},-\frac{18}{7}\end{array}\right)$
• C. $\left(\begin{array}{c}\frac{11}{7},\frac{18}{7}\end{array}\right)$
• D. $\left(\begin{array}{c}-\frac{11}{7},\frac{18}{7}\end{array}\right)$

Answer 1

The correct option is C $\left(\begin{array}{c}\frac{11}{7},\frac{18}{7}\end{array}\right)$

Using the section formula, if a point $(x,y)$ divides the
line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{my2+n{y}_1}{m+n})$
Substituting $(x_1,y_1)=(1,2)$ and $(x_2,y_2)=(2,3)$ and $m=4,n=3$ in the section formula, we get
$R=(\frac{4\left(2\right)+3\left(1\right)}{4+3},\frac{4\left(3\right)+3\left(2\right)}{4+3})=(\frac{11}{7},\frac{18}{7})$