Question

In what ratio is the line segment AB joining A $(6,3)$ and B$(-4,5)$ is divided by the $y$-axis?

• A. $2:3$
• B. $3:2$
• C. $4:3$
• D. $3:4$

Answer 1

The correct option is B $3:2$

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

Substituting $(x_1,y_1)=(6,3)$ and $(x_2,y_2)=(-4,5)$ in the section formula, we get the point $(\frac{m(-4)+n\left(6\right)}{m+n},\frac{m\left(5\right)+n\left(3\right)}{m+n})=(\frac{-4m+6n}{m+n},\frac{5m+3n}{m+n})$

As the point lies on $y$ - axis, $x$ -coordinate $=0$.
$\frac{-4m+6n}{m+n}=0$

$4m=6n$

or

$m:n=3:2$