Question
In what ratio does $y$-axis divide the line segment joining the points $(-4,7)$ and $(3,-7)$ ?
✓
Answer
Let the $y$-axis cut the join of $A(-4,7)$ and $B(3,-7)$ at the point $p$ in the ratio $k: 1$. Then,
By section formula,
Coordinates of p $=\Big(\frac{\text{k}\times3+1\times(-4)}{\text{k}+1},\frac{\text{k}\times(-7)+1\times7}{\text{k}+1}\Big)$
$=\Big(\frac{3\text{k}-4}{\text{k}+1},\frac{-7\text{k}+7}{\text{k}+1}\Big)$
But p lies on y-axis. So, its abscissa is 0.
$\therefore\ \frac{3\text{k}-4}{\text{k}+1}=0$
$\Rightarrow\ 3\text{k}-4=0$
$\Rightarrow\ 3\text{k}=4$
$\Rightarrow\ \text{k}=\frac{4}{3}$
So, the required ratio is$ 4 : 3.$
By section formula,
Coordinates of p $=\Big(\frac{\text{k}\times3+1\times(-4)}{\text{k}+1},\frac{\text{k}\times(-7)+1\times7}{\text{k}+1}\Big)$
$=\Big(\frac{3\text{k}-4}{\text{k}+1},\frac{-7\text{k}+7}{\text{k}+1}\Big)$
But p lies on y-axis. So, its abscissa is 0.
$\therefore\ \frac{3\text{k}-4}{\text{k}+1}=0$
$\Rightarrow\ 3\text{k}-4=0$
$\Rightarrow\ 3\text{k}=4$
$\Rightarrow\ \text{k}=\frac{4}{3}$
So, the required ratio is$ 4 : 3.$
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