Question

Find the ratio in which Y-axis divides the line segment joining the points $(5,-6)$ and $(-1,-4)$.

Hint:

When finding ratios of division by an axis, always apply the section formula with $x=0$ (for Y-axis) or $y=0$ (for X-axis).

Answer 1

Let the points be $A(5,-6)$ and $B(-1,-4)$. Suppose the Y-axis divides $AB$ at point $P(0,y)$ in the ratio $m:n$. Using the section formula for the $x$-coordinate: \[ x_P = \frac{mx_2 + nx_1}{m+n} \] Here, $x_P = 0$, $x_1 = 5$, $x_2 = -1$. \[ 0 = \frac{m(-1) + n(5)}{m+n} \] \[ 0 = \frac{-m + 5n}{m+n} \] \[ -m + 5n = 0 $\Rightarrow$ m = 5n \] Thus, the ratio is $\; m:n = 5:1$. Check for $y$-coordinate: \[ y_P = \frac{m y_2 + n y_1}{m+n} = \frac{5(-4) + 1(-6)}{6} = \frac{-20 - 6}{6} = -\frac{26}{6} = -\frac{13}{3} \] Since $-\frac{13}{3}$ lies between $-6$ and $-4$, the point lies on the segment. \[ \boxed{\text{The Y-axis divides the line segment in the ratio } 5:1} \]