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(A) Let the ratio in which the $y$-axis divides the line segment be $k:1$.By the section formula,the coordinates of the point dividing the segment joi

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Ratio $5:1$,Point $(0, -13/3)$

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(A) Let the ratio in which the $y$-axis divides the line segment be $k:1$.By the section formula,the coordinates of the point dividing the segment joining $(x_1, y_1) = (5, -6)$ and $(x_2, y_2) = (-1, -4)$ in the ratio $k:1$ are given by:$\left( \frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1} \right) = \left( \frac{k(-1) + 5}{k+1}, \frac{k(-4) + (-6)}{k+1} \right) = \left( \frac{-k+5}{k+1}, \frac{-4k-6}{k+1} \right)$.Since this point lies on the $y$-axis,its $x$-coordinate (abscissa) must be $0$.Therefore,$\frac{-k+5}{k+1} = 0$,which implies $-k+5 = 0$,so $k = 5$.Thus,the ratio is $5:1$.Now,substitute $k=5$ into the $y$-coordinate expression:$y = \frac{-4(5) - 6}{5+1} = \frac{-20-6}{6} = \frac{-26}{6} = -\frac{13}{3}$.Hence,the point of intersection is $(0, -13/3)$.

Find the ratio in which the $y$-axis divides the line segment joining the points $(5, -6)$ and $(-1, -4)$. Also,find the point of intersection.

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