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Find the ratio in which the segment joining the points $(2, -5)$ and $(5, 3)$ is divided by $x$-axis. Also, find coordinates of the point on $x$-axis.
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Solution:
Let the required ratio be $k:1$.
Since given line segment is divided by $x$-axis
therefore $\frac{3k - 5}{k + 1} = 0 \Rightarrow k = \frac{5}{3}$
Hence required ratio is $5 : 3$
$x = \frac{25 + 6}{8} = \frac{31}{8}$
The required point is $\left(\frac{31}{8}, 0\right)$
Let the required ratio be $k:1$.
Since given line segment is divided by $x$-axis
therefore $\frac{3k - 5}{k + 1} = 0 \Rightarrow k = \frac{5}{3}$
Hence required ratio is $5 : 3$
$x = \frac{25 + 6}{8} = \frac{31}{8}$
The required point is $\left(\frac{31}{8}, 0\right)$