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Question
The line segment joining (\mathrm{A}(2, 3)) and (\mathrm{B}(6, -5)) is intercepted by the x-axis at the point k. Find the ratio in which k divides AB. Also, write the co-ordinates of the point k.
Sum
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Solution
Let the x-axis cut AB at $$k$$ in the ratio $$m : 1$$.
Since $$k$$ lies on the x-axis, its y-coordinate is 0.
$$\frac{m \times (-5) + 1 \times 3}{m + 1} = 0$$
$$\Rightarrow -5m + 3 = 0$$
$$\Rightarrow m = \frac{3}{5}$$
Hence the x-axis divides AB in the ratio $$\dfrac{3}{5} : 1 = 3 : 5$$.
The x-coordinate of $$k$$ is $$x = \frac{\frac{3}{5} \times 6 + 1 \times 2}{\frac{3}{5} + 1}$$
$$= \frac{\frac{18}{5} + 2}{\frac{8}{5}}$$
$$= \frac{\frac{28}{5}}{\frac{8}{5}}$$
$$= \frac{28}{8}$$
$$= \frac{7}{2}$$
Hence $$k\left(\frac{7}{2}, 0\right)$$.
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