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प्रश्न
In what ratio is the segment joining the points \(\mathrm{A}(6,5)\) and \(\mathrm{B}(-3, 2)\) divided by the y-axis? Find the point at which the y-axis cuts AB.
योग
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उत्तर
Let the y-axis cut AB in the ratio $$m : 1$$ at the point $$(0, y)$$.
Since the x-coordinate is 0,
$$\frac{m \times (-3) + 1 \times 6}{m + 1} = 0$$
$$\Rightarrow -3m + 6 = 0$$
$$\Rightarrow m = 2$$
Hence the y-axis divides AB in the ratio $$2 : 1$$.
The y-coordinate of the point of cut is $$y = \frac{2 \times 2 + 1 \times 5}{2 + 1}$$
$$= \frac{4 + 5}{3}$$
$$= \frac{9}{3}$$
$$= 3$$
Hence the y-axis cuts AB at $$(0, 3)$$.
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