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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.

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Question

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.

Sum
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Solution

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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Chapter 13: Section and Mid-point Formulae - EXERCISE 13 [Page 210]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 13 Section and Mid-point Formulae
EXERCISE 13 | Q 14. | Page 210
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