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In what ratio does the point (\mathrm{P}(a, –1)) divide the join of (\mathrm{A}(1, –3)) and (\mathrm{B}(6, 2)). Hence, find the value of a.

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Question

In what ratio does the point \(\mathrm{P}(a, -1)\) divide the join of \(\mathrm{A}(1, -3)\) and \(\mathrm{B}(6, 2)\). Hence, find the value of a.

Sum
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Solution

Let P divide AB in the ratio $$k : 1$$.

Using the y-coordinate of P which is $$-1$$:

$$\frac{k \times 2 + 1 \times (-3)}{k + 1} = -1$$

$$\Rightarrow 2k - 3 = -k - 1$$

$$\Rightarrow 3k = 2$$

$$\Rightarrow k = \frac{2}{3}$$

Hence the required ratio is $$\dfrac{2}{3} : 1 = 2 : 3$$.

Now the x-coordinate gives the value of $$a$$:

$$a = \frac{2 \times 6 + 3 \times 1}{2 + 3}$$

$$= \frac{12 + 3}{5}$$

$$= \frac{15}{5}$$

$$= 3$$

Hence $$a = 3$$.

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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 12. | Page 210
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