हिंदी

The line segment joining the points (\mathrm{A}(4, –3)) and (\mathrm{B}(4, 2)) is divided by the point P such that (\mathrm{AP}: \mathrm{AB} = 2 : 5). Find the co-ordinates of P.

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प्रश्न

The line segment joining the points \(\mathrm{A}(4, -3)\) and \(\mathrm{B}(4, 2)\) is divided by the point P such that \(\mathrm{AP}: \mathrm{AB}=2:5\). Find the co-ordinates of P.

\[[\textbf{Hint: } \frac{AP}{AB} = \frac{2}{5} \Rightarrow \frac{AP}{AP + PB} = \frac{2}{5} \Rightarrow 3AP = 2PB \Rightarrow \frac{AP}{PB} = \frac{2}{3} \Rightarrow AP : PB = 2 : 3\text{.]}\]

योग
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उत्तर

Since $$AP : AB = 2 : 5$$, we get $$AP : PB = 2 : 3$$ (as shown in the hint). So $$m : n = 2 : 3$$.

$$x=\frac{2 \times 4 + 3 \times 4}{2 + 3}$$

$$=\frac{8 + 12}{5}$$

$$=\frac{20}{5}$$

$$= 4$$

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

$$=\frac{4 - 9}{5}$$

$$=\frac{-5}{5}$$

$$= -1$$

Hence the co-ordinates of P are $$P(4, -1)$$.

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अध्याय 13: Section and Mid-point Formulae - EXERCISE 13 [पृष्ठ २०९]

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आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 13 Section and Mid-point Formulae
EXERCISE 13 | Q 1. (ii) | पृष्ठ २०९
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