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Question Number 209307 by efronzo1 last updated on 06/Jul/24

Commented by mr W last updated on 06/Jul/24

$∣ Q R ∣< \frac{8}{3} - 1 = \frac{5}{3}$ $\frac{Q R}{P R} = \frac{2}{1} \Rightarrow \frac{P Q}{Q R} = \frac{\sqrt{5}}{2}$ $∣ P Q ∣= \frac{\sqrt{5}}{2} \times Q R < \frac{\sqrt{5}}{2} \times \frac{5}{3} = \frac{5 \sqrt{5}}{6} < \sqrt{5}$ $\Rightarrow i m p o s s i b l e t h a t ∣ P Q ∣= \sqrt{5}!$ $\Rightarrow q u e s t i o n i s w r o n g!$
Commented by mr W last updated on 06/Jul/24

Commented by efronzo1 last updated on 07/Jul/24


	Answered by A5T last updated on 07/Jul/24

	$F o r t h e c o r r e c t e d q u e s t i o n : y = {(\frac{2}{3})}^{x + 1} + \frac{8}{3}$ $P (p_{1}, p_{2}), Q (q_{1}, q_{2})$ $\sqrt{{(q_{1} - p_{1})}^{2} + {(q_{2} - p_{2})}^{2}} = \sqrt{5}$ $y - 2 x = k \Rightarrow p_{2} - 2 p_{1} = q_{2} - 2 q_{1}$ $\Rightarrow 2 (p_{1} - q_{1}) = p_{2} - q_{2} . . . (i)$ $\Rightarrow \sqrt{{(p_{1} - q_{1})}^{2} + 4 {(p_{1} - q_{1})}^{2}} = \sqrt{5}$ $\Rightarrow 5 {(p_{1} - q_{1})}^{2} = 5 \Rightarrow (p_{1} - q_{1}) = - 1$ $p_{2} = \frac{2^{p_{1} + 3}}{3^{p_{1} + 3}} + 1; q_{2} = \frac{2^{q_{1} + 1}}{3^{q_{1} + 1}} + \frac{8}{3}$ $\Rightarrow p_{2} = \frac{2^{q_{1}} \times 4}{3^{q_{1}} \times 9} + 1; q_{2} = \frac{2^{q_{1}} \times 2}{3^{q_{1}} \times 3} + \frac{8}{3}$ $p_{2} = \frac{2 q_{2}}{3} - \frac{7}{9} \Rightarrow p_{2} - q_{2} = \frac{- q_{2}}{3} - \frac{7}{9}$ $\Rightarrow - 2 + \frac{7}{9} = \frac{- 11}{9} = \frac{- q_{2}}{3} \Rightarrow q_{2} = \frac{11}{3} \Rightarrow p_{2} = \frac{15}{9}$ $\Rightarrow q_{1} + 1 = 0 \Rightarrow q_{1} = - 1 \Rightarrow p_{1} = - 2$ $\Rightarrow k = p_{2} - 2 p_{1} = \frac{15}{9} + 4 = \frac{17}{3}$
	Answered by mr W last updated on 07/Jul/24

	$y_{1} = {(\frac{2}{3})}^{x_{1} + 3} + 1 = 2 x_{1} + k . . . (i)$ $y_{2} = {(\frac{2}{3})}^{x_{1} + 1} + \frac{8}{3} = 2 x_{2} + k . . . (i i)$ $P Q = \sqrt{5} \Rightarrow x_{2} - x_{1} = 1$ ${(\frac{2}{3})}^{x_{2} + 1} + \frac{8}{3} - {(\frac{2}{3})}^{x_{1} + 3} - 1 = 2 (x_{2} - x_{1})$ $[{(\frac{2}{3})}^{x_{2} - x_{1} - 2} - 1] {(\frac{2}{3})}^{x_{1} + 3} = \frac{1}{3}$ $[{(\frac{2}{3})}^{- 1} - 1] {(\frac{2}{3})}^{x_{1} + 3} = \frac{1}{3}$ ${(\frac{2}{3})}^{x_{1} + 3} = \frac{2}{3}$ $\Rightarrow x_{1} + 3 = 1 \Rightarrow x_{1} = - 2$ ${(\frac{2}{3})}^{- 2 + 3} + 1 = 2 (- 2) + k \Rightarrow k = \frac{17}{3} ✓$
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