if M is a point on the line y=x and points P(0,1),Q(2,0) are such that PM+PQ is minimum then find P


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Question Number 148993 by gsk2684 last updated on 02/Aug/21

$i f M i s a p o i n t o n t h e l i n e y = x a n d$ $p o i n t s P (0, 1), Q (2, 0) a r e s u c h t h a t$ $P M + P Q i s m i n i m u m t h e n f i n d P$
Commented by mr W last updated on 02/Aug/21

$i t h i n k t h e q u e s t i o n s h o u l d b e$ $f i n d M s u c h t h a t P M + Q M i s$ $m i n i m u m .$
Commented by gsk2684 last updated on 02/Aug/21

$y e s$ $t y p o g r a p h i c a l e r r o r$ $s o l u t i o n p l e a s e$
Commented by bramlexs22 last updated on 02/Aug/21

$if PM + MQ minimum \Rightarrow M ?$ $let f (x) = \sqrt{x^{2} + {(x - 1)}^{2}} + \sqrt{{(x - 2)}^{2} + x^{2}}$ $f {(x)}_{\min} if points M, P and Q colinear$ $\Rightarrow so we get m_{PQ} = m_{MQ}$ $\Rightarrow \frac{1 - 0}{0 - 2} = \frac{x - 0}{x - 2}$ $\Rightarrow - \frac{1}{2} = \frac{x}{x - 2}$ $\Rightarrow - x + 2 = 2 x; x = \frac{2}{3}$ $thus M (\frac{2}{3}, \frac{2}{3})$
Commented by bramlexs22 last updated on 02/Aug/21

Commented by gsk2684 last updated on 04/Aug/21

$i h a v e f o u n d t h a t f e e t o f p e r p e n d i c u l a r s$ $f r o m g i v e n p o i n t s o n t o t h e g i v e n$ $l i n e$ $a n d u s i n g t h e s e r i g h t a n g l e d t r i a n g l e s$ $c a n i g o f u r t h e r t o f i n d M s u c h t h a t$ $P M + Q M t o b e m i n i m u m$ $h e l p m e$
	Answered by iloveisrael last updated on 02/Aug/21

	$M (x, x) \Rightarrow PM + PQ = \sqrt{x^{2} + {(x - 1)}^{2}} + \sqrt{2^{2} + {(- 1)}^{2}}$ $= \sqrt{5} + \sqrt{{2 x}^{2} - 2 x + 1}$ $minimum when 4 x - 2 = 0; x = \frac{1}{2}$ $then M (\frac{1}{2}, \frac{1}{2})$ $why P ?$
	Commented by gsk2684 last updated on 02/Aug/21

	$t h a n k y o u$
	Answered by Kamel last updated on 02/Aug/21

	$i f M i s a p o i n t o n t h e l i n e y = x a n d$ $p o i n t s P (0, 1), Q (2, 0) a r e s u c h t h a t$ $P M + P Q i s m i n i m u m t h e n f i n d P$ $P M + Q M m i n i m u m \Rightarrow M \in (P Q) / M (x, x)$ $(P Q) : y = a x + 1 \Rightarrow (P Q) : y = - \frac{1}{2} x + 1$ $M \in (P Q) \Leftrightarrow x = y = - \frac{1}{2} x + 1 \Rightarrow x = y = \frac{2}{3}$
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