A man starts from the point P (-3,4) and will reach the point Q (0,1) touching the line 2x+y=7 at - Maths - Straight Lines

Question

A man starts from the point P (-3,4) and will reach the point Q
(0,1) touching the line 2x+y=7 at R Find R on the line so that he
will travel in the shortest distance (ONURGENTBASIS)

Shubham Chourasiya · Accepted Answer

$P (- 3, 4) a n d Q (0, 1) L e t t h e p o int R b e (x, y) N o w P R + R Q i s \min i m u m B y d i s \tan c e f o r m u l a \sqrt{(x - (- 3))^{2} + {(y - 4)}^{2}} + \sqrt{(x - 0)^{2} + {(y - 1)}^{2}} i s \min i m u n$ The point is lying on 2x+y=7 , so y=7-2x Replacing y by x we get , $\sqrt{(x + 3)^{2} + (7 - 2 x - 4)^{2}} + \sqrt{x^{2} + (7 - 2 x - 1)^{2}} \Rightarrow \sqrt{(x + 3)^{2} + (- 2 x + 3)^{2}} + \sqrt{x^{2} + (- 2 x + 6)^{2}} \Rightarrow \sqrt{x^{2} + 9 + 6 x + 4 x^{2} + 9 - 12 x} + \sqrt{x^{2} + {4 x}^{2} - 24 x^{} + 36} L \Rightarrow \sqrt{{5 x}^{2} + 18 - 6 x} + \sqrt{{5 x}^{2} - 24 x^{} + 36}$ Now Differentiating it w r t to x ,we get $L' = \frac{10 x - 6}{2 \sqrt{5 x^{2} - 6 x + 18}} + \frac{10 x - 24}{2 \sqrt{5 x^{2} - 24 x + 36}} = 0 (e q u a t i n g t o 0 t o f i n d t h e c r i t i c a l p o int) \Rightarrow \frac{10 x - 6}{2 \sqrt{5 x^{2} - 6 x + 18}} = - \frac{10 x - 24}{2 \sqrt{5 x^{2} - 24 x + 36}} \Rightarrow \frac{5 x - 3}{\sqrt{5 x^{2} - 6 x + 18}} = \frac{12 - 5 x}{\sqrt{5 x^{2} - 24 x + 36}} S q u a r i n g b o t h s i d e, w e g e t \frac{(5 x - 3)^{2}}{5 x^{2} - 6 x + 18} = \frac{(12 - 5 x)^{2}}{5 x^{2} - 24 x + 36} \Rightarrow \frac{(5 x - 3)^{2}}{5 x^{2} - 6 x + 18} = \frac{(12 - 5 x)^{2}}{5 x^{2} - 24 x + 36} \Rightarrow \frac{25 x^{2} - 30 x + 9}{5 x^{2} - 6 x + 18} = \frac{25 x^{2} - 120 x + 144}{5 x^{2} - 24 x + 36} \Rightarrow 5 - \frac{81}{5 x^{2} - 6 x + 18} = 5 - \frac{36}{5 x^{2} - 24 x + 36} \Rightarrow \frac{81}{5 x^{2} - 6 x + 18} = \frac{36}{5 x^{2} - 24 x + 36} \Rightarrow \frac{9}{5 x^{2} - 6 x + 18} = \frac{4}{5 x^{2} - 24 x + 36} o n s o l v i n g w e g e t, 25 x^{2} - 192 x + 252 = 0 \Rightarrow x = 6 a n d 84 / 50$ L'' >0 for 84/50 So the value is minimum for x=84/50 and y= 91/25

A man starts from the point P (-3,4) and will reach the point Q (0,1) touching the line 2x+y=7 at R. Find R on the line so that he will travel in the shortest distance.(ONURGENTBASIS)