Question

If P, Q and R are three points with co-ordinates (1, 4), (2, 6) and (m, 2m –1) respectively, then the value of m for which PR + RQ is minimum is: (A) 17/8 (B) 5/2 (C) 31/14 (D) 27/10

Answer 1

Given : P, Q and R are three points with co-ordinates (1, 4), (2, 6) and (m, 2m –1) respectively

To Find : the value of m for which PR + RQ is minimum  

(A) 17/8 (B) 5/2 (C) 31/14 (D) 27/10

Solution:

P, Q and R are three points with co-ordinates (1, 4), (2, 6) and (m, 2m –1)

PR = √(m - 1)² + (2m - 1-4)²  =  √(m - 1)² + (2m - 5)²

=  √m² -2m + 1  +4m² -20m +25

= √5m² -22m + 26

QR = √(m - 2)² + (2m - 1-6)²  =  √(m - 2)² + (2m - 7)²

=  √m² -4m + 4  +4m² -28m +49

= √5m² -32m + 53

PR + RQ = √5m² -22m + 26 + √5m² -32m + 53

Z = √5m² -22m + 26 + √5m² -32m + 53

dZ/dm =  (10m - 22)/2√5m² -22m + 26  +  (10m - 32)/2√5m² -32m + 53

=> dZ/dm = (5m - 11)/ √5m² -22m + 26  +  (5m - 16)/ √5m² -32m + 53

dZ/dm = 0

=>  (5m - 11)/ √5m² -22m + 26  +  (5m - 16)/ √5m² -32m + 53 = 0

=>  (5m - 11) √5m² -32m + 53  +   (5m - 16) √5m² -22m + 26  = 0

=> (5m - 11) √5m² -32m + 53 = - (5m - 16) √5m² -22m + 26  

Squaring both sides

=>  (5m - 11)²(5m² -32m + 53) = (5m - 16)² (5m² -22m + 26 )

on solving

m = 27/10

m = 27/10 will give minimum

Another way :

m , 2m - 1 Hence line y = 2x - 1

P & Q are both on same side of 2x - 1

Hence R will be on perpendicular bisector of PQ

Slope of PQ = (6 - 4 )/(2 - 1) = 2

Slope of perpendicular bisector = -1/2

mid point of of PQ =  1.5 , 5

Slope of  perpendicular bisector   = (2m - 1 - 5)/(m - 1.5)  = -1/2

=> 4m - 12 = -m + 1.5

=> 5m = 13.5

=> m = 13.5/5

=> m = 27/10

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