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
Answers
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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