1) Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A.Find the coordinates of P and Q
Answers
Given : P and Q are the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A.
To Find : coordinates of P and Q
Solution:
A ( 2, -2)
B ( -7 , 4)
P and Q points of trisection
=> AP = PQ = QB = AB/3
=> AP : PB = 1 : 2 (∵ PB = PQ + QB)
=> AQ : QB = 2 : 1 (∵ AQ = AP + PQ)
AP : PB = 1 : 2
=> P divided AB in 1 : 2 ratio
A ( 2, -2) , B ( -7 , 4)
=> P = ( 1*(-7) + 2(2)) /(1 + 2) , ( 1*4 + 2(-2))/(1 + 2)
=> P = (-7 + 4)/3 , (4-4)/3
=> P = ( -3/3 , 0/3)
=> P = - 1, 0
AQ : QB = 2 : 1
=> Q divided AB in 2 : 1 ratio
A ( 2, -2) , B ( -7 , 4)
=> Q = (2*(-7) + 1(2)) /(2 + 1) , ( 2*4 +1(-2))/(2+ 1)
=> Q = (-14 + 2)/3 , (8-2)/3
=> Q = ( -12/3 , 6/3)
=> Q = - 4 , 2
Hence P =(-1, 0) , Q = (-4 , 2)
A ( 2, -2) P =(-1, 0) , Q = (-4 , 2) , B ( -7 , 4)
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