find the points of trisection of the line joining A (2,-3) B (4, 5)
Answers
Answer:
Step-by-step explanation:
Hence P divides AB internally in the ratio 1 : 2 and Q divides AB internally in the ratio 2 : 1
By the section formula, the required points are
AP = 1
PQ = 1
QB = 1
Section formula internally = (lx₂ + mx₁)/(l + m) , (ly₂ + my₁)/(l + m)
P divides the line segment in the ratio 1:2
l = 1 m = 2
A(4,-1) and B(-2,-3)
= [(1(-2) + 2(4)]/(1+2) , [(1(-3) + 2(-1)]/(1+2)
= (-2+8)/3 , (-3-2)/3
= 6/3 , -5/3
= P (2 , -5/3)
Q divides the line segment in the ratio 2:1
l = 2 m = 1
= [(2(-2) + 1(4)]/(2+1) , [(2(-3) + 1(-1)]/(2+1)
= (-4+4)/3 , (-6-1)/3
= 0/3 , -7/3
= Q (0 , -7/3)
Example 2 :
Find the points of trisection of the line segment joining the points A (2 , -2) and B (-7 , 4).
Solution :
Let P and Q are the points of the trisection of the line segment joining the points A and B
Here AP = PQ = QB
Answer:
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