Please answer the question in the attachment
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Answers
Step-by-step explanation:
Given that P and Q cut the line in three equal sections.
So, P divides AB internally in the ratio 1:2 and Q divides AB internally in 2:1.
(i) Coordinate of P:
AP : PB = 1 : 2
Given that,
(x₁,y₁) = (-2,0),(x₂,y₂) = (0,8),m = 1, n = 2.
By section formula, we have
P = (mx₂ + nx₁/m + n, my₂ + ny₁/m + n)
= (-4/3, 8/3)
(ii) Coordinate of Q:
AQ : QB = 2 : 1
Given that,
(x₁,y₁) = (-2,0) and (x₂,y₂) = (0,8). m = 2, n = 1
Q = (mx₂ + nx₁/m + n, my₂ + ny₁/m + n)
= (-2/3, 16/3).
Therefore:
Coordinate of P = (-4/3, 8/3)
Coordinate of Q = (-2/3,16/3)
Hope it helps!
AP/PB = AP/2AP (P-Q-B)
= 1/2
m : n = 1 : 2
BY SECTION FORMULA FOR INTERNAL DIVISION
x = mx2 + nx1 /m + n , y = my2 +ny1 / m + n
= 1(0) + 2(-2) / 1 +2 , = 1(8) + 2(0) / 1 + 2
= -4/3 , = 8/3
P = (-4/3,8/3)
Q is the midpoint of PB
BY midpoint formula
Q = {x1 + x2 /2 , y1 + y2 / 2}
= (-4/3+ 0 / 2 , 8/3 + 8 / 2 }
Q = (-2/3 , 16/3)
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