Question

A(20,0) B(10,-20) are two fixed points. Find the coordinates of point p in AB such that 3PB=AB also find the coordinates of Q in AB such that AB=6AQ.

Answer 1

Step-by-step explanation:

Given, 3PB =AB

⇒ABPB=31

⇒AB-PBPB=3-11

⇒APPB=21

Using section formula,

Coordinates of P are

P(x,y)=P(2×10+1×202+1,2×(-20)+1×02+1)

=P(403,-403)

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Answer 2

Step-by-step explanation:

Given, 3PB = AB

AB/PB = 1/3

AB - PB/PB = 3 - 1/1 = 2/1

AP/PB = 2/1

Using section formula coordinates of P are

P(x, y) =

P(2*10 + 1*20)/2+1, 2*-20 + 1*0/2+1)

= P(40/3, - 40/3)

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