A(20, 0) and B(10, -20) are two fixed points.Fond the coordinates of the point P in AB such that : 3PB =AB .Also , find the coordinates of some other point Q in AB such that AB=6AQ
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Answer:
p(x,y)=(25/2,-25/2)
Step-by-step explanation:
3AP=AB
3/1=AB/PB
m:n=3:1
x=mx2+nx1/m+n y=my2+ny1/m+n
=3(-20)+1(0)/3+1
=3(10)+1(20)/3+1 =-25/2
=30+20/4
=25/2
P(x,y)=(25/2,-25/2)
likewise taking AB=6AQ;AB/AQ=6/1 THEREFORE m:n=6:1
x=mx2+nx1/m+n y=my2+ny1/m+n
=6(10)+1(20)/6+1 =6(-20)+1(0)/6+1
=80/7 -120/7
Q(x,y)= (80/7,-120/7)
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