Find the coordinates of the point dividing the line segment
joining the points A(8,0) and B(0,12) internally in the
ratio 2:3
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Answer:Let P & Q be the points of trisection. Then AP:PB=1:2 & AQ:QB=2:1
(i) Let P divides AB in ratio 1:2
Hence m
1
=1;m
2
=2;x
1
=2;y
1
=−3;x
2
=−4,y
2
=−6
P(x,y)=P(
m
1
+m
2
m
1
x
2
+m
2
x
1
,
m
1
+m
2
m
1
y
2
+m
2
y
1
)=P(
1+2
1(−4)+2(2)
,
1+2
1(−6)+2(−3)
)
=P(
3
−4+4
,
3
−6−6
)=P(0,0)
(ii) Let Q divides AB in ratio 2:1
Here m
1
=2,m
2
=1,x
1
=2,y
1
=−3,x
2
=−4,y
2
=−6
=Q(
2+1
2(−4)+1(2)
,
2+1
2(−6)+1(−3)
)
=Q(
3
−8+2
,
3
−12−3
)
=Q(−2,−5)
∴ Hence the coordinates of points of trisection are (0,0) & (−2,−5).
Step-by-step explanation:
Attachments:
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