find coordinates of point which divides the line segment joining P (2,-3)and Q (-4,6) into 3 equal parts .
Answers
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
BRAINLEST MARK KR DO NA
Step-by-step explanation:
=>Let P & Q be the points of trisection.
Then AP:PB=1:2 & AQ:QB=2:1
Condition-1 :
=>Let P divides AB in ratio 1:2
Hence,
m=1 , m'=2
x=2 , y= -3
x'= -4 , y'=6
P(x,y)
=P[(mx'+m'x) /(m+m') , (my'+m'y) /(m+m')]
=P[(1)(-4)+(2)(2) /(1)+(2) , (1)(6)+(2)(-3) /(1)+(2)]
=P[(-4+4) /3 , (6-6) /3]
=P(0 , 0)
Condition-2 :
=>Let Q divides AB in ratio 2:1
Hence,
m=2 , m'=1
x=2 , y= -3
x'= -4 , y'=6
Q(x,y)
=Q[(mx'+m'x) /(m+m') , (my'+m'y) /(m+m')]
=Q[(2)(-4)+(1)(2) /(2+1) , (2)(6)+(1)(-3) /(2+1)]
=Q[(-8+2) /3 , (12-3) /3]
=Q( -2 , 3)
∴ Hence the coordinates of points of trisection are (0,0) & (-2,3)