if the points A ( k+1 , 2k ) , B ( 3k, 2k+ 3 ) and C ( 5k - 1 , 5k ) are collinear , then find the value of k.
vishagh:
A bit busy cant answer i ll just tell u how
Take the cross product of all ,this shud be equal to 0(since collinear) get the value of k
I didn't understand
k=3/2
K =6
Answers
Answered by
28
given,
A( k+1,2k ), B( 3k,2k+3 ) C( 5k-1,5k )
slope of AB =slope of BC
=>2k+3-2k/3k-(k+1)=5k-(2k+3)/5k-1-3k
=>3/3k-k-1=5k-2k-3/2k-1
=>3/2k-1=3k-3/2k-1
=>3=3k-3
=>6=3k
=>k=6/3=2
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A( k+1,2k ), B( 3k,2k+3 ) C( 5k-1,5k )
slope of AB =slope of BC
=>2k+3-2k/3k-(k+1)=5k-(2k+3)/5k-1-3k
=>3/3k-k-1=5k-2k-3/2k-1
=>3/2k-1=3k-3/2k-1
=>3=3k-3
=>6=3k
=>k=6/3=2
HOPE IT IS HELPFUL
pls mark it as brainliest
pls mark it as brainliest
if it is helpful to u
thanks a lot !
Answered by
37
colinear means , area of triangle form by this points = 0
area of traingle = 1/2 {(k +1)(2k +3-5k) +3k (5k -2k) +(5k -1)(2k -2k-3)}.
0 ={(k +1)(3-3k) + 9k^2 -15k +3 }
0 =3-3k^2 + 9k^2 -15k +3
6k^2 -15k +6 =0
2k^2 -5k +2 =0
k ={5 +_3}/4 = 2 , 1/2
k = 2 , 1/2
area of traingle = 1/2 {(k +1)(2k +3-5k) +3k (5k -2k) +(5k -1)(2k -2k-3)}.
0 ={(k +1)(3-3k) + 9k^2 -15k +3 }
0 =3-3k^2 + 9k^2 -15k +3
6k^2 -15k +6 =0
2k^2 -5k +2 =0
k ={5 +_3}/4 = 2 , 1/2
k = 2 , 1/2
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