. Find the value of k for which the points (3k-1, k-2), (k, k-7) and(k-1, -k-2) are collinear.
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K=3
A (3k-1 , k-2)
B (k, k-7)
C (k-1,-k-2)
Given that A, B and C
Slope of AB =(k-2) -(k-7) (3k-1) -k
=5/2k-1
Slope of AC =(k-2) -(-k-2)/(3k-1)-(k-1)
2k/2k=1
Hence, Slope of AB should be 1
5/2k-1=1
2k-1=5
2k=6
k=6/2
k=3
jaideepberi:
the answer is right but you made it way more complicated
just use the area of triangle formula and equate it to 0
nhiii asnn ha
this method has not been taught in CBSE class 10 :))))))))))))
oo really yr but muja vo method nhi aata jo tum bol raha ho
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