find the value of k for which the points (3 k - 1, k - 2) , (k, k - 7) and (K - 1, - K - 2) are collinier
plz provide each and every step
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Answer:
k = 3
Step-by-step explanation:
A = ( 3k-1, k-2 )
B = ( k , k-7 )
C = ( k-1, -k-2 )
Slope of line AB:
[ (k-7) - (k-2) ] / [ k - (3k-1) ]
= -5 / ( -2k + 1 )
Slope of line AC:
[ (-k-2) - (k-2) ] / [ (k-1) - (3k-1) ]
= -2k / -2k
= 1
A, B, C collinear
<=> line AB and line AC are the same line
<=> slope of AB = slope of AC
<=> -5 / ( -2k + 1 ) = 1
<=> -5 = -2k + 1
<=> 2k = 6
<=> k = 3
Anonymous:
Hello. Hope this helps you. Plz mark it brainliest. Have a good day!
soory
but i wanted the method in which area of triangle formed by 3 points is zero
Ouch! How to make a simple problem look difficult! Oh well, we gotta do what we gotta do. Good luck!
ya
ok bye
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