For what values of k are the points (8,1), (3,-2k) and (k,-5) collinear?
SUPER URGENT PLS!!!
MisterMeow:
I am not a maths student but ... I guess that's about making an equation :3
Which one is correct? :/
um; ncert won't go for fraction answer so ; guess ya ... Iam going for beginner
Answers
Answered by
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arΔABC = 0 where A(8,1), B(3,-2k) and C(k,-5)
arΔABC= 0 = [8 (-2k+5) + 3(-5-1) + k(1+ 2k)] (ar is zero because collinear pts)
0 = -16k + 40 - 18 + k + 2k²
2k² - 15k + 22 = 0
2k² - 4k - 11k + 22 = 0
2k(k-2) -11(k-2) = 0
(2k-11) (k-2) = 0
∴k= 11/2 or 2
arΔABC= 0 = [8 (-2k+5) + 3(-5-1) + k(1+ 2k)] (ar is zero because collinear pts)
0 = -16k + 40 - 18 + k + 2k²
2k² - 15k + 22 = 0
2k² - 4k - 11k + 22 = 0
2k(k-2) -11(k-2) = 0
(2k-11) (k-2) = 0
∴k= 11/2 or 2
Can u elaborate the ans pls?
I mean with full formulas nd steps
Karthiks answer is wrong because he has put the wrong values
okay wait
Ok thanq
Is it clear now?
Yes thanq dear
mark as brainliest if it helped you
Sure
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