Find the value of k for which the points ( 3k - 1, k - 2 ), ( k, k - 7 ) and ( k - 1, - k - 2 ) are collinear.
manumanu24643:
3K × 9=3rest to the power
3K ×9=3rest to the power 13
Answers
Answered by
91
hope it helps............
it's easier than solving it by area of triangle formula
it's easier than solving it by area of triangle formula
Attachments:
hey but if the slope are equal then it is possible that they lie on parallel lines
so this method is not 100% true
i know that this method is taught in icse school
...I'm in cbse board
Whatever u cbse students study,but ultimately questions will only come from ncert books.
So cbse stand nowhere in comparison to icse.(first check our syllabi)
Answered by
130
Given points are A(3k - 1, k - 2), B(k,k-7) and C(k - 1, -k - 2).
Here, (x₁,y₁) = (3k - 1, k - 2), (x₂,y₂) = (k, k - 7) and (x₃,y₃) = (k - 1, -k - 2).
Given that they are collinear.
Area of ΔABC = 0
⇒ [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 0
⇒ [(3k - 1)(k - 7 + k + 2) + k(-k - 2 - k + 2) + (k - 1)(k - 2 - k + 7)] = 0
⇒ [(3k - 1)(2k - 5) + k(-2k) + 5(k - 1)] = 0
⇒ 6k^2 - 15k - 2k + 5 - 2k^2 + 5k - 5 = 0
⇒ 4k^2 - 12k = 0
⇒ 4k(k - 3) = 0
⇒ k = 3.
Therefore, the value of k = 3.
Hope it helps!
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