if points (k,3),(6,-2) and (-3,4) are collineat,then the value of k is
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7
Hi !
(k,3) = (x₁,y₁)
(6,-2) = (x₂,y₂)
(-3,4)= (x₃,y₃)
As they are collinear ,
x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0
k(-2-4) + 6( 4 -3) + -3(3 -(-2)) = 0
k(-6) + 6(1) + -3(5) = 0
-6k + 6 - 15 = 0
-6k - 9 = 0
-6k = 9
k = 9/-6
k = -3/2
(k,3) = (x₁,y₁)
(6,-2) = (x₂,y₂)
(-3,4)= (x₃,y₃)
As they are collinear ,
x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) = 0
k(-2-4) + 6( 4 -3) + -3(3 -(-2)) = 0
k(-6) + 6(1) + -3(5) = 0
-6k + 6 - 15 = 0
-6k - 9 = 0
-6k = 9
k = 9/-6
k = -3/2
mahir10:
thanks coreect
^_^
give this ans. to
if the points A(7,-2),B(5,1) and C(3,2k) are collinear then the value of k is
Answered by
4
Given points are A(k,3), B(6,-2),C(-3,4).
Given that ABC are collinear.
The formula is:
x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0
k(-2-4) + 6(4-3) + -3(3+2) = 0
k(-6) + 6(1) + (-3)(5) = 0
-6k + 6 - 15 = 0
-6k - 9 = 0
-6k = 9
k = 9/-6
k = -3/2.
Hope this helps!
Given that ABC are collinear.
The formula is:
x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = 0
k(-2-4) + 6(4-3) + -3(3+2) = 0
k(-6) + 6(1) + (-3)(5) = 0
-6k + 6 - 15 = 0
-6k - 9 = 0
-6k = 9
k = 9/-6
k = -3/2.
Hope this helps!
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