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For what value of k are the points (8,1) ,(3,-2k),and (k,-5) collinear |
| Answer» According to the question, A(8, 1), B(3, - 2k) and C(k, -5) are collinear.If three points are collinear then Area of triangle is zero{tex}\\therefore{/tex}\xa08(-2k + 5) + 3(-5 -1) + k(1 + 2k) = 0{tex}\\Rightarrow{/tex}\xa0-16k + 40 + 3{tex}\\times{/tex}(-6) + k + 2k2\xa0= 0{tex}\\Rightarrow{/tex}\xa0-16k + 40 - 18 + k + 2k2\xa0= 0{tex}\\Rightarrow{/tex}\xa02k2 -15k + 22 = 0{tex}\\Rightarrow{/tex}\xa02k2 - 11k - 4k + 22 = 0{tex}\\Rightarrow{/tex}\xa0{tex}k(2k - 11) - 2(2k - 11) = 0{/tex}{tex}\\Rightarrow{/tex}{tex}\xa0(2k -11) (k - 2) = 0{/tex}{tex}\\Rightarrow{/tex}\xa02k - 11 = 0 or k - 2 = 0{tex}\\Rightarrow{/tex}\xa0{tex}k = \\frac { 11 } { 2 }{/tex}or k = 2 | |