Question

If the points A(6,10) , B(7,k) , C(8,-10) are collinear points, find the value of K​

Answer 1

$Let \: A(x_{1},y_{1}) = (6,10) , B(x_{2},y_{2}) =(7,k)\\and \: C(x_{3},y_{3}) = ( 8,-10) \:are \: three \:points$

$Area \: of \: the \: \triangle ABC = 0$

$\blue{ \because three \: points \: are \: collinear ) }$

$\frac{1}{2}| x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| = 0$

$\implies | x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})| = 0$

$\implies |6[k-(-10)+7(-10-10)+8(10-k)|=0$

$\implies |6(k+10)+7(-20)+80-8k| = 0$

$\implies |6k+60-140+80-8k|=0$

$\implies | -2k +140 - 140 | = 0$

$\implies |-2k| = 0$

$\implies k = 0$

Therefore.,

$\red{ Value \: of \: k } \green { = 0 }$

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Answer 2

Answer:

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Step-by-step explanation:

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