Question

Find the value of K if (K,1), (5,5) and (10,7) are collinear.

Answer 1

I think this will be the correct one.....

Answer 2

$Let \: A(x_{1},y_{1}) = (k,1), \\B(x_{2},y_{2}) = (5,5), and \\</p><p>C(x_{3},y_{3}) = (10,7)\:are \:three \: collinear \}points$

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

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

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

$\implies |k(5-7)+5(7-1)+10(1-5)|=0$

$\implies |-2k+5\times 6+10(-4)|=0$

$\implies |-2k+30-40|=0$

$\implies |-2k-10|=0$

$\implies -2k = 10$

$\implies k = \frac{10}{-2}$

$\implies k = -5$

Therefore.,

$\red{ Value \:of \:k } \green {= -5}$

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