Question

If the straight lines $fraction numerator straight x minus 1 over denominator straight k end fraction space equals fraction numerator straight y minus 2 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator 3 end fraction space and space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y minus 3 over denominator straight k end fraction space equals fraction numerator straight z minus 1 over denominator 2 end fraction$ intersect at a point, then the integer k is equal to

-5

2

5

-5

Solution

-5

fraction numerator straight x minus 1 over denominator straight k end fraction space equals space fraction numerator straight y minus 2 over denominator 2 end fraction space equals space fraction numerator straight z minus 3 over denominator 3 end fraction space and space fraction numerator straight x minus 2 over denominator 3 end fraction space equals space fraction numerator straight y minus 3 over denominator straight k end fraction space equals space fraction numerator straight z minus 1 over denominator 2 end fraction Since space lines space intersect space in space straight a space point open vertical bar table row straight k 2 3 row 3 straight k 2 row 1 1 cell negative 2 end cell end table close vertical bar space equals space 0

∴ 2k² + 5k − 25 = 0
k = − 5, 5/2

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