# if a graph of 2x+ky=10k ,intersect at x axis at (2,0)find k

Please find below the solution to the asked query :

$2\mathrm{x}+\mathrm{ky}=10\mathrm{k}\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{point}\left(2,0\right)\phantom{\rule{0ex}{0ex}}2\left(2\right)+\mathrm{k}\left(0\right)=10\mathrm{k}\phantom{\rule{0ex}{0ex}}4=10\mathrm{k}\phantom{\rule{0ex}{0ex}}\mathrm{k}=\frac{2}{5}=0.4\mathrm{ANS}...$

$2\mathrm{x}+\mathrm{ky}=10\mathrm{k}\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{point}\left(2,0\right)\phantom{\rule{0ex}{0ex}}2\left(2\right)+\mathrm{k}\left(0\right)=10\mathrm{k}\phantom{\rule{0ex}{0ex}}4=10\mathrm{k}\phantom{\rule{0ex}{0ex}}\mathrm{k}=\frac{2}{5}=0.4\mathrm{ANS}...$

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