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Question

# Find the value of k if $\left(\frac{2}{5},\frac{1}{3}\right)$ , $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$ are collinear points .

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Solution

## Given: Points $\left(\frac{2}{5},\frac{1}{3}\right)$ , $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$ are collinear. So, slope of the line joining $\left(\frac{2}{5},\frac{1}{3}\right)$ and $\left(\frac{1}{2},k\right)$ = slope of the line joining $\left(\frac{1}{2},k\right)$ and $\left(\frac{4}{5},0\right)$ $\mathrm{i}.\mathrm{e}.,\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{{y}_{3}-{y}_{2}}{{x}_{3}-{x}_{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{k-\frac{1}{3}}{\frac{1}{2}-\frac{2}{5}}=\frac{0-k}{\frac{4}{5}-\frac{1}{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{k-\frac{1}{3}}{\frac{1}{10}}=\frac{-k}{\frac{3}{10}}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{10}\left(k-\frac{1}{3}\right)=\frac{1}{10}\left(-k\right)\phantom{\rule{0ex}{0ex}}⇒3\left(k-\frac{1}{3}\right)=-k\phantom{\rule{0ex}{0ex}}⇒3k-1=-k\phantom{\rule{0ex}{0ex}}⇒3k+k=1\phantom{\rule{0ex}{0ex}}⇒4k=1\phantom{\rule{0ex}{0ex}}⇒k=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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