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Question

Find the equation of the line satisfying the given conditions, giving it in slope-intercept form if possible.Through $(1,6)$, perpendicular to $3 x+5 y=1$

Solution

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Two lines are perpendicular if their slope are negative reciprocal. Change the given equation into slope-intercept form. $y=mx+b$

\begin{aligned} 3x+5y&=1\\ 5y&=-3x+1\\ \dfrac{5y}{5}&=\dfrac{-3x}{5}+\dfrac{1}{5}\\ y&=-\dfrac{3}{5}x+\dfrac{1}{5} \end{aligned}

The slope in the equation is $-\dfrac{3}{5}$.

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