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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)(1,6), perpendicular to 3x+5y=13 x+5 y=1

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

3x+5y=15y=3x+15y5=3x5+15y=35x+15\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 35-\dfrac{3}{5}.

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