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Section 3.2 : Lines
11. Find the equation of the line through \(\left( { - 7,2} \right)\) and is perpendicular to the line \(3x - 14y = 4\).
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Start SolutionFirst, we need to get the slope of our new line, i.e. the line through the point \(\left( { - 7,2} \right)\). We know this line is parallel to the line \(3x - 14y = 4\) and the slopes must be negative reciprocals of each other.
Therefore, all we need to do is put the equation of the second line into slope-intercept form and get its slope.
\[\begin{align*}3x - 14y & = 4\\ 14y & = 3x - 4\\ y & = \frac{3}{{14}}x - \frac{2}{7}\hspace{0.25in}:\hspace{0.25in}{m_2} = \frac{3}{{14}}\end{align*}\]So, the new line must have slope of,
\[{m_1} = - \frac{1}{{{m_2}}} = - \frac{1}{{{}^{3}/{}_{{14}}}} = - \frac{{14}}{3}\] Show Step 2Now, we have both the slope of the new line as well as a point through the new line so we can use the point-slope form of the line to write down the equation of the new line.
\[y = 2 - \frac{{14}}{3}\left( {x - \left( { - 7} \right)} \right) = 2 - \frac{{14}}{3}\left( {x + 7} \right) = - \frac{{14}}{3}x - \frac{{92}}{3}\]