How do you find the slope of the line parallel to and perpendicular to $4 x - 3 y = 17$ $4x-3y=17$ ?

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Answer 1 · 2017-08-31T10:51:41

Slope of the parallal line is $\frac{4}{3}$ $4/3$ and slope of the perpendicular line is $- \frac{3}{4}$ $-3/4$

$4 x - 3 y = 17 or 3 y = 4 x - 17 or y = \frac{4}{3} x - \frac{17}{3}$ $4x-3y =17 or 3y = 4x-17 or y = 4/3x -17/3$

Slope of the line is $m = \frac{4}{3} [y = m x + c]$ $m=4/3 [y=mx+c]$

Slope of parallel lines are equal.

Slope of the parallal line is $m_{1} = \frac{4}{3}$ $m_1=4/3$

Product of slopes of perpendicular lines is $m_{2} \cdot m = - 1$ $m_2*m=-1$

$m_{2} = - \frac{1}{\frac{4}{3}} = - \frac{3}{4}$ $m_2 = -1/(4/3) = -3/4$

Slope of the perpendicular line is $m_{2} = - \frac{3}{4}$ $m_2=-3/4$

How do you find the slope of the line parallel to and perpendicular to 4x−3y=174x-3y=17?