Question

find the slope of a line perpendicular to each given line.

1. $y=\frac{6}{5}x - 4$
2. $y=-\frac{8}{3}x + 3$
3. $y=2x + 3$
4. $y=\frac{1}{2}x + 1$

Explanation:

Step1: Recall perpendicular slope rule

Perpendicular slope = negative reciprocal of original slope ($m_{\perp} = -\frac{1}{m}$)

Step2: Solve for 11) $y=\frac{6}{5}x-4$

Original slope $m=\frac{6}{5}$, so $m_{\perp}=-\frac{1}{\frac{6}{5}} = -\frac{5}{6}$

Step3: Solve for 12) $y=-\frac{8}{3}x+3$

Original slope $m=-\frac{8}{3}$, so $m_{\perp}=-\frac{1}{-\frac{8}{3}} = \frac{3}{8}$

Step4: Solve for 13) $y=2x+3$

Original slope $m=2$, so $m_{\perp}=-\frac{1}{2}$

Step5: Solve for 14) $y=\frac{1}{2}x+1$

Original slope $m=\frac{1}{2}$, so $m_{\perp}=-\frac{1}{\frac{1}{2}} = -2$

Answer:

1. $-\frac{5}{6}$
2. $\frac{3}{8}$
3. $-\frac{1}{2}$
4. $-2$