Find the line perpendicular to y=3x-7 that goes through (3,5)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (3,5) \\\\ y = 3x-7 \qquad m = 3 \\ m_{\text{perpendicular}} = -\frac{1}{m} &=& -\frac{1}{3} \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-5}{x-3} &=& -\frac{1}{3} \\ y-5 &=& -\frac{1}{3}\cdot (x-3) \\ y-5 &=& -\frac{x}{3}+1 \\ y &=& -\frac{x}{3}+1+5 \\ y &=& -\frac{x}{3}+6 \\ \end{array}\\\)
Find the line perpendicular to y=3x-7 that goes through (3,5)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (3,5) \\\\ y = 3x-7 \qquad m = 3 \\ m_{\text{perpendicular}} = -\frac{1}{m} &=& -\frac{1}{3} \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-5}{x-3} &=& -\frac{1}{3} \\ y-5 &=& -\frac{1}{3}\cdot (x-3) \\ y-5 &=& -\frac{x}{3}+1 \\ y &=& -\frac{x}{3}+1+5 \\ y &=& -\frac{x}{3}+6 \\ \end{array}\\\)