+26403 Find the line perpendicular to y=-8/3 x+1 that goes through (8,9)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (8,9) \\\\ y = -\frac83 x +1 \qquad m = -\frac83 \\ m_{\text{perpendicular}} = -\frac{1}{-\frac83} &=& \frac38 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-9}{x-8} &=& \frac38 \\ y-9 &=& \frac38\cdot (x-8) \\ y-9 &=& \frac38 x -3 \\ y &=& \frac38 x -3 +9 \\ y &=& \frac38 x + 6 \\ \end{array}\\ \)
+26403 Find the line perpendicular to y=-8/3 x+1 that goes through (8,9)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (8,9) \\\\ y = -\frac83 x +1 \qquad m = -\frac83 \\ m_{\text{perpendicular}} = -\frac{1}{-\frac83} &=& \frac38 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-9}{x-8} &=& \frac38 \\ y-9 &=& \frac38\cdot (x-8) \\ y-9 &=& \frac38 x -3 \\ y &=& \frac38 x -3 +9 \\ y &=& \frac38 x + 6 \\ \end{array}\\ \)
