+26401 Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)
\(\begin{array}{rcl} 8x-2y &=& 4 \qquad | \qquad \cdot (-1) \\ -8x+2y &=& -4 \\ 2y &=& 8x - 4 \\ y &=& \frac{8x-4}{2} \\ y &=& \frac{8x}{2} - \frac{4}{2}\\ y &=& 4x - 2 \\ \end{array}\)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (-12,-1) \\\\ y &=& 4x - 2 \qquad m = 4 \\ m_{\text{perpendicular}} &=& -\frac{1}{ 4 } \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-(-1)}{x-(-12)} &=& -\frac{1}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } \cdot (x+12) \\ y+1 &=& -\frac{1}{ 4 } x -\frac{12}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } x -3 \\ y &=& -\frac{1}{ 4 } x -3 -1 \\ y &=& -\frac{1}{ 4 } x -4 \\ \end{array}\)
+26401 Find the line perpendicular to 8x-2y=4 that goes through (-12,-1)
\(\begin{array}{rcl} 8x-2y &=& 4 \qquad | \qquad \cdot (-1) \\ -8x+2y &=& -4 \\ 2y &=& 8x - 4 \\ y &=& \frac{8x-4}{2} \\ y &=& \frac{8x}{2} - \frac{4}{2}\\ y &=& 4x - 2 \\ \end{array}\)
\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (-12,-1) \\\\ y &=& 4x - 2 \qquad m = 4 \\ m_{\text{perpendicular}} &=& -\frac{1}{ 4 } \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-(-1)}{x-(-12)} &=& -\frac{1}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } \cdot (x+12) \\ y+1 &=& -\frac{1}{ 4 } x -\frac{12}{ 4 } \\ y+1 &=& -\frac{1}{ 4 } x -3 \\ y &=& -\frac{1}{ 4 } x -3 -1 \\ y &=& -\frac{1}{ 4 } x -4 \\ \end{array}\)
