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find the line perpendicular to y=2/5x-3 that goes through (4,4)

 Nov 16, 2015

Best Answer 

 #1
avatar+26403 
+10

find the line perpendicular to y=2/5x-3 that goes through (4,4)

 

\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (4,4) \\\\ y = \frac25 x -3 \qquad m = \frac25 \\ m_{\text{perpendicular}} = -\frac{1}{\frac25} &=& -\frac52 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-4}{x-4} &=& -\frac52 \\ y-4 &=& -\frac52\cdot (x-4) \\ y-4 &=& -\frac52 x +10 \\ y &=& -\frac52 x +10 +4 \\ y &=& -\frac52 x + 14 \\ \end{array}\\\)

laugh

 Nov 16, 2015
 #1
avatar+26403 
+10
Best Answer

find the line perpendicular to y=2/5x-3 that goes through (4,4)

 

\(\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (4,4) \\\\ y = \frac25 x -3 \qquad m = \frac25 \\ m_{\text{perpendicular}} = -\frac{1}{\frac25} &=& -\frac52 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-4}{x-4} &=& -\frac52 \\ y-4 &=& -\frac52\cdot (x-4) \\ y-4 &=& -\frac52 x +10 \\ y &=& -\frac52 x +10 +4 \\ y &=& -\frac52 x + 14 \\ \end{array}\\\)

laugh

heureka Nov 16, 2015
 #2
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+5

Thanks For The Help!! 😊

 Nov 16, 2015

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