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Find the line perpendicular to y=3x-7 that goes through (3,5)

 Nov 16, 2015

Best Answer 

 #1
avatar+23299 
+25

Find the line perpendicular to y=3x-7 that goes through (3,5)

 

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

laugh

.
 Nov 16, 2015
 #1
avatar+23299 
+25
Best Answer

Find the line perpendicular to y=3x-7 that goes through (3,5)

 

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

laugh

heureka Nov 16, 2015
 #2
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0

That Is Wrong 

 Nov 16, 2015
edited by Guest  Nov 16, 2015
 #3
avatar+28182 
+12

heureka has given the correct answer to the question that was asked.

 

If you think it is wrong you need to say why.

 Nov 16, 2015

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