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find the equation of a line, in general form that is parallel to hte line 3x-5y-15=0 and passes threw the point (3,-2)

Guest Mar 6, 2017
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 #1
avatar+76946 
+5

3x-5y-15=0   rearrange to solve for y

 

3x - 15  = 5y   divide both sides by 5

 

(3/5)x - 3   = y       the slope is  (3/5)

 

And since parallel lines have the same slope.....the equation of a parallel line passing through  (3, -2)  is

 

y = (3/5) (x - 3) - 2

 

y = (3/5)x - 9/5 - 2

 

y = (3/5)x - 9/5 - 105

 

y = (3/5)x - 19/5

 

 

cool cool cool

CPhill  Mar 6, 2017
 #2
avatar+18612 
0

find the equation of a line, in general form that is parallel to hte line 3x-5y-15=0 and passes threw the point (3,-2)

 

Formula:

\(\begin{array}{|rcll|} \hline ax+by+c &=& 0 \quad & | \quad \qquad \text{parallel and passes through the point } P\binom{x_y}{y_p} \\\\ ax+by-(ax_p+by_p) &=& 0 \\ \hline \end{array}\)

 

The equation:

\(\begin{array}{|rcll|} \hline 3x-5y-15 &=& 0\quad & | \quad a=3 \quad b= -5 \\\\ ax+by-(ax_p+by_p) &=& 0 \quad & |\quad x_p = 3 \quad y_p = -2 \\ 3x-5y-[3\cdot 3-5\cdot (-2)] &=& 0 \\ 3x-5y-(9+10) &=& 0 \\ \mathbf{3x-5y-19} & \mathbf{=} & \mathbf{0} \qquad \text{or } \qquad \mathbf{y = \frac35\cdot x -\frac{19}{5} }\\ \hline \end{array} \)

 

laugh

heureka  Mar 6, 2017

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