Question 187114
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The first thing to do is use the fact that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1 \parallel L_2 \ \ \Leftrightarrow\ \ m_1 = m_2]


to determine that the slope of the line for which you must derive an equation is equal to the slope of your given line, *[tex \Large y = \frac{5}{6}x + b]. (I put the <i>b</i> in there because you left something off of your given equation.  It won't make any difference to the solution of this problem, but I just like to be neat).


Now you need to use the point-slope form of the equation of a line, the slope number that you get by inspection of your given equation, and the coordinates of your given point to derive your equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y - y_1 = m(x - x_1) ]


Where *[tex \Large m = \frac{5}{6}] and *[tex \Large x_1 = 1] and *[tex \Large y_1 = -5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y - (-5) = \frac{5}{6}(x - 1) ]


Finally, all you need to do is rearrange this equation into


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = mx + b]


form and you are done.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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