Question 261846
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The *[tex \Large y]-intercept of a line represented by a two-variable equation is the point *[tex \Large \left(0,\,b\right)] where *[tex \Large b] is the value of *[tex \Large y] when *[tex \Large x\ =\ 0].  So your first step is to substitute 0 for *[tex \Large x] in your equation and calculate the resulting value of *[tex \Large y] so that you can determine the point *[tex \Large \left(0,\,b\right)].

Once you have done that, you will have two points on the desired line -- the one you just derived and the given point *[tex \Large \left(5,\,-1\right)]

Having two points on your desired line you can then use the two-point form of the equation of a line to derive an equation for the desired line.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \left(\frac{y_1\ -\ y_2}{x_1\ -\ x_2}\right)(x\ -\ x_1) ]

where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)]
are the coordinates of the two points.

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