Question 267252
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Since you are given two points, you can proceed one of two ways.  You can either use the two-point form of the equation of a line:


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


Or you can use the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


and then apply the calculated slope to the point-slope form of the equation of a line:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(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 your given points.


Let's presume that *[tex \Large P_1\left(x_1,y_1\right)\ =\ \left(5,3\right)] and *[tex \Large P_2\left(x_2,y_2\right)\ =\ \left(4,-6\right)]


It doesn't really matter which you select to be 1 and which you select to be 2 as long as you are consistent throughout any given problem.


First calculate the slope:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{(3)\ -\ (-6)}{(5)\ -\ (4)}\ =\ \frac{9}{1}\ =\ 9]


Now substitute values in the point-slope form:


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ (4)\ =\ (9)(x\ -\ (5))]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 4\ =\ 9(x\ -\ 5)]


From there you can go to slope-intercept form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 9x\ -\ 41]


Or to standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9x\ -\ y\ =\ 41]



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