Question 183889
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It is difficult to determine exactly what you are asking.  First of all, you don't have an equation anywhere -- there is no equals sign.  But if you actually meant *[tex \Large y = x + 1], then what I think you meant were two points, namely (3,4) and (5,1) are not both points on that line.  (3, 4) is a point on the line, but (5, 1) is not because *[tex \Large 1 \neq 5 + 1].


If you are asking to derive the equation for the line that passes through the points (3,4) and (5,1), then proceed in this fashion:


The two-point form of the equation of a straight line is:


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


where *[tex \Large P_1:(x_1,y_1)] and *[tex \Large P_2:(x_2,y_2)] are the two given points.  It doesn't matter which of your points you call number 1 and which you call number 2 as long as you are consistent when you make the substitutions:


*[tex \LARGE \text{          }\math y - 4 = \frac{4 - 1}{3 - 5}(x - 3)]


*[tex \LARGE \text{          }\math y - 4 = \frac{3}{-2}(x - 3)]


*[tex \LARGE \text{          }\math -2y + 8 = 3(x - 3)]


*[tex \LARGE \text{          }\math -2y + 8 = 3x - 9]


*[tex \LARGE \text{          }\math -3x - 2y = - 17]


*[tex \LARGE \text{          }\math 3x + 2y = 17]



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