Question 179972
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The first thing you need is the distance formula.


The distance, <i>D</i>, between two points *[tex \Large P_1:(x_1,y_1)] and *[tex \Large P_2:(x_2,y_2)] is given by:


*[tex \Large D = sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}]


In this problem you are given the distance, namely *[tex \Large sqrt{106}] and one of the points, namely *[tex \Large P_1:(0,0)].  Furthermore, you know that for any point *[tex \Large (x,y)] on the given line, *[tex \Large y = x + 4], hence the second point is *[tex \Large P_2:(x,x + 4)].


Now substitute what we know into the distance formula:


*[tex \Large sqrt{106} = sqrt{(0 - x)^2 + (0 - (x + 4))^2}]


Square both sides:


*[tex \Large 106 = (0 - x)^2 + (0 - (x + 4))^2]


Simplify:


*[tex \Large 106 = x^2 + x^2 + 8x + 16]


*[tex \Large  2x^2 + 8x + 16 = 106]


*[tex \Large  x^2 + 4x + 8 = 53]


*[tex \Large  x^2 + 4x - 45 = 0]


Leaving you with a factorable quadratic that you should be able to solve easily.  Note that you will ultimately have two answers to the original problem.



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