Question 986379
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You are basically being asked to graph the hypotenuse of a right triangle where one leg measures 500 and the hypotenuse measures 1200.  Hence the other leg of the triangle must measure *[tex \Large \sqrt{1200^2\ -\ 500^2}].


You can start your line at the origin, and the second point of your line would be


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt{1200^2\ -\ 500^2},\,500\right)]


Use this point and the origin, *[tex \Large (0,0),], in 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) ]


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


The smallest *[tex \Large x]-value is obviously zero, and the largest is *[tex \Large \sqrt{1200^2\ -\ 500^2}].  You can write out the interval notation for yourself.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \