Question 505818
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Consider the three equations in pairs.  For each pair, set the RHSs equal to each other, solve for *[tex \Large x] and then substitute the value of *[tex \Large x] back into either equation to calculate the value of *[tex \Large y] for the point of intersection for the pair of equations that you selected.  Once you have determined all three points of intersection, that is the three vertices of the triangle, use the distance formula 3 times to find the measure of each of the sides of the triangle.  If the triangle is an isosceles right triangle, then the measure of two of the sides will be equal and the measure of the third side will be the square root of two times the measure of either of the equal measure sides.

Distance Formula:

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

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.

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