Question 181747
<font face="Times New Roman" size="+2">


You have three pairs of equations, equations 1 and 2, equations 2 and 3, and equations 1 and 3.  Solve each system for the point of intersection.  The three solutions will be your vertices.


If it is a right triangle, then two of the sides will be perpendicular.  Perpendicular lines have slopes that are negative reciprocals of each other. That is to say:


*[tex \LARGE \text{          }\math L_1 \perp L_2 \ \Leftrightarrow \ m_1 = \frac {-1}{m_2}]


Put all three of your equations in slope-intercept form (*[tex \Large y = mx + b]).  If any pair of slope numbers has the negative reciprocal arrangement, then those two lines are perpendicular and the triangle is a right triangle.  Otherwise, it is not a right triangle.





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