Question 199383
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Use the formula for the slope of a line passing through two points, namely:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m = \frac{y_1 - y_2}{x_1 - x_2} ]


Where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the two points.  Do this twice: once for segment PQ and once for segment QR.


Since:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  L_1 \perp L_2 \ \ \Leftrightarrow\ \ m_1 = -\frac{1}{m_2} \text{ and } m_1, m_2 \neq 0]


If


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m_{PQ} = -\frac{1}{m_{QR}}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ PQ \perp QR]


and PQR is a right triangle by definition.


The triangle is isosceles if the measure of PQ is equal to the measure of QR.  Use the 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 endpoints of the segment to calculate the measures of PQ and QR.  If they are equal, the triangle is isosceles; otherwise not.


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