Question 198823
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Since the triangle is bounded by the coordinate axes, the vertexes must be the points *[tex \Large \left(0,0\right)], *[tex \Large \left(x_1,0\right)], and *[tex \Large \left(0,y_1\right)]


To calculate *[tex \Large y_1] set *[tex \Large x = 0] in *[tex \Large ax + by = 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_1 = \frac{6}{b}]


Likewise


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1 = \frac{6}{a}]


Giving us vertices of *[tex \Large \left(0,0\right)], *[tex \Large \left(\frac{6}{a}\right)], and *[tex \Large \left(0,\frac{6}{b}\right)]


The measure of the length of the side of the triangle coincident with the *[tex \Large y]-axis is just the *[tex \Large y]-coordinate of the *[tex \Large y]-intercept, namely *[tex \Large \frac{6}{b}], likewise the measure of the other leg of the triangle is *[tex \Large \frac{6}{a}]


The area of a triangle is the base (either of the legs) times the height (the other leg) divided by 2, and we are given that this is equal to 6, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\left(\frac{6}{a}\right)\left(\frac{6}{b}\right)}{2} = 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{6}{a}\right)\left(\frac{6}{b}\right) = 12]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{36}{ab} = 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{ab} = 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ab = 3]


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