Question 184612
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Since the <i>x</i> axis forms one side of the triangle, use that as the base.


Find the <i>x</i>-intercepts of each of the two equations by substituting 0 for <i>y</i> in each of them.  Then the length of the base of the triangle will be the magnitude of the difference between these <i>x</i>-intercepts.


Next find the point of intersection of the two lines represented by the two given equations by solving the 2-equation, 2-variable linear system.  This one lends itself readily to solution by elimination if you go for eliminating the <i>x</i> term.  Once you have that point of intersection, the <i>y</i>-coordinate will be the height of the triangle.


Now that you know the length of the base and the height, just use the triangle area formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A = \frac{bh}{2}]


{{{drawing(
500, 500, -5, 10, -5, 10,

graph(
500, 500, -5, 10, -5, 10,
y=-( 2/ 3)x+(14 /3 ),
y=( 4/ 5)x+(16 /5 )
))}}}


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