Question 313871
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 3x\ =\ -2]


Substitute 0 for *[tex \LARGE x], then *[tex \LARGE y\ =\ -2], hence *[tex \LARGE y]-intercept is *[tex \LARGE (0,-2)].


Substitute 0 for *[tex \LARGE y], then *[tex \LARGE x\ =\ \frac{2}{3}], hence *[tex \LARGE x]-intercept is *[tex \LARGE (\frac{2}{3},0)].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 3x\ =\ 2]


Substitute 0 for *[tex \LARGE x], then *[tex \LARGE y\ =\ 2], hence *[tex \LARGE y]-intercept is *[tex \LARGE (0,2)].


Substitute 0 for *[tex \LARGE y], then *[tex \LARGE x\ =\ \frac{2}{3}], hence *[tex \LARGE x]-intercept is *[tex \LARGE (\frac{2}{3},0)].  Note that this is the same as the *[tex \LARGE x]-intercept for the first equation, hence it is the point of intersection of the two lines represented by the two equations and therefore the third vertex of the triangle.


The other two vertexes are the *[tex \LARGE y]-intercepts of the two equations.  The base of the triangle is therefore that segment of the *[tex \LARGE y]-axis between the two intercept points, while the altitude of the triangle is the segment between the origin and the point of intersection of the two lines.


You can either use the distance formula:


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


or simple inspection to determine that the measure of the base of the triangle is 4 and the measure of the altitude is *[tex \LARGE \frac{2}{3}].


Use these values in the formula for the area of a triangle:


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


where *[tex \LARGE b] represents the measure of the base and *[tex \LARGE h] represents the measure of the height or altitude.



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