Question 234536
<font face="Garamond" size="+2">


The first thing to do is plot the three points and draw the triangle so that you can see what you have to work with.


You can tell from the fact that two of the points have identical *[tex \Large x]-coordinates that one of the sides of the triangle is a vertical line segment.  You can also see that two of the points have identical *[tex \Large y]-coordinates, so another side of the triangle is a horizontal line segment.


{{{drawing(
500, 500, -5, 10, -5, 10,
grid(1),
blue(line(0,9,0,-4),
line(0,9,5,-4),
line(0,-4,5,-4)),
green(circle(0,9,.15),
circle(0,-4,.15),
circle(5,-4,.15)),
red(locate(0.4,9,A(0,9)),
locate(0.4,-4.4,B(0,-4)),
locate(5.4,-4.4,C(5,-4)))
)}}}



Having one side vertical and another side horizontal tells us that this is a right triangle.  We can say the base is either one of these two sides and the height is the other.


In order to compute the area, use the formula:


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


where *[tex \Large b] is the measure of the base and *[tex \Large h] is the measure of the height.  Now all we need are those measurements.


Given the coordinates of two points you can find the distance between them using 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 two points.


Let's call line segment BC the base.  B has coordinates *[tex \Large \left(0,-4\right)] and C has coordinates *[tex \Large \left(5,-4\right)].  Substitute these values into the distance formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d_{BC} = sqrt{(0\ -\ 5)^2 + ((-4) - (-4))^2}\ =\ \sqrt{5^2+0^2}\ =\ \sqrt{25}\ =\ 5]


hence the measure of the base is 5.


Do the same process to calculate the measure of the height, that is line segment AB.  Then use these values in the Area Formula to calculate your area.


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