Answer:
1) It is a right triangle.
[tex]Area=25.93\ units^2[/tex]
2) It is a right triangle.
[tex]Area=4\ units^2[/tex]
The points of each triangle are plotted in the images attached.
Step-by-step explanation:
1) The points [tex]A=(-5,3); B= ((6,0); C= (5,5)[/tex] are plotted in the first image attached.
Knowing the points of the triangle, you can find the slope of [tex]AC[/tex] and [tex]BC[/tex] with this formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Then:
[tex]m_{AC}=\frac{5-3}{5-(-5)}=\frac{2}{10}=\frac{1}{5}[/tex]
[tex]m_{BC}=\frac{5-0}{5-6}=\frac{5}{-1}=-5[/tex]
Since the slopes of the sides [tex]AC[/tex] and [tex]BC[/tex] are negative reciprocals, they are perpendicular; therefore IT IS A RIGHT TRIANGLE.
Find the length of [tex]AC[/tex] and [tex]BC[/tex] in order to calculate the area of the triangle:
[tex]AC=\sqrt{(-5-5)^2+(3-5)^2}=10.19\ units\\\\BC=\sqrt{(5-6)^2+(5-0)^2}=5.09\ units[/tex]
The area is:
[tex]A=\frac{AC*BC}{2}=\frac{(10.19\ units)(5.09\ units)}{2}=25.93\ units^2[/tex]
2) The points [tex]A=(4,-3); B=(4,1); C=(2,1)[/tex] are plotted in the second image attached.
By definition horizontal and vertical lines are perpendicular, therefore IT IS A RIGHT TRIANGLE.
You can observe in the figure that the lenghts of the sides [tex]AB[/tex] and [tex]BC[/tex] are:
[tex]AB=4\ units[/tex]
[tex]BC=2\ units[/tex]
Therefore, the area is:
[tex]A=\frac{AB*BC}{2}=\frac{(4\ units)(2\ units)}{2}=4\ units^2[/tex]
