Question 749136
Calculate the area of the triangle with the following vertices:

6.  (-3, -3), (0, -3), (1, 1)


7. (8, -2), (-0, -2), (2, -5)


8. .(-9, 0), (-1, 2), (-5, 4)


9. (3, -7), (6, 4), (-2, -3)
===============================


10. A(-5,-7), B(-1,-7), C(-3,-4)

A...B...C...A
-5 -1 -3 -1
-7 -7 -4 -7
------------------
Get the sum of the diagonal products starting at the upper left.
-5*-7 + -1*-4 + -3*-7 = 35 + 4 + 21 = 60
-----
Get the sum of the diagonal products starting at the lower left.
-7*-1 + -7*-3 + -4*-1 = 7 + 21 + 4 = 32
============
Area = 1/2 the difference = (60 - 32)/2
Area = 14 sq units.
Do the others the same way.
Do the parallelograms below the same way, too.  It works for any # of points.
-------------------
Find the area of the parallelogram with the following vertices:

11. (-2, 3), (5, 8), (3, 3), and (0, 8)


12. (-2, 7), (-4, 4), (-11, 4), and (-9, 7)


13. (-6, 6), (-6, 3), (-12, 3), and (-12, 6)


14. (12, -3), (5, -6), (5, -3), and (12, -6)


15. (-2, 2), (-6, 9), (-13, 9), and (-9, 2)

 

Find the equation of the line that passes through the following points. Put your equation into slope-intercept form:

16.  (4, 25), (8, 61)
-------------------
Using determinants:
|x +y 1|
|4 25 1| = 0
|8 61 1|
Note:  + is for alignment
x*(25-61) - y*(4-8) + 1*(244-200) = 0
-36x + 4y + 44 = 0
y = 9x - 11
===============
Do the others the same way.
17. (-8, -1), (0, -1)

18. (3, -41), (1, -9)

19. (-6, -70), (4, 50)

20. (9, -98), (0, 19)

I'll be glad to check your work if you email via the Thank You note.
PS  Don't put a space after the comma in the points.