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How do you find the area of a triangle whose vertices are (2,-2), (8,5), (6,-10)?

Answer 1

I fount #38# ua

I would use a matrix method involving the Determinant of a square matrix. The area of the triangle is #+-1/2# the determinant of the matrix formed by the coordinates of the vertecies of the triangle and a column of #1#, i.e.: Area#=+-1/2*det##((2,-2,1),(8,5,1),(6,-10,1))=# #=+-1/2(-72)=# changing sign through #+-# you get: Area#=76/2=38# units of area

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Answer 2

Jameson Allen

To find the area of a triangle given its vertices (x1, y1), (x2, y2), and (x3, y3), you can use the formula:

Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Using the coordinates (2, -2), (8, 5), and (6, -10), plug the values into the formula:

Area = 0.5 * |2(5 - (-10)) + 8((-10) - (-2)) + 6((-2) - 5)|

Calculate the values:

Area = 0.5 * |2(15) + 8(-8) + 6(-7)| = 0.5 * |30 - 64 - 42| = 0.5 * |-76|

Take the absolute value:

Area = 0.5 * 76 = 38 square units

So, the area of the triangle is 38 square units.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.