# How do you use determinants to find the area of a triangle?

If someone could give me the formula?

or step by step

well the given vertices are A(1,1), B(6,6), C(2,10)

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• Anonymous

Given: A triangle with vertices (1, 1), (6, 6), and (2, 10).

Find the area of the triangle using determinants.

The area (a) of a triangle = (1/2) base (b) times height (h)

a = (±1/2) b h

In determinants the answer can be negative or positive and area must be a positive therefore use ± if needed.

Form a matrix with the x-values being the first column, the corresponding y-values being the second column, and the third column has all 1's

Points: (1, 1), (6, 6), and (2, 10)

area = ±1/2 times the matrix determinants

| 1 1 1 |

| 6 6 1 | (±1/2)

|2 10 1|

area = (±1/2)(uphill diagonals - downhill diagonals)

area = (±1/2) times [ see next line ]

[(1)(6)(1)+(1)(1)(2)+(1)(6)(10) -(1)(6)(2) -(1)(10)(1) -(1)(1)(6)]

area = (±1/2)(6 + 2 + 60 - 12 - 10 - 6)

area = (±1/2)(68 - 28)

area = (±1/2)(40)

area = 20 square units ANSWER

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CHECK

After much computations of right triangles; I was able to create the triangle side lengths and calculate the area manually.

Sides of triangel are 7.07, 9.06, and 5.66

The height of the triangle is 4.42 if 9.06 is chosen as the base.

area = 1/2 (base) (height)

area = 1/2(9.06)(4.42)

area = 20.02

and it checks out.

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NOTE: The diagonals are written as follows

For the matrix

| a b c|

| d e f |

| g h i |

Downhill Diagonals - Uphill Diagonals =

aei + bfg + cdh - ceg - fha - ibd

GOOD LUCK