What is the area of a triangle whose vertices are d(3, 3) , e(3, −1) , and f(−2, −5) ?

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units²

The area of the triangle is 10 sq units

SOLUTION:

Given, we have to find the area of a triangle whose vertices are D(3, 3), E(3, −1) and F(−2, −5)

We know that,

Where, are vertices of the triangle.

Here in our problem,

Now, substitute the above values in the formula.

366

step-by-step explanation:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)