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Triangle IJK, with vertices I(-9,-8), J(-5,-6), and K(-7,-3), is drawn on the coordinate grid below.

Sagot :

MrRoyal

Answer:

[tex]Area= 8[/tex]

Step-by-step explanation:

Given

[tex]I = (-9,-8)[/tex]

[tex]J = (-5,-6)[/tex]

[tex]K = (-7,-3)[/tex]

The complete part of the question is to calculate the area of IJK

The area is calculated using:

[tex]Area = \frac{1}{2}|x_1y_2 - x_2y_1 + x_2y_3 - x_3y_2 + x_3y_1 - x_1y_3 |[/tex]

Where:

[tex]I = (-9,-8)[/tex] --- [tex](x_1,y_1)[/tex]

[tex]J = (-5,-6)[/tex] --- [tex](x_2,y_2)[/tex]

[tex]K = (-7,-3)[/tex] --- [tex](x_3,y_3)[/tex]

So, we have:

[tex]Area = \frac{1}{2}|(-9*-6) - (-5*-8) + (-5*-3) - (-7*-6) + (-7*-8) - (-9*-3) |[/tex]

Solve the brackets

[tex]Area = \frac{1}{2}|54 - 40 + 15- 42 + 56 - 27 |[/tex]

[tex]Area = \frac{1}{2}|16|[/tex]

The absolute value of 16 is 16

So:

[tex]Area = \frac{1}{2} * 16[/tex]

[tex]Area= 8[/tex]

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