FOR THE TRIANGLE WHOSE VERTICES ARE (-6,-4) (1,-3) AND (5,-3) SOLVE THE AREA OF THE TRIANGLE BY TAKING A COUNTER CLOCKWISE ORDER FROM A
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You didn't label a point A.
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Not clear, but you can find the area.
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A(-6,-4) B(1,-3) AND C(5,-3)
Add point D(-6,-3)
AD = 1
DC = 11
Area of ADC = b*h/2 = 5.5 sq units
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AD = 1
DB = 7
Area of ADB = 3.5 sq units
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Area of ABC = 5.5 - 3.5 = 2 sq units
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A different method:
A(-6,-4) B(1,-3) AND C(5,-3)
A B C A
-6 1 5 -6
-4 -3 -3 -4
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Add the diagonal products starting at the upper left:
-6*-3 + 1*-3 + 5*-4 = 18 -3 -20 = -5
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Add the diagonal products starting at the lower left:
-4*1 + -3*5 + -3*-6 = -4 -15 +18 = -1
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The area is 1/2 the difference:
4/2 = 2 sq units
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This method works for all polygons, any # of sides.