Find the area of the triangle whose vertices are: (–5, –1),

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136.

Find the area of the triangle whose vertices are:
(–5, –1), (3, –5), (5, 2)


Let the given points be A(-5, -1), B(3, -5) and C(5, 2).
Here, we have
x1 = -5, y1 = -1
x2 = 3, y2 = -5
and    x3 = 5, y3 = 2
Now, Area of ∆ABC

equals 1 half left square bracket straight x subscript 1 left parenthesis straight y subscript 2 minus straight y subscript 3 right parenthesis plus straight x subscript 2 left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis plus straight x subscript 3 left parenthesis straight y subscript 1 minus straight y subscript 2 right parenthesis right square bracket
equals space 1 half left square bracket left parenthesis negative 5 right parenthesis left curly bracket negative 5 minus 2 right curly bracket plus left parenthesis 3 right parenthesis left curly bracket 2 minus left parenthesis negative 1 right parenthesis right curly bracket space plus space left parenthesis 5 right parenthesis space left curly bracket left parenthesis negative 1 right parenthesis minus left parenthesis negative 5 right parenthesis right curly bracket right square bracket
equals 1 half left square bracket 35 plus 9 plus 20 right square bracket space equals space 32 space square space units.

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