Question

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

Answer 1

Answer:

area of ∆=1/2[X1(y2-y1)+X2(y3-y1)+x3(y1-y2)

✷applied when we have vertices✷

so know we hàve the pts of vertices

(-5,-1) (x1,y1)

,(3,-5) (X2,y2)

(5,2) (x3,y3)

put these in the formula

☞area of ∆= 1/2{-5(3-5) + 3(2+1) +5(-1+5)}

= 1/2{ 10 +9 + 20}

= 39/2

so the area of ∆= 39/2

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Answer 2

$\huge\mathfrak\blue{Answer:}$

$\huge\orange{32 \: units {}^{2}}$

Step-by-step explanation:

Given:

$Co-ordinates \: of \: A = (-5 , -1)$

$Co-ordinates \: of \: B = (3 , -5)$

$Co-ordinates \: of \: C = (5 , 2)$

According to the Q,

$> > Area \: of \: ABC$

Using Area of Triangle Formula:

$> > \frac{1}{2} [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]$

$> > \frac{1}{2} [ - 5( - 5 - 2) + 3(2 + 1) + 5( -1 + 5)]$

$> > \frac{1}{2} [( - 5 \times - 7) + (3 \times 3) + (5 \times 4)]$

$> > \frac{1}{2} ( 35 + 9 + 20)$

$> > \frac{1}{2} \times 64$

$\orange {> > 32 \: units {}^{2}}$

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