Question

find the perimeter of the triangle whose vertices are (5,2) (4,7) (7,-4)
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Answer 1

Answer:

2 sq. units

Step-by-step explanation:

Given: (5,2) (4,7) (7,-4)

Let us assume A(x1, y1) = (5,2)

Let us assume B(x2,y2) = (4,7)

Let us assume C(x3, y3) = (7,-4)

Area of triangle

= ½ {x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)}

Now,

Area of given triangle

= ½ [ { 5 (7 - (-4) ) } + 4 { (-4 -2) + 7(2 - 7) } ]

= ½ {55 - 24 - 35}

= 4/2

= 2 sq. units