find the perimeter of triangle whose vertices are (0,0),(a,0),(0,b)
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Let the vertices of the triangle be A(0,0) B(a,0) C(0,b)
Hence
AB = √[(x₂-x₁)² + (y₂-y₁)²]
AB = √(a-0)² + (0-0)²
AB = √a² = a
BC = √[(x₂-x₁)² + (y₂-y₁)²]
BC = √(a-0)² + (0-b)²
BC = √a²+b²
AC = √[(x₂-x₁)² + (y₂-y₁)²]
AC = √(0-0)² + (0-b)²
AC = √b² = b
Hence,
Perimeter = AB+BC+AC
= a + b + √a²+b²
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