Prove thst A(-2,-1),B(1,0)C(4,3) D(1,2)are the vertices of a parallelogram and its area
Answers
Step-by-step explanation:
by distance form:-
by distance form:-AB=√(-2-1)^2 + (-1-0)^2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so that
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10BD=√(1-1)^2+(2-0)^2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10BD=√(1-1)^2+(2-0)^2=2
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10BD=√(1-1)^2+(2-0)^2=2AC=BD
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10BD=√(1-1)^2+(2-0)^2=2AC=BDFor area
by distance form:-AB=√(-2-1)^2 + (-1-0)^2=√9+1=√10so thatBC=√3^2+3^2=√18=3√2DC=√3^2+1^2=√10AD=√3^2+3^2=√18=3√2AB=DC=√10AD=BC=3√2AC=√(1+2)^2+(3+1)^2=√36+4=√90=2√10BD=√(1-1)^2+(2-0)^2=2AC=BDFor area 1\2[x^1 (1^2-1^3)+x^2(-1-1^2)+x^3(-1a-1^1)+xq(-1,-12).