Show that the points (1,7),(4,2),(-1,-1)&(-4,4) are the vertices of a square.Also find the area of the square.
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first find length of each side
AB=√(3^2+(-5)^2)=√34
BC=√(-5)^2+(-3)^2)=√34
CD=√(-3)^2+(5)^2=√34
AD=√(-5)^2+(-3)^2=√34
here we see ,
AB=BC=CD =AD
all sides are equal
now find
AC=√(-2)^2+(-8)^2=√68
here ,
AC^2=BC^2+AB^2
so, angle are also 90°
hence ABCD is square
area of square=(side length) ^2=(√34)^2
=34 sq unit
AB=√(3^2+(-5)^2)=√34
BC=√(-5)^2+(-3)^2)=√34
CD=√(-3)^2+(5)^2=√34
AD=√(-5)^2+(-3)^2=√34
here we see ,
AB=BC=CD =AD
all sides are equal
now find
AC=√(-2)^2+(-8)^2=√68
here ,
AC^2=BC^2+AB^2
so, angle are also 90°
hence ABCD is square
area of square=(side length) ^2=(√34)^2
=34 sq unit
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