show that the four points (0,1),(6,7),(-2,3)&(8,3) are the vertices of rectangle

show that the four points (0,1),(6,7),(-2,3)&(8,3) are the vertices of

1.	show that the four points (0,1),(6,7),(-2,3)&(8,3) are the vertices of rectangle
Answer» Let A (0 - 1), B (6, 7), C (-2, 3) and D (8, 3) be the given points. Then,AD =\xa0{tex} \\sqrt { ( 8 - 0 ) ^ { 2 } + ( 3 + 1 ) ^ { 2 } } = \\sqrt { 64 + 16 } = 4 \\sqrt { 5 }{/tex}BC =\xa0{tex} \\sqrt { ( 6 + 2 ) ^ { 2 } + ( 7 - 3 ) ^ { 2 } } = \\sqrt { 64 + 16 } = 4 \\sqrt { 5 }{/tex}AC =\xa0{tex} \\sqrt { ( - 2 - 0 ) ^ { 2 } + ( 3 + 1 ) ^ { 2 } } = \\sqrt { 4 + 16 } = 2 \\sqrt { 5 }{/tex}and, BD =\xa0{tex} \\sqrt { ( 8 - 6 ) ^ { 2 } + ( 3 - 7 ) ^ { 2 } } = \\sqrt { 4 + 16 } = 2 \\sqrt { 5 }{/tex}Therefore, AD = BC and AC = BDSo, ADBC is a parallelogramNow, AB =\xa0{tex} \\sqrt { ( 6 - 0 ) ^ { 2 } + ( 7 + 1 ) ^ { 2 } } = \\sqrt { 36 + 64 }{/tex}= 10and, CD =\xa0{tex} \\sqrt { ( 8 + 2 ) ^ { 2 } + ( 3 - 3 ) ^ { 2 } }{/tex}= 10Clearly, AB2\xa0= AD2\xa0+ DB2\xa0and CD2\xa0= CB2\xa0+ BD2Hence, ADBC is a rectangle.Area of rectangle ADBC =\xa0{tex} A D \\times D B = ( 4 \\sqrt { 5 } \\times 2 \\sqrt { 5 } ){/tex}sq. units = 40 sq. units.