If A(0,-1) B(6,7) C(-2,3) And D(8,3) …

Let A (0 - 1), B (6, 7), C (-2, 3) and D (8, 3) be the given points. Then,
AD = {tex} \sqrt { ( 8 - 0 ) ^ { 2 } + ( 3 + 1 ) ^ { 2 } } = \sqrt { 64 + 16 } = 4 \sqrt { 5 }{/tex}
BC = {tex} \sqrt { ( 6 + 2 ) ^ { 2 } + ( 7 - 3 ) ^ { 2 } } = \sqrt { 64 + 16 } = 4 \sqrt { 5 }{/tex}
AC = {tex} \sqrt { ( - 2 - 0 ) ^ { 2 } + ( 3 + 1 ) ^ { 2 } } = \sqrt { 4 + 16 } = 2 \sqrt { 5 }{/tex}
and, BD = {tex} \sqrt { ( 8 - 6 ) ^ { 2 } + ( 3 - 7 ) ^ { 2 } } = \sqrt { 4 + 16 } = 2 \sqrt { 5 }{/tex}
Therefore, AD = BC and AC = BD
So, ADBC is a parallelogram
Now, AB = {tex} \sqrt { ( 6 - 0 ) ^ { 2 } + ( 7 + 1 ) ^ { 2 } } = \sqrt { 36 + 64 }{/tex}= 10
and, CD = {tex} \sqrt { ( 8 + 2 ) ^ { 2 } + ( 3 - 3 ) ^ { 2 } }{/tex}= 10
Clearly, AB² = AD² + DB² and CD² = CB² + BD²
Hence, ADBC is a rectangle.
Area of rectangle ADBC = {tex} A D \times D B = ( 4 \sqrt { 5 } \times 2 \sqrt { 5 } ){/tex}sq. units = 40 sq. units.