Question

Show hat A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

Answer 1

The given vertices are  A(1,2), B(4,3),C(6,6) and D(3,5) .

`AB = sqrt((1-4)^2+(2-3)^2) = sqrt((-3)^2 +(-1)^2) `

`= sqrt(9+1) = sqrt(10) `

`BC = sqrt((4-6)^2 +(3-6)^2) = sqrt((-2)^2 +(-3)^2)`

`= sqrt(4+9) = sqrt(13)`

`CD = sqrt((6-3) ^2 +(6-5)^2) = sqrt((3)^2 +(1)^2) `

`= sqrt(9+1) = sqrt(10)`

`AD = sqrt((1-3)^2 +(2-5)^2 ) = sqrt((-2)^2 +(-3)^2)`

`= sqrt(4+9) = sqrt(13) `

`∵ AB =  CD = sqrt(10) " units and"  BC= AD = sqrt(13)   units `

Therefore, ABCD is a parallelogram

`AC = sqrt((1-6)^2 +(2-6)^2 )= sqrt((-5)^2 +(-4)^2)`

`= sqrt(25+16) = sqrt(41) `

`BD = sqrt((4-3)^2 +(3-5)^2 ) = sqrt((1)^2 +(-2)^2) `

`= sqrt(1+4) = sqrt(5) `

Thus, the diagonal AC and BD are not equal and hence ABCD is not a rectangle