Question

P (a,a-1) Q (a-3,a+1) R (-2a-1,_a-1) S (-a,-2a-1) are vertices of rectangle PQRS. Find the value of 'a' and hence find the co-ordinates of vertices of rectangle PQRS

Answer 1

Since it is a parallelogram (rectangle), we have the following property:
$x_1+x_3=x_2+x_4\\ Hence \: a+ (-2a-1)-(-a)=a-3 \\ Solving, \\ -1+3=a\\ a=2$