The points \(A(-1,-2),\ B(4,3),\ C(2,5),\ D(-3,0)\) in that order form a rectangle.
True.
Step 1: A quadrilateral is a rectangle if:
So we will check (i) midpoints of diagonals \(AC\) and \(BD\), and (ii) lengths \(AC\) and \(BD\).
Step 2: Find the midpoint of diagonal \(AC\).
\(A(-1,-2),\ C(2,5)\)
Midpoint formula: \(M = \left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right)\)
\(M_{AC} = \left(\tfrac{-1+2}{2},\tfrac{-2+5}{2}\right) = \left(\tfrac{1}{2},\tfrac{3}{2}\right)\)
Step 3: Find the midpoint of diagonal \(BD\).
\(B(4,3),\ D(-3,0)\)
\(M_{BD} = \left(\tfrac{4+(-3)}{2},\tfrac{3+0}{2}\right) = \left(\tfrac{1}{2},\tfrac{3}{2}\right)\)
Step 4: Compare midpoints.
\(M_{AC} = \left(\tfrac{1}{2},\tfrac{3}{2}\right)\) and \(M_{BD} = \left(\tfrac{1}{2},\tfrac{3}{2}\right)\).
Since the midpoints are equal, diagonals \(AC\) and \(BD\) bisect each other. Hence \(ABCD\) is a parallelogram.
Step 5: Find the length of diagonal \(AC\).
Distance formula: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\)
\(AC = \sqrt{(2-(-1))^2 + (5-(-2))^2} = \sqrt{3^2 + 7^2} = \sqrt{9+49} = \sqrt{58}\)
Step 6: Find the length of diagonal \(BD\).
\(BD = \sqrt{((-3)-4)^2 + (0-3)^2} = \sqrt{(-7)^2 + (-3)^2} = \sqrt{49+9} = \sqrt{58}\)
Step 7: Compare diagonal lengths.
\(AC = \sqrt{58}\) and \(BD = \sqrt{58}\).
So the diagonals are equal in length.
Final Conclusion: Since the diagonals bisect each other and are equal, \(ABCD\) is a rectangle. Therefore, the statement is True.