Question

The points A(-1, -2), B(4,3), C(2,5) and D(-3,0) in that order form which structure?

• A. Square
• B. Rectangle
• C. Parallelogram
• D. Rhombus

Answer 1

The correct option is B

Rectangle

Distance between A(-1,-2) and B(4, 3)
$AB=\sqrt{(4+1{)}^2+(3+2{)}^2}=\sqrt{5^2+5^2}=\sqrt{25+25}=5\sqrt{2}$

Distance between C(2,5) and D(-3,0)
$CD=\sqrt{(-3-2{)}^2+(0-5{)}^2}=\sqrt{(-5{)}^2+(-5{)}^2}=\sqrt{25+25}=5\sqrt{2}$

$\left[\begin{array}{c}\because \text{distance between the points}(x_1,y_1)and(y_2,y_2)\\ d=\sqrt{(x_2-x_1{)}^2+(x_2-y_1{)}^2}\end{array}\right]\text{Distance between A(-1,-2) and D(-3,0),}AD=\sqrt{(-3+1{)}^2+(0+2{)}^2}=\sqrt{(-2{)}^{+}{2}^2}=\sqrt{4+4}=2\sqrt{2}$
Similarly, BC = $2\sqrt{2}$
We know that, in a rectangle, opposite sides are equal and diagonals are equal and they bisect each other.
Since, AB = CD and AD = BC
Distance between A(-1,-2) and C(2,5)
$AC=\sqrt{(2+1{)}^2+(5+2{)}^2}=\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{58}$
And distance between D(-3,0) and B(4,3),
$DB=\sqrt{(4+3{)}^2+(3-0{)}^2}=\sqrt{7^2+3^2}=\sqrt{49+9}=\sqrt{58}$
Since, diagonals AC and BD are equal.
Hence, the points A(-1,-2), B(4,3), C(2,5) and D(-3,0) form a rectangle.