Question

Question 3
What type of quadrilateral do the points A(2,-2), B(7,3) C(11,-1) and D(6,-6) taken in that order form?

Answer 1

To find the type of quadrilateral, we find the length of all four sides as well as two diagonals and see whatever condition of quadrilateral is satisied by these sides as well as diagonals.

Now, using distance formula between two points,
$AB=\sqrt{(7-2{)}^2+(3+2{)}^2}=\sqrt{(5{)}^2+(5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\left[\begin{array}{c}\because \text{distance between the points}(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]BC=\sqrt{(11-7{)}^2+(-1-3{)}^2}=\sqrt{(4{)}^2+(-4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}CD=\sqrt{(6-11{)}^2+(-6+1{)}^2}=\sqrt{(-5{)}^2+(-5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}AndDA=\sqrt{(2-6{)}^2+(-2+6{)}^2}=\sqrt{(-4{)}^2+(4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}Distance,AC=\sqrt{(11-2{)}^2+(-1+2{)}^2}=\sqrt{(9{)}^2+(1{)}^2}=\sqrt{81+1}=\sqrt{82}AndBD=\sqrt{(6-7{)}^2+(-6-3{)}^2}=\sqrt{(-1{)}^2+(-9{)}^2}=\sqrt{1+81}=\sqrt{82}$
Here, we see that the sides AB = CD and BC = DA
Also, diagonals are equal, i.e, AC = BD
which shows the quadrilateral is a rectangle.