Question

Name the type of quadrilateral formed, if any by the following points and give reasons for your answer.
(i)$$(-1,-2),(1,0),(-1,2),(-3,0)$$
(ii)$$(-3,5),(3,1),(0,3),(-1,-4)$$
(iii)$$(4,5),(7,6),(4,3),(1,2)$$

Answer 1

Hint: Here we can use distance formula to calculate distance between the two points. After calculating all the distances required. So, that we can verify the shape of the quadrilateral formed using those specific points.

Formula used:
Distance between the two points = $$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$$ i.e. Distance Formula.

Complete step-by-step answer:
(i) The points given in question are $$A=\left(-1,-2\right)$$ , $$B=\left(1,0\right)$$ , $$C=\left(-1,2\right)$$ , and $$D=\left(-3,0\right)$$ Then,
Distance between two points can be calculate by formula $$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$$ .
AB = $$\sqrt{{\left(1+1\right)}^2+{\left(0+2\right)}^2}$$
$$\Rightarrow \sqrt{{\left(2\right)}^2+{\left(2\right)}^2}$$
 $$\Rightarrow \sqrt{4+4}$$
$$\Rightarrow \sqrt{8}=2\sqrt{2}$$
BC = $$\sqrt{{\left(-1-1\right)}^2+{\left(2+0\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-2\right)}^2+{\left(2\right)}^2}$$
 $$\Rightarrow \sqrt{4+4}$$
$$\Rightarrow \sqrt{8}=2\sqrt{2}$$
CD = $$\sqrt{{\left(-3+1\right)}^2+{\left(0-2\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}$$
 $$\Rightarrow \sqrt{4+4}$$
$$\Rightarrow \sqrt{8}=2\sqrt{2}$$
DA = $$\sqrt{{\left(-1+3\right)}^2+{\left(0+2\right)}^2}$$
$$\Rightarrow \sqrt{{\left(2\right)}^2+{\left(2\right)}^2}$$
 $$\Rightarrow \sqrt{4+4}$$
$$\Rightarrow \sqrt{8}=2\sqrt{2}$$
AC = $$\sqrt{{\left(-1+1\right)}^2+{\left(-2-2\right)}^2}$$
$$\Rightarrow \sqrt{{\left(0\right)}^2+{\left(-4\right)}^2}$$
 $$\Rightarrow \sqrt{0+16}$$
$$\Rightarrow \sqrt{16}=\sqrt{4}$$
BD = $$\sqrt{{\left(-3-1\right)}^2+{\left(0-0\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-4\right)}^2+{\left(0\right)}^2}$$
 $$\Rightarrow \sqrt{16+0}$$
$$\Rightarrow \sqrt{16}=4$$
As, all the four sides that are AB, BC, CD and DA are equal and the diagonals AC and BD are equal.
Hence, Quadrilateral ABCD is a square.

(ii) The points given in question are $$A=\left(-3,5\right)$$ , $$B=\left(3,1\right)$$ , $$C=\left(0,3\right)$$ , and $$D=\left(-1,-4\right)$$ Then,
Distance between two points can be calculate by formula $$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$$ .
AB = $$\sqrt{{\left(3+3\right)}^2+{\left(1-5\right)}^2}$$
$$\Rightarrow \sqrt{{\left(6\right)}^2+{\left(-4\right)}^2}$$
 $$\Rightarrow \sqrt{36+16}$$
$$\Rightarrow \sqrt{52}=2\sqrt{13}$$
BC = $$\sqrt{{\left(0-3\right)}^2+{\left(3-1\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-3\right)}^2+{\left(2\right)}^2}$$
 $$\Rightarrow \sqrt{9+4}$$
$$\Rightarrow \sqrt{13}$$
CD = $$\sqrt{{\left(-1-0\right)}^2+{\left(-4-3\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-1\right)}^2+{\left(-7\right)}^2}$$
 $$\Rightarrow \sqrt{1+49}$$
$$\Rightarrow \sqrt{50}=5\sqrt{2}$$
DA = $$\sqrt{{\left(-3+1\right)}^2+{\left(5+4\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-2\right)}^2+{\left(9\right)}^2}$$
 $$\Rightarrow \sqrt{4+81}$$
$$\Rightarrow \sqrt{85}$$
Here $$AB\ne BC\ne CD\ne DA$$
Hence, It is just a quadrilateral.

(iii) The points given in question are $$A=\left(4,5\right)$$ , $$B=\left(7,6\right)$$ , $$C=\left(4,3\right)$$ , and $$D=\left(1,2\right)$$ Then,
Distance between two points can be calculate by formula $$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$$ .
AB = $$\sqrt{{\left(7-4\right)}^2+{\left(6-5\right)}^2}$$
$$\Rightarrow \sqrt{{\left(3\right)}^2+{\left(1\right)}^2}$$
 $$\Rightarrow \sqrt{9+1}$$
$$\Rightarrow \sqrt{10}$$
BC = $$\sqrt{{\left(4-7\right)}^2+{\left(3-6\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}$$
 $$\Rightarrow \sqrt{9+9}$$
$$\Rightarrow \sqrt{18}=3\sqrt{2}$$
CD = $$\sqrt{{\left(1-4\right)}^2+{\left(2-3\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-3\right)}^2+{\left(-1\right)}^2}$$
 $$\Rightarrow \sqrt{9+1}$$
$$\Rightarrow \sqrt{10}$$
DA = $$\sqrt{{\left(1-4\right)}^2+{\left(2-5\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-3\right)}^2+{\left(-3\right)}^2}$$
 $$\Rightarrow \sqrt{9+9}$$
$$\Rightarrow \sqrt{18}=3\sqrt{2}$$
AC = $$\sqrt{{\left(4-4\right)}^2+{\left(3-5\right)}^2}$$
$$\Rightarrow \sqrt{{\left(0\right)}^2+{\left(-2\right)}^2}$$
 $$\Rightarrow \sqrt{0+4}$$
$$\Rightarrow \sqrt{4}=2$$
BD = $$\sqrt{{\left(1-7\right)}^2+{\left(2-6\right)}^2}$$
$$\Rightarrow \sqrt{{\left(-6\right)}^2+{\left(-4\right)}^2}$$
 $$\Rightarrow \sqrt{36+16}$$
$$\Rightarrow \sqrt{52}=2\sqrt{13}$$
Here AB=CD, BC=DA. But AC $$\ne$$ BD
Hence the pairs of opposite sides are equal but diagonal are not equal. So, it is a parallelogram.

Note: We can solve these types of questions with the help of distance formula. Once distance between all the points are calculated we can verify the shape using inner diagonals and outer sides like if all sides and both diagonals are equal that means it is a square.