Question

show that the points (1 ,7) ,(4, 2), (-1, -1) and (-4, 4) taken in order are the vertics of a square. Also find its area

Answer 1

Answer:

Step-by-step explanation:

A (1 ,7) ,   B(4, 2),  C(-1, -1) and D(-4, 4)

Find the distance of sides using distance formula.

Distance = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

A (1 ,7) ,   B(4, 2)

$AB = \sqrt{(4-1)^{2}+(2-7)^{2}}\\\\ =\sqrt{3^{2}+(-5)^{2}}\\\\ \\ = \sqrt{9+25}=\sqrt{34} units\\\\\\\\BC = \sqrt{(-1-4)^{2}+(-1-2)^{2}}\\\\ =\sqrt{(-5)^{2}+(-3)^{2}} = \sqrt{25+9}=\sqrt{34} units\\\\\\CD = \sqrt{(-4-[-1])^{2}+(4-[-1])^{2}}\\\\=\sqrt{(-4+1)^{2}+(4+1)^{2}}\\\\=\sqrt{(-3)^{2}+(5)^{2}}\\\\=\sqrt{9+25} = \sqrt{34} units\\\\\\AD = \sqrt{(-4-1)^{2}+(4-7)^{2}}\\\\=\sqrt{(-5)^{2}+(-3)^{2}} =\sqrt{25+9} = \sqrt{34} units$

AB = BC = CD = AD = √34 units. so , ABCD is  a square.

Area of square = side * side

                          = √34 * √34

                          = 34 square units