Question

show that the points (1,7), (4,2), (-1,-1) ans (-4,4) are the vertices of a square​

Answer 1

Answer:

Given points are (1,7),(4,2),(-1,-1),(-4,4)

Let the points are A,B,C,D.

Distance formula = √(x₂-x₁)²+(y₂-y₁)²

                   AB = √(4-1)²+(2-7)²

                         = √9+25

                         = √34

                    BC = √(-1-4)²+(-1-2)²

                          = √25+9

                          =  √34

                     CD = √(-4-(-1))²+(4-(-1))²

                           = √9+25

                           =  √34

                      DA = √(1-(-4))²+(7-4)²

                            = √25+9

                            = √34

We know that the all sides of the square are equal.

Here by the distance formula, we got the for sides equal .

So,from this we can say that these points are the vertices of a square.

Answer 2

Answer:

Step-by-step explanation:

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

Slope $m_{AB}$ = $\frac{2-7}{4-1} = -\frac{5}{3}$ and $m_{BC} = \frac{-1-2}{-1-4} = \frac{3}{5}$

$m_{AB} m_{BC} = - 1$ ⇒ AB⊥BC

$m_{CD} = \frac{-1-4}{-1+4} = -\frac{5}{3}$ and $m_{AD} = \frac{4-7}{-4-1} = \frac{3}{5}$ ⇒ AD⊥CD

AB = √[(1-4)²+(7-2)²] = √34

BC = √[(-1-4)²+(-1-2)²] = √34

CD = √[(-4+1)²+(4+1)²] = √34

AD = √[(-4-1)²+(4-7)²] = √34

Thus, ABCD is a square.