Question

The coordinates of the extremities of one diagonal
of a square are (1, 1) and (-2,-1). Find the coordi-
nates of its other vertices.​

Answer 1

So the other two vertices of the square ABCD is (0.5 , −1.5) and (−1.5 ,1.5 )

Step-by-step explanation:

Given:

Here ABCD is a square and A(1,1) and C(−2,−1) are the given vertices. Let the coordinates of vertex B be (x,y).

And

⇒AB = BC                           [1]

⇒$AB^2=BC^2$

⇒$(x-1)^2+(y-1)^2=(x+2)^2+(y+1)^2$

⇒$x^2+1-2x+y^2+1-2y=x^2+4+4x+y^2+1+2y$

⇒6x + 4y = −3                                    [2]

⇒6x = - 3 - 4y

⇒$x= \frac{-3-4y}{6}$             [3]    

By using pythagoras theorem

⇒ $AC^2=AB^2+BC^2$        

⇒ $AC^2=2AB^2$       [from 1]

⇒ $(-2-1)^2+(-1-1)^2=2(x-1)^2+2(y-1)^2$

⇒ $9+4=2x^2+2-4x+2y^2+2-4y$

⇒ $2x^2+2y^2-4x-4y=9$        

⇒ $2(\frac{-3-4y}{6} )^2+2y^2-4(\frac{-3-4y}{6} )^2-4y=9$   [from 3]

⇒ $2(\frac{9+16y^2+24y}{36} )+2y^2+2+\frac{8y}{3}-4y=9$

⇒ $9+16y^2+24y+36y^2+48y-72y=126$

⇒ $52y^2=126-9$

⇒ $52y^2=117$

⇒ $y=\sqrt{\frac{117}{52} }$

⇒ y = ±1.5

If y = +1.5 put in equation 2

⇒ 6x+4y=−3

⇒ 6x+4(1.5)=−3

⇒ 6x+6=−3

⇒ 6x=−9

⇒ $x= \frac{-9}{6}$

⇒ x = - 1.5

If y= −1.5

⇒ 6x+4(−1.5)=−3

⇒ 6x−6=−3

⇒ 6x=3

⇒ $x=\frac{3}{6}$

⇒ x = 0.5

So the other two vertices of the square ABCD is (0.5 , −1.5) and (−1.5 ,1.5 )