Question

If two opposite vertices of a square are (3,4) and (1,-1) , find the coordinates of the other two vertices

Answer 1

(4,3)&(-1,1)...
I guess this the answer...

Answer 2

Vertices of a square are $(\frac{9}{2},\frac{1}{2})$ and $(-\frac{1}{2},\frac{5}{2})$

Step-by-step explanation:

Given : If two opposite vertices of a square are (3,4) and (1,-1).

To find : The coordinates of the other two vertices ?

Solution :

Let ABCD be a square.  

A(3, 4) and C(1. -1)

Let coordinates of a point B(x, y)

AB = BC

Squaring on both sides,

$AB^2= BC^2$

Using distance formula, $d^2=(y_2-y_1)^2+(x_2-x_1)^2$

$(x - 3)^2 + (y - 4)^2 = (x - 1)^2 + (y + 1)^2$

$x^2+9-6x+y^2+16-8y=x^2+1-2x+y^2+1+2y$

$x^2+9-6x+y^2+16-8y-x^2-1+2x-y^2-1-2y=0$

$4x+10y=23$

Length of hypotenuse is given by,

$H=\sqrt2 s$

$AC=\sqrt2 AB$

Squaring both side,

$AC^2=2AB^2$

Again use distance formula,

$(3 - 1)^2 + (4 + 1)^2=2[(x - 3)^2 + (y - 4)^2]$

$(2)^2 + (5)^2=2[x^2]$

$\frac{29}{2}=(x - 3)^2 + (y - 4)^2$

Solving this we get, $x=\frac{9}{2},-\frac{1}{2}$

When $x=\frac{9}{2}$

$y=\frac{1}{2}$

When $x=-\frac{1}{2}$

$y=\frac{5}{2}$

Vertices of a square are $(\frac{9}{2},\frac{1}{2})$ and $(-\frac{1}{2},\frac{5}{2})$

#Learn more

Straight lines

https://brainly.in/question/7177958, Answered by Sprao534