Question

the third vertex of an equilateral triangle whose other two vertices are (1,1) and (-1,-1) respectively, is :

Answer 1

hey mate
here's the solution....

Answer 2

Hence the third vertex of an equilateral triangle is (√31,√-31).

Step-by-step explanation:

GIVEN:

Let tquilateral triangle ABC, A(x,y),   B(1,1), C(-1,-1)

To find: Co-ordinates of A, x & y

By distance formula, we calculate AB, BC & AC. Then we equate them to get x & y.

Distance between 2 points

$\sqrt{( x_1-x_2)^2 + ( y_1-y_2)^2}$   where$(x_1, y_1) &( x_2,y_2)$ are coordinates of 2 given points

So, in the given triangle ABC,

$BC=\sqrt{(1-(-1))^2+(1-(-1))^2}$

$BC=\sqrt{(2)^2+(2)^2}$

$BC=\sqrt{8}$                                [1]

$AB=\sqrt{(x-1)^2+(y-1)^2}$          

$AB=\sqrt{(x^2-2x+1+y^2-2y+1}$

$AB=\sqrt{x^2+y^2-2x-2y+2}$        [2]

$AC=\sqrt{(x-(-1))^2+(y-(-1))^2}$

$AC=\sqrt{(x+1)^2+(y+1)^2}$

$AC=\sqrt{x^2+2x+1+y^2+2y+1}$

$AC=\sqrt{x^2+y^2+2x+2y+2}$         [3]

Let's equate (1) & (3) AC = BC

⇒$AC^2 =BC^2$

$(\sqrt{x^2+y^2+2x+2y+2})^2=(\sqrt{8})^2$  

$x^2+y^2+2x+2y+2=(8)^2$

$x^2+y^2+2x+2y+2=64$

$x^2+y^2+2x+2y=64-2$

$x^2+y^2+2x+2y=62$                          [4]

Now, we equate(2) & (3) AB = AC

⇒$AB^2 = AC^2$

$(\sqrt{x^2+y^2-2x-2y+2})^2=(\sqrt{x^2+y^2+2x+2y+2})^2$

$x^2+y^2-2x-2y+2=x^2+y^2+2x+2y+2$

$x^2+y^2-2x-2y+2=x^2+y^2+2x+2y+2$

-2y-2y=2x+2x

-4y=4x

y=-x                                                                   [5]

Now, put the value of y in equation (4) , we get,

$x^2+y^2+2x+2y=62$  

$x^2+(-x)^2+2x+2(-x)=62$

$x^2+x^2+2x-2x=62$  

$2x^2=62$

$x^2=31$

$x=\sqrt{31}$

Now, get y by eq (5)

y=-x

$y=-(\sqrt{31} )$

$y=-\sqrt{31}$

Hence the third vertex of an equilateral triangle is (\sqrt{31},-\sqrt{31}).