Question

Find the third vertex of an equilateral triangle whose other two vertices are (1, 1) and
1, -1).

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Answer 1

Answer:

(1±√3,0)

Step-by-step explanation:

There are many methods to do this, but I will choose the simplest one.

In equilateral triangle, all sides are equal.

Using Distance Formula,

Distance from (1,1) and (1,-1)

$=\sqrt{(-1-1)^2+(1-1)^2}\\=\sqrt{(-2)^2+0^2}\\=\sqrt4\\=2$

Let the third vertex be (x,y)

Distance from (1,1) and (x,y)=2 units

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

Also, distance from (1,-1)= 2 units

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

So, we can say that

$\sqrt{x^2+y^2-2x-2y+2}=\sqrt{x^2+y^2-2x+2y+2}\\x^2+y^2-2x-2y+2=x^2+y^2-2x+2y+2\\-2y=2y\\4y=0\\y=0$.......(iii)

Using the value of y, we can find x from (ii)

$2=\sqrt{x^2+0^2-2x+2(0)+2}\\4=x^2-2x+2\\x^2-2x-2=0$

Using Quadratic Formula,

$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-2)}}{2(1)}\\\\=\frac{2\pm\sqrt{4+8}}{2}\\\\=\frac{2\pm\sqrt{12}}{2}\\\\=\frac{2\pm2\sqrt3}{2}\\\\=1\pm\sqrt3$

Thus, the third vertex if the triangle is (1±√3,0)

Hope it helps!

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