Question

If two vertices of an equilateral triangle b
quilateral triangle be (0, 0), (3. 13), find the third vertex.

Answer 1

Step-by-step explanation:

√(3-0)+(13-0)

√3+13

√16

Answer 2

correct question

If two vertices of an equilateral triangle b

quilateral triangle be (0, 0), (3. $\sqrt{3}$), find the third vertex.

solution

we know in an equilateral triangle all the sides are equal .

first let us find the length of side bc by distance formula

BC = $\sqrt{(3-0)^{2} +(\sqrt{3} -0)^{2} }$

   =$\sqrt{9+3}$

  =$2\sqrt{3}$ units

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

      =$\sqrt{x^{2} +y^{2} }$

⇒$\sqrt{x^{2} +y^{2} }$=$2\sqrt{3}$  

⇒$x^{2} +y^{2}$= 12       ........(1)

AC= $\sqrt{(x-3)^{2}+(y-\sqrt{3} )^{2} }$=$2\sqrt{3}$

    ⇒$\sqrt{x^{2}+y^{2}-6x-2\sqrt{3}y+12 }$=$2\sqrt{3}$

    ⇒ 12 -6x-2$\sqrt{3}$y+12= 12           ......using eqn (1)and squaring both sides

     ⇒6x+2$\sqrt{3}$y= 12      .......(2)

now finding the equation of line AB by using slope and the point B

y - 0 = tan(60°)(x-0)

⇒y= $\sqrt{3}$x     putting this is eqn (2) we get ,

       6x +2$\sqrt{3}$*$\sqrt{3}$x=12

    ⇒6x+6x=12

         12x=12

            x=1

⇒ y = $\sqrt{3}$*1

 y=$\sqrt{3}$

hence the third vertex is (0,$\sqrt{3}$)