Question

if (3,2) and(-3,2) are two vertices of an equilateral triangle which contains the origin,find the third vertex.​

Answer 1

Answer:

Coordinate of third vertex is ( 0 , -3.2)

Step-by-step explanation:

Given: Coordinates of two vertices of a equilateral triangle ABC

           A( 3 , 2 )  and B( -3 , 2 )

           ΔABC contain origin point i.e., ( 0 , 0 )

To find: Coordinate of third vertex C.

Let coordinates of 3rd vertex be ( x , y )

All sides of equivalent triangle are equal in length.

Distance formula is given by,

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So, we have

CB = AB

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

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

$\sqrt{(x+3)^2+(y-2)^2}=\sqrt{36}$

Squaring both sides,

$(x+3)^2+(y-2)^2=36$ ........................... (1)

AC = AB

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

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

$\sqrt{(x-3)^2+(y-2)^2}=\sqrt{36}$

Squaring both sides,

$(x-3)^2+(y-2)^2=36$ ........................... (2)

Subtract eqn (2) from eqn (1)

$(x+3)^2+(y-2)^2-((x-3)^2+(y-2)^2)=36-36$

$(x+3)^2+(y-2)^2-(x-3)^2-(y-2)^2=0$

$(x+3)^2-(x-3)^2=0$

$x^2+9+6x-(x^2+9-6x)=0$

$x^2+9+6x-x^2-9+6x=0$

12x = 0

x = 0

put this vale in eqn(1).

we get

$(0+3)^2+(y-2)^2=36$

$9+(y-2)^2=36$

y² + 4 - 4x = 36 - 9

y² - 4x + 4 - 27 = 0

y² - 4x -23 = 0

using quadratic formula,

$y=\frac{-(-4)\pm\sqrt{(-4)^2-4\times1\times(-23)}}{2\times1}$

$y=\frac{4\pm\sqrt{16+92}}{2}$

$y=\frac{4\pm\sqrt{108}}{2}$

$y=\frac{4\pm10.39}{2}$

$y=\frac{4+10.39}{2}\;\;and\:\:y=\frac{4-10.39}{2}$

y = 7.195 and y = -3.195

Since triangle contain origin

⇒ Negative value is accepted

⇒ Coordinates of C( 0 , -3.195 )

Therefore, Coordinate of third vertex is ( 0 , -3.2)