Question

If two vertices of an equilateral triangle are (3,0) and (6,0) find the third vertex

Answer 1

Answer:

Step-by-step explanation:

Given  

If two vertices of an equilateral triangle are (3,0) and (6,0) find the third vertex

Let Δ ABC be an equilateral triangle. So AB = BC = CA

AB^2 = BC^2 = CA^2

(3 – 6)^2 + (0 – 0)^2 = (x – 6)^2 + (y – 0)^2 = (x – 3)^2 + (y – 0)^2

(- 3)^2 = (x – 6)^2 + y^2 = (x – 3)^2 + y^2

9 = (x – 6)^2 + y^2 = (x – 3)^2 + y^2

(x – 6)^2 + y^2 = (x – 3)^2 + y^2

So (x – 6)^2  = (x – 3)^2  

So by taking square root we get

x – 6 = ± (x – 3)

x – 6 = - (x – 3)

       = - x + 3

We get 0 = 3 is not possible.

So x + x = 3 + 6

2x = 9

x = 9/2

x = 4.5

Again we have

(x – 6)^2 + y^2 = 9

(4.5 – 6)^2 + y^2 = 9

(-1.5)^2 + y^2 = 9

2.25 + y^2 = 9

y ^2 = 6.75

y = ± 2.598 = ± 2.6

Now coordinates of point C are (4.5 , 2.6) , (4.5 , - 2.6)