# Two vertices of a triangle are (4, 3) and (3, 8). If the triangle is equilateral what is the third vertex?

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### 1 Answer

As the triangle is equilateral, the length of all the sides is the same.

The distance between two points (x1, y1) and (x2, y2) is given by sqrt[(x1 - x2)^2 + (y1 - y2)^2].

The vertices given to us are (4, 3) and (3, 8). The distance between these points is sqrt[(4 - 3)^2 + (8 - 3)^2] = sqrt[1 + 25] = sqrt 26

Let the vertex to be determined be (x, y). It is at a distance of sqrt 26 from (4, 3) and (3, 8)

=> sqrt[ (x - 4)^2 + (y - 3)^2] = sqrt[(x - 3)^2 + (y - 8)^2]= sqrt 26

=> (x - 4)^2 + (y - 3)^2 = 26...(1)

and (x - 3)^2 + (y - 8)^2 = 26...(2)

Now we have to solve the two equations for x and y.

(1) - (2)

=> (x - 4 - x + 3)(x - 4 + x - 3) + (y - 3 - y + 8)(y - 3 + y - 8) = 0

=> -1(2x - 7) + 5(2y - 11) = 0

=> -2x + 7 + 10y - 55 = 0

=> x = 5y - 24

substitute in (1)

(5y - 24 - 4)^2 + (y - 3)^2 = 26

=> (5y - 28)^2 + (y - 3)^2 = 26

=> 25y^2 - 280y + 784 + y^2 - 6y + 9 = 26

=> 26y^2 - 286y + 767 = 0

y1 = (11 - sqrt 3)/2

y2 = (11 + sqrt 3)/2

x1 = 5y - 24 = 55/2 - 5*(sqrt 3)/2 - 24 = 7/2 - 5*(sqrt 3)/2

x2 = 55/2 + 5*(sqrt 3)/2 - 24 = 7/2 + 5*(sqrt 3)/2

**The required third vertex can be ((7 + 5*(sqrt 3))/2, (11 + sqrt 3)/2) or ((7- 5*(sqrt 3))/2, (11 - sqrt 3)/2)**