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

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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)