An equilateral triangle on the coordinate plane has vertices at (-1,2) and (4,2). What are the possible coordinates of the third vertex rounded to the nearest tenth?

Answer 1

It is an equilateral triangle; Points are A(-1,2) and B(4,2) ; Line AB is || to x- axis hence mid point of AB is (1.5,2) , distance AB is 5[ no need to use distance formula here]; Hence point C , the third vertex will be on the perpendicular bisector of AB => x value of third vertex is 1.5; Let the third vertex is C(1.5, k); By distance formula we get AC = SQRT[(-1-1.5)^2 + (2-k)^2)] which is 5; Solving this equation we get k as 6.33 and -2.33;

Hence the vertex C is (1.5, 6,33) or (1.5, -2.33)

Answer 2

The coordinates given are a straight line 5 units in length so all sides are 5 units.

You know that the midpoint of the 2 points is x = 1.5, So then you need to setup the distance formula with an unknown value for y that satisfies the distance formula.

You find that the possible candidates for y are 6.33 or -2.33 & both are valid for this problem.

Distance formula = SQRT( (x₂-x₁)² + (y₂-y₁)² ) = 5 where the x's are known and one set of y values are known, you solve for the unknown y values.