Two vertices of an equilateral triangle are (10,-4) and (0,6 find the third vertex)

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Geometry An equilateral triangle has three sides of equal length. If two vertices of an equilateral triangle are (0,4) and (0,0) find the third vertex. How many of these triangles are possible?

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Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$. How can one locate the third vertex?

Maybe someone could give me the easy way please.

My attempt:

• Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1$. So M(5,1)$.

• Find the equation of the perpendicular bisector. Slope of $AB = 10/-10 = -1$, Slope $m$ of the perpendicular bisector $= +1$ $$ \Rightarrow y-1 = 1*(x-5)\Rightarrow y = x-4 .$$

• There are $2$ vertices, $C$ and $D$, both on the line $y = x-4$. The altitude of the triangle is $10$. Find the $2$ points on $y = x-4$ which is $10$ units distance from $M(5,1)$: $$d^2 = (\text{difference of } x)^2 + (\text{difference of } y)^2 \Rightarrow 100 = (x-5)^2 + (y-1)^2$$ Sub for $y = x-4$: $$ 100 = (x-5)^2 + (x-5)^2\Rightarrow (x-5)^2 = 50 \Rightarrow x-5 = \pm\sqrt{50}$$ Thus $x = 5 + \sqrt{50}$, $y = 1 + \sqrt{50}$ and so the Vertices are $C$: $x = 5 - \sqrt{50}$, $y = 1 - \sqrt{50}$ --> Vertex $D$

Two vertices of an equilateral triangle are (10,-4) and (0,6). Find the third vertex ------------- Label the points A(10,-4) and B(0,6) Find the length of AB and the midpoint, label it M --- Length of AB = sqrt(diffy^2 + diffx^2) = sqrt(100+100) = 10sqrt(2) --- Find the average of x & y separately. (10+0)/2 = 5 (-4+6)/2 = 1 M(5,1) ----- Find the equation of the perpendicular bisector. Slope of AB = 10/-10 = -1 Slope m of the perpendicular bisector = +1 --> y-1 = 1*(x-5) y = x-4 ------- There are 2 vertices, C & D, both are on the line y = x-4 -- The altitude of the triangle is 10. Find the 2 points on y = x-4 10 units distance from M(5,1) --- d^2 = diffx^2 + diffy^2 100 = (x-5)^2 + (y-1)^2 Sub for y 100 = (x-5)^2 + (x-5)^2 (x-5)^2 = 50 x-5 = �sqrt(50) --- x = 5 + sqrt(50), y = 1 + sqrt(50) --> Vertex C x = 5 - sqrt(50), y = 1 - sqrt(50) --> Vertex D