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The x-axis contains the base of an equilateral triangle RST. The origin is at S. Vertex T has coordinates (2h, 0) and the y-coordinate of R is g, with g > 0.


Enter the coordinates for the midpoint of ST:


Enter the x- coordinate of R:

Verified Answer :

pkmnmushy

For the first part remember that an equilateral triangle is a triangle in which all three sides are equal & all three internal angles are each 60°. So x-coordinate of R is in the middle of ST = (1/2)(2h-0) = h
And for the second  since this is an equilateral triangle the x coordinate of point R is equal to the coordinate of the midpoint of ST, which you figured out in the previous answer. Hope this works for you

Answer:

Coordinates of Mid Point of ST is ( h , 0 ) and x-coordinate of R is   \(\pm\sqrt{4h^2-g^2}\).

Step-by-step explanation:

Given: Coordinate of S ( 0 , 0 )  ,  Coordinate of T ( 2h , 0 )

           y-coordinate of R = g

Coordinates of Mid point of ST = \((\frac{2h+0}{2},\frac{0+0}{2})\)

                                                    = ( h , 0 )

let x-coordinate of point R be x

Distance of RS = Distance of ST

\(\sqrt{(x-0)^2+(g-0)^2}=\sqrt{(2h-0)^2+(0-0)^2}\)

\(\sqrt{x^2+g^2}=\sqrt{(2h)^2}\)

Squaring both sides, we get

\(x^2+g^2=(2h)^2\)

\(x^2=(2h)^2-g^2\)

\(x=\pm\sqrt{4h^2-g^2}\)

So, x-coordinate of R is \(\pm\sqrt{4h^2-g^2}\)

Therefore, Coordinates of Mid Point of ST is ( h , 0 ) and x-coordinate of R is   \(\pm\sqrt{4h^2-g^2}\).

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