Question

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:

Answer 1

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   $\pm\sqrt{4h^2-g^2}$.

Answer 2

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