Answer:
coordinates of H = (1, 0)
Coordinates of G = ( - 3.6, -2) or (5.6, -2)
Step-by-step explanation:
E(5,3) and F (2,-1) are two vertices of a square EFGH and H is in the x-axis.
Let the coordinates of H is (x, 0) and G is (a, b).
The length of side EF is
[tex]EF = \sqrt{(5 -2)^2 + (3 +1)^2} = 5[/tex]
So,
[tex]EH = \sqrt{(5 -x)^2 + (3 -0)^2} = 5\\\\(5 -x)^2+ 9 = 25\\\\5 - x = 4\\\\x = 1[/tex]
And
[tex]GH = \sqrt{(a -x)^2+ b^2} = 5\\\\(a -1)^2+ b^2 = 25\\\\a^2 + b^2 + 1 - 2 a = 25\\\\a^2 + b^2 - 2a = 24 .... (1)[/tex]
Now
[tex]FG = \sqrt{(a -2)^2+ (b +1)^2} = 5\\\\(a -2)^2+ (b+1)^2 = 25\\\\a^2 + b^2 + 2b - 2 a = 20\\ ..... (2)[/tex]
Solving (1) and (2)
b = - 2 ,
[tex]a = \frac{2 \pm\sqrt{4 + 80}}{2}\\\\a = \frac{2 \pm 9.2}{2}\\\\a = - 3.6, 5.6[/tex]
