Answer:
The possible coordinates for R are contained in the following expression:
[tex]R(x,y) = (-5.5, -2\pm 6.1)[/tex]
Explanation:
From statement we notice that line segment PQ is parallel to x-axis and by Geometry we know that an equilateral triangles have three side with equal length and three angles of 60°. In consequence, the height of the triangle is parallel to the y-axis and the possible coordinates for point R are contained in this equation:
[tex]R(x,y) = (x_{M,PQ}, y_{M,PQ} \pm l_{PQ}\cdot \sin 60^{\circ})[/tex] (1)
Where:
[tex]x_{M, PQ}[/tex], [tex]y_{M,PQ}[/tex] - Coordinates of the midpoint of the line segment PQ.
[tex]l_{PQ}[/tex] - Side length of the line segment PQ.
The length of the line segment PQ is determined by the Pythagorean Theorem:
[tex]l_{PQ} = \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}[/tex] (2)
If we know that [tex]P(x,y) = (-9, -2)[/tex] and [tex]Q(x,y) = (-2, -2)[/tex], then the length of the line segment PQ is:
[tex]l_{PQ} = \sqrt{[-2-(-9)]^{2}+[(-2)-(-2)]^{2}}[/tex]
[tex]l_{PQ} = 7[/tex]
And the coordinates of the midpoint are, respectively:
[tex]M(x,y) = \frac{1}{2}\cdot (-9, -2) + \frac{1}{2}\cdot (-2, -2)[/tex]
[tex]M(x,y) = \frac{1}{2}\cdot (-11, -4)[/tex]
[tex]M(x,y) = \left(-\frac{11}{2}, -2 \right)[/tex]
Lastly, the possible coordinates of vertex R are, respectively:
[tex]R(x,y) = \left( -\frac{11}{2},-2\pm 7\cdot \sin 60^{\circ} \right)[/tex]
[tex]R(x,y) = (-5.5, -2\pm 6.1)[/tex]
