Question

Victor draws one side of equilateral APQR on the coordinate plane at points P(-9, -2) and Q (-2,-2). What are the two possible coordinates of vertex R? Round to the nearest tenth.​

Answer 1

Answer:

The possible coordinates for R are contained in the following expression:

$R(x,y) = (-5.5, -2\pm 6.1)$

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:

$R(x,y) = (x_{M,PQ}, y_{M,PQ} \pm l_{PQ}\cdot \sin 60^{\circ})$ (1)

Where:

$x_{M, PQ}$, $y_{M,PQ}$ - Coordinates of the midpoint of the line segment PQ.

$l_{PQ}$ - Side length of the line segment PQ.

The length of the line segment PQ is determined by the Pythagorean Theorem:

$l_{PQ} = \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}$ (2)

If we know that $P(x,y) = (-9, -2)$ and $Q(x,y) = (-2, -2)$, then the length of the line segment PQ is:

$l_{PQ} = \sqrt{[-2-(-9)]^{2}+[(-2)-(-2)]^{2}}$

$l_{PQ} = 7$

And the coordinates of the midpoint are, respectively:

$M(x,y) = \frac{1}{2}\cdot (-9, -2) + \frac{1}{2}\cdot (-2, -2)$

$M(x,y) = \frac{1}{2}\cdot (-11, -4)$

$M(x,y) = \left(-\frac{11}{2}, -2 \right)$

Lastly, the possible coordinates of vertex R are, respectively:

$R(x,y) = \left( -\frac{11}{2},-2\pm 7\cdot \sin 60^{\circ} \right)$

$R(x,y) = (-5.5, -2\pm 6.1)$