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

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

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

Step-by-step 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) = (1)/(2)\cdot (-9, -2) + (1)/(2)\cdot (-2, -2)`

`M(x,y) = (1)/(2)\cdot (-11, -4)`

`M(x,y) = \left(-(11)/(2), -2 \right)`

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

`R(x,y) = \left( -(11)/(2),-2\pm 7\cdot \sin 60^(\circ) \right)`

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