A ball game between Dylan and Mya is about to begin. Using sidewalk chalk, they draw their “arena” on the ground. The arena can be modeled by the equation x/2 + y/2 = 81 . Dylan has taken his position at (-5, -2) and Mya has taken her position at (6, 4). Each graph unit represents 1 meter.
what are the points
How far apart are Dylan and Mya (to two decimal places)?Jordyn joins the game so that she forms a triangle with Dylan and Mya. Her exact coordinates are NOT important, but we do know that from Jordyn’s position, the angle formed to Dylan and Mya is 76°. From Dylan’s position, the angle formed to Mya and Jordyn is 50°.
Draw a diagram below with this information and the information from Tasks 1 and 2.
Use this diagram to find the distances from Jordan to Dylan and Jordyn to Mya (to two decimal places).
Remember: DO NOT assign Jordyn coordinates. Place Jordyn somewhere so that the angles above make sense.
To find the distance between Dylan and Mya, we can use the distance formula. The coordinates of Dylan are (-5, -2) and the coordinates of Mya are (6, 4).
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Plugging in the values, we get:
d = sqrt((6 - (-5))^2 + (4 - (-2))^2)
= sqrt(11^2 + 6^2)
= sqrt(121 + 36)
= sqrt(157)
≈ 12.53 meters
So, Dylan and Mya are approximately 12.53 meters apart.
To find the distances from Jordyn to Dylan and Jordyn to Mya, we can use the law of cosines. Let's assume Jordyn's position as (x, y).
From Dylan's position, the angle formed to Mya and Jordyn is 50°. This means that the angle between the vectors (6, 4) - (-5, -2) and (x, y) - (-5, -2) is 50°.
From Mya's position, the angle formed to Dylan and Jordyn is 76°. This means that the angle between the vectors (-5, -2) - (6, 4) and (-5, -2) - (x, y) is 76°.
Using the dot product formula, the cosine of an angle between two vectors is given by:
cos(angle) = (A dot B) / (|A| * |B|)
Let's calculate the dot product and magnitudes of the vectors.
For the vector (6, 4) - (-5, -2):
A = (6 - (-5), 4 - (-2)) = (11, 6)
|A| = sqrt(11^2 + 6^2) = sqrt(121 + 36) = sqrt(157)
For the vector (x, y) - (-5, -2):
B = (x - (-5), y - (-2)) = (x + 5, y + 2)
|B| = sqrt((x + 5)^2 + (y + 2)^2)
For the vector (-5, -2) - (6, 4):
C = (-5 - 6, -2 - 4) = (-11, -6)
|C| = sqrt((-11)^2 + (-6)^2) = sqrt(121 + 36) = sqrt(157)
For the vector (-5, -2) - (x, y):
D = (-5 - x, -2 - y)
|D| = sqrt((-5 - x)^2 + (-2 - y)^2)
Using the dot product formula for the 50° angle:
cos(50°) = (11 * (x + 5) + 6 * (y + 2)) / (sqrt(157) * sqrt((x + 5)^2 + (y + 2)^2))
Using the dot product formula for the 76° angle:
cos(76°) = (-11 * (-5 - x) + (-6) * (-2 - y)) / (sqrt(157) * sqrt((-5 - x)^2 + (-2 - y)^2))
To make sense of the diagram, we can place Jordyn somewhere near the midpoint of Dylan and Mya. Since we don't need the exact coordinates, we can estimate Jordyn's position as (0.5, 1).
Using these estimated coordinates, we can calculate the distances from Jordyn to Dylan and Jordyn to Mya using the distance formula.
Distance from Jordyn to Dylan:
d1 = sqrt((0.5 - (-5))^2 + (1 - (-2))^2)
= sqrt(5.5^2 + 3^2)
≈ 6.27 meters
Distance from Jordyn to Mya:
d2 = sqrt((0.5 - 6)^2 + (1 - 4)^2)
= sqrt((-5.5)^2 + (-3)^2)
≈ 6.27 meters
So, the distances from Jordyn to Dylan and Jordyn to Mya are both approximately 6.27 meters.