9514 1404 393
Answer:
A. x^2 +(y -7)^2 = 625
B. S'(39, 20)
C. P'(19, 35)
D. C'(56.5, 17.5)
Explanation:
Part A: The infinite number of points with those coordinates will be the solutions to the equation of a circle centered at Q with radius 25. The standard form equation of a circle with center (h, k) and radius r is ...
(x -h)^2 +(y -k)^2 = r^2
For (h, k) = (0, 7) and r=25, the equation for the points of interest is ...
x^2 +(y -7)^2 = 625
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Part B: It is useful to recognize ΔROQ as a right triangle with sides that are the Pythagorean triple (7, 24, 25). (You can figure side RO using the Pythagorean theorem if you're not familiar with this triple.) This means the coordinates of R are (24, 0) and the horizontal distance to the line x=39 is 39-24=15 units.
If we name point X(39, 0), then triangle RXS' has sides RX = 15 and RS' = 25. We recognize these as a multiple of the Pythagorean triple (3, 4, 5), so we know that XS' = 20, (Again, you can figure this using the Pythagorean theorem if you're not familiar with the relevant triples.)
The coordinates of S' are (39, 20).
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Part C:
Point S' is 15 right and 20 up from R. A vector at right angles to RS' will have the coordinates swapped, and one of them negated. Point P' will be 20 left and 15 up from point S', so ...
P' = S' +(-20, 15) = (39, 20) +(-20, 15) =
P' = (19, 35)
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Part D: After rotation, the center (C) of the square is the midpoint of RP', so is ...
C = (R +P')/2 = ((24, 0) +(19, 35))/2 = (43, 35)/2 = (21.5, 17.5)
Reflection over the line x=39 is accomplished by the transformation ...
(x, y) ⇒ (2·39 -x, y)
C(21.5, 17.5) ⇒ C'(78 -21.5, 17.5) = C'(56.5, 17.5)