Step-by-step explanation:
Let imagine a circle with a center 5,3.
Let our circle have a radius, that passes through 1,2.
We need to find the length of r or distance from 5,3 to 1,2.
[tex]d = \sqrt{(3 - 2) {}^{2} + (5 - 1) {}^{2} } [/tex]
[tex]d = \sqrt{1 {}^{2} + {4}^{2} } [/tex]
[tex]d = \sqrt{17} [/tex]
So the radius is sqr root of 17.
So now we have a circle that has a radius of sqr root of 17 and a center 5,3. Let have a line passing through 5,3 and 1,2 as well.
The slope is 1/4 so
[tex]y = \frac{1}{4} x + \frac{7}{4} [/tex]
Look at first photo
Now,we are rotating 90 degrees So we need to find a line that is perpendicular to our orginal line and that pass through the center (5,3)
Negative reciprocal slope of 1/4 is -4 and our point is 5,3 so
[tex]y - 3 = - 4(x - 5)[/tex]
[tex]y = - 4x + 23[/tex]
The point that intersects the circle will be the new image after we rotate (1,2) 90 degrees, which is the 2nd photo.
To find the next rotated point of (2,3), the distance of the preimage (2,3) and new image (5,3) is 3 so that leaves
(5,0)
So our new rotated points are
(6,-1)
(5,0)