Triangle XYZ has vertices X(1, 1) , Y(3, 4) and Z(4, 3) . Complete the vertices of the image of XYZ after a rotation of 270° clockwise about X.

Answer 1

IF you were rotating around the origin, then for every 90° you rotate, you switch the values of "x" and "y" around. Additionally, you need to adjust the sign of the "x" and "y" as appropriate for the quadrant you end up in. So, for instance, IF you were rotating the point (2, 3) around the origin clockwise 90°, you would be starting in Q1 and ending in Q4. You would switch the "x" and "y" values once to get (3, 2) but Q4 has positive "x" values but negative "y" values so the point would need to be (3, -2). Rotating it another 90° clockwise would put the point in Q3 where it would become (-2, -3) because we switched the "x" and "y" again and points in Q3 have both negative "x" and negative "y" values. And so forth.

Since you are rotating around a different point than the origin, the easiest thing to do is to create a new "fake" origin at the point you are rotating around. In this case that is point "X". That would temporarily change the points Y and Z because the new origin is 1 unit closer to them in both the "x" and "y" directions. So the new temporary location for point Y is (2, 3) and for point Z it is (3, 2). Now rotate by swapping the "x" and "y" values 3 times (because you are rotating 90°, 3 times). The new Y would be (-3, 2) and the new Z would be (-2, 3).

But now you need to move them back to the original coordinate plane by adding 1 to each value (you subtracted 1 from each to move them to the temporary coordinate plane). So Y' is really at (-2, 3) and Z' is at (-1, 4)

Answer 2

Triangle XYZ has vertices X(1, 1) , Y(3, 4) and Z(4, 3) . Complete the vertices of the image of XYZ after a rotation of 270° clockwise about X.

Rotation is a little tricky.

The first step you'd want to take is translate X back to the origin, you always want the point you are rotating about at the origin. To get X to the origin we will translate down 1 and left 1. Remember this! We will have to translate back the same amount.

X(1-1,1-1) = X(0,0)

Y(3-1,4-1) = Y(2,3)

Z(4-1,3-1) = Z(3,2)

Now the formula for clockwise rotation around the origin is

x′=x⋅cos(θ)−y⋅sin(θ)

y'=x ⋅sin(θ)+y⋅cos(θ)

Where x' is our new rotated x value and y' is our new rotated y value.

You can come up with these formulas on your own with some Pythagorean Theory magic, for counter clockwise rotations sine and cosine will have their signs reversed. For the purposes of your question I will tell you

cos(270 degrees) = 0

sin(270 degrees) = 1

So after simplifying we end up with the formulas

x′= −y

y'= x

However, at the very beginning of the problem we translated everything down 1 and left 1. We have to make sure to undo this translation shifting up 1 and right 1, we can do this by adding 1 to our new x' and new y' values.

Our new formulas become

x'= -y + 1

y'= x + 1

From here we can plug our values in X(0,0), Y(2,3), Z(3,2)

X'( x', y' )

X'( [-y+1], [x+1] )

X'( [-0+1], [0+1] )

X'( 1,1)

We see X remains unchanged at (1,1), this is expected as it is the point you are rotating around.

Y'( x', y' )

Y'( [-y+1], [x+1] )

Y'( [-3+1], [2+1] )

Y'( -2,3)

Y is rotated to (-2,3)

Z'( x', y' )

Z'( [-y+1], [x+1] )

Z'( [-2+1], [3+1] )

Z'( -1,4)

Z is rotated to (-1,4)

X'(1,1), Y'(-2,3), and Z'(-1,4)