QUESTION IMAGE
Question
quinton tried to transform triangle fgh according to the rule ((x, y) \to (-y, x)). which best describes his attempt?
correct. he transformed the triangle according to the rule ((x, y) \to (-y, x))
incorrect. he transformed the triangle according to the rule ((x, y) \to (y, -x))
incorrect. he transformed the triangle according to the rule ((x, y) \to (-x, -y))
incorrect. he transformed the triangle according to the rule ((x, y) \to (-y, -x))
Step1: Identify Coordinates
First, find the coordinates of triangle FGH and its image \( F'G'H' \). Let's assume:
- \( F(3, 2) \), \( G(1, 2) \), \( H(5, 4) \) (from the graph)
- \( F'(-2, 3) \), \( G'(-2, 1) \), \( H'(-4, 5) \) (from the image)
Step2: Apply Rule \( (x, y) \to (-y, x) \)
Check transformation for \( F(3, 2) \):
\( (-y, x) = (-2, 3) \), which matches \( F' \).
For \( G(1, 2) \):
\( (-y, x) = (-2, 1) \), matches \( G' \).
For \( H(5, 4) \):
\( (-y, x) = (-4, 5) \), matches \( H' \). Wait, no—wait, the correct rule was supposed to be \( (x, y) \to (-y, x) \)? Wait, no, the problem says Quinton tried to use \( (x, y) \to (-y, x) \), but let's re - check. Wait, maybe I misread. Wait, the options: the correct one is "Incorrect. He transformed the triangle according to the rule \( (x, y) \to (-y, x) \)"? No, wait, the first option says "Correct. He transformed the triangle according to the rule \( (x, y) \to (-y, x) \)". Wait, no, let's re - evaluate.
Wait, actually, when we apply \( (x,y)\to(-y,x) \) to \( F(3,2) \): \( x = 3,y = 2 \), so \( -y=-2,x = 3 \), so \( (-2,3) \), which is \( F' \). For \( G(1,2) \): \( -y=-2,x = 1 \), so \( (-2,1) \), which is \( G' \). For \( H(5,4) \): \( -y=-4,x = 5 \), so \( (-4,5) \), which is \( H' \). Wait, but the problem says "Quinton tried to transform triangle FGH according to the rule \( (x, y) \to (-y, x) \). Which best describes his attempt?" Wait, the first option is "Correct. He transformed the triangle according to the rule \( (x, y) \to (-y, x) \)". But wait, maybe I made a mistake in coordinates. Wait, maybe the original triangle has different coordinates. Let's re - check the graph.
Looking at the graph, the original triangle FGH: Let's assume the x - axis and y - axis. Let's take G: G is at (1,2), F at (3,2), H at (5,4). The image \( G' \) is at (-2,1), \( F' \) at (-2,3), \( H' \) at (-4,5). Wait, when we apply \( (x,y)\to(-y,x) \) to G(1,2): \( -y=-2,x = 1 \), so (-2,1) which is \( G' \). To F(3,2): \( -y=-2,x = 3 \), so (-2,3) which is \( F' \). To H(5,4): \( -y=-4,x = 5 \), so (-4,5) which is \( H' \). So actually, he did it correctly? But the options: Wait, no, maybe the problem is that the rule he was supposed to use was different? Wait, no, the question is "Which best describes his attempt?" and the first option is "Correct. He transformed the triangle according to the rule \( (x, y) \to (-y, x) \)".
Wait, maybe I messed up the coordinates. Let's re - assign coordinates properly. Let's look at the grid:
For the original triangle (lower triangle):
- G: Let's say x = 1, y = 2 (since it's 1 unit right on x - axis, 2 units up on y - axis)
- F: x = 3, y = 2 (3 units right, 2 units up)
- H: x = 5, y = 4 (5 units right, 4 units up)
For the image triangle (upper triangle, in the negative x - region):
- \( G' \): x=-2, y = 1 (wait, no, the y - axis on the image: wait, the grid has x and y axes. Wait, maybe the y - axis is flipped. Wait, maybe the coordinates are:
Original F: (3,2) (x = 3, y = 2)
Image F': (-2,3) (x=-2, y = 3)
Applying \( (x,y)\to(-y,x) \): for F(3,2), - y=-2, x = 3, so (-2,3), which matches F'.
For G(1,2): - y=-2, x = 1, so (-2,1), which matches G' (if G' is (-2,1)).
For H(5,4): - y=-4, x = 5, so (-4,5), which matches H' (if H' is (-4,5)).
So the transformation \( (x,y)\to(-y,x) \) maps F to F', G to G', H to H'. So Quinton's attempt was correct? But the option says "Correct. He transformed the triangle according to the rule \( (x, y) \to (-y, x) \)". Wait, but the other options:
- "Incorrect. He transformed the triangle according to the ru…
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Correct. He transformed the triangle according to the rule \((x, y) \to (-y, x)\)