Question
Triangle JKL is rotated 45° counterclockwise using the origin as the center of rotation.
On a coordinate plane, triangle J K L has points (negative 2, 0), (0, negative 4), (negative 3, negative 5).
Which graph shows the location of triangle J’K’L’?
On a coordinate plane, triangle J prime K prime L prime has points (negative 1.5, negative 1.5), (3, negative 3), (1.2, negative 5.7).
On a coordinate plane, triangle J prime K prime L prime has points (Negative 1.5, 1.8), (negative 3, negative 3), (negative 6, negative 1.2).
On a coordinate plane, triangle J prime K prime L prime has points (negative 4, negative 1), (0, negative 2.2), (negative 1.2, negative 5).
On a coordinate plane, triangle J prime K prime L prime has points (0, negative 2), (4, 0), (5, negative 3).
On a coordinate plane, triangle J K L has points (negative 2, 0), (0, negative 4), (negative 3, negative 5).
Which graph shows the location of triangle J’K’L’?
On a coordinate plane, triangle J prime K prime L prime has points (negative 1.5, negative 1.5), (3, negative 3), (1.2, negative 5.7).
On a coordinate plane, triangle J prime K prime L prime has points (Negative 1.5, 1.8), (negative 3, negative 3), (negative 6, negative 1.2).
On a coordinate plane, triangle J prime K prime L prime has points (negative 4, negative 1), (0, negative 2.2), (negative 1.2, negative 5).
On a coordinate plane, triangle J prime K prime L prime has points (0, negative 2), (4, 0), (5, negative 3).
Answers
GPT-4o mini
To rotate points in the coordinate plane by 45° counterclockwise about the origin, you can use the following transformation formulas:
For a point \((x, y)\):
- The new x-coordinate is given by: \(x' = x \cdot \cos(45°) - y \cdot \sin(45°)\)
- The new y-coordinate is given by: \(y' = x \cdot \sin(45°) + y \cdot \cos(45°)\)
Since \(\cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2}\), we can simplify the transformations to:
- \(x' = \frac{\sqrt{2}}{2}(x - y)\)
- \(y' = \frac{\sqrt{2}}{2}(x + y)\)
Now, let's apply this transformation to each vertex of the triangle JKL:
1. **Point J (-2, 0)**:
- \( x' = \frac{\sqrt{2}}{2}(-2 - 0) = -\sqrt{2} \approx -1.414\)
- \( y' = \frac{\sqrt{2}}{2}(-2 + 0) = -\sqrt{2} \approx -1.414\)
- So J' is approximately \((-1.414, -1.414)\) or \((-1.5, -1.5)\) when rounded.
2. **Point K (0, -4)**:
- \( x' = \frac{\sqrt{2}}{2}(0 + 4) = 2\sqrt{2} \approx 2.828\)
- \( y' = \frac{\sqrt{2}}{2}(0 - 4) = -2\sqrt{2} \approx -2.828\)
- So K' is approximately \((2.828, -2.828)\) or \((3, -3)\) when rounded.
3. **Point L (-3, -5)**:
- \( x' = \frac{\sqrt{2}}{2}(-3 + 5) = \sqrt{2} \approx 1.414\)
- \( y' = \frac{\sqrt{2}}{2}(-3 - 5) = -4\sqrt{2} \approx -5.656\)
- So L' is approximately \((1.414, -5.656)\) or \((1.2, -5.7)\) when rounded.
Now we have the points J'\((-1.5, -1.5)\), K'\((3, -3)\), and L'\((1.2, -5.7)\).
Among the options provided, the graph that matches these new coordinates is the one with the points:
\((-1.5, -1.5)\), \((3, -3)\), and \((1.2, -5.7)\).
Thus, the correct graph to show the location of triangle J’K’L’ would be the first option given.
For a point \((x, y)\):
- The new x-coordinate is given by: \(x' = x \cdot \cos(45°) - y \cdot \sin(45°)\)
- The new y-coordinate is given by: \(y' = x \cdot \sin(45°) + y \cdot \cos(45°)\)
Since \(\cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2}\), we can simplify the transformations to:
- \(x' = \frac{\sqrt{2}}{2}(x - y)\)
- \(y' = \frac{\sqrt{2}}{2}(x + y)\)
Now, let's apply this transformation to each vertex of the triangle JKL:
1. **Point J (-2, 0)**:
- \( x' = \frac{\sqrt{2}}{2}(-2 - 0) = -\sqrt{2} \approx -1.414\)
- \( y' = \frac{\sqrt{2}}{2}(-2 + 0) = -\sqrt{2} \approx -1.414\)
- So J' is approximately \((-1.414, -1.414)\) or \((-1.5, -1.5)\) when rounded.
2. **Point K (0, -4)**:
- \( x' = \frac{\sqrt{2}}{2}(0 + 4) = 2\sqrt{2} \approx 2.828\)
- \( y' = \frac{\sqrt{2}}{2}(0 - 4) = -2\sqrt{2} \approx -2.828\)
- So K' is approximately \((2.828, -2.828)\) or \((3, -3)\) when rounded.
3. **Point L (-3, -5)**:
- \( x' = \frac{\sqrt{2}}{2}(-3 + 5) = \sqrt{2} \approx 1.414\)
- \( y' = \frac{\sqrt{2}}{2}(-3 - 5) = -4\sqrt{2} \approx -5.656\)
- So L' is approximately \((1.414, -5.656)\) or \((1.2, -5.7)\) when rounded.
Now we have the points J'\((-1.5, -1.5)\), K'\((3, -3)\), and L'\((1.2, -5.7)\).
Among the options provided, the graph that matches these new coordinates is the one with the points:
\((-1.5, -1.5)\), \((3, -3)\), and \((1.2, -5.7)\).
Thus, the correct graph to show the location of triangle J’K’L’ would be the first option given.