Question

graph the image of δxyz after the following sequence of transformations: translation 7 units left and 17 units up rotation 270° counterclockwise around the origin

Explanation:

To solve this, we first identify the coordinates of the vertices of \(\triangle XYZ\). From the graph:

• \(X\): Let's assume the coordinates. Looking at the grid, \(X\) seems to be at \((-8, -12)\), \(Y\) at \((-4, -9)\), \(Z\) at \((-8, -5)\) (we'll confirm the grid steps: each square is 1 unit, so moving left/right and up/down by 1 unit per square).

Step 1: Translation 7 units left and 17 units up

The translation rule for a point \((x, y)\) is \((x - 7, y + 17)\).

• For \(X(-8, -12)\):

New \(X_1\): \((-8 - 7, -12 + 17) = (-15, 5)\)

• For \(Y(-4, -9)\):

New \(Y_1\): \((-4 - 7, -9 + 17) = (-11, 8)\)

• For \(Z(-8, -5)\):

New \(Z_1\): \((-8 - 7, -5 + 17) = (-15, 12)\)

Step 2: Rotation \(270^\circ\) counterclockwise around the origin

The rule for a \(270^\circ\) counterclockwise rotation around the origin is \((x, y) \to (y, -x)\) (derived from rotation matrices: \(270^\circ\) CCW is equivalent to \(90^\circ\) CW, and the transformation is \((x, y) \mapsto (y, -x)\)).

• For \(X_1(-15, 5)\):

New \(X_2\): \((5, -(-15)) = (5, 15)\)

• For \(Y_1(-11, 8)\):

New \(Y_2\): \((8, -(-11)) = (8, 11)\)

• For \(Z_1(-15, 12)\):

New \(Z_2\): \((12, -(-15)) = (12, 15)\)

Graphing the Final Image

Plot the points \(X_2(5, 15)\), \(Y_2(8, 11)\), and \(Z_2(12, 15)\) on the coordinate plane and connect them to form the transformed triangle.

(Note: If the initial coordinates were misread, adjust based on the actual grid. For example, if \(X\) is at \((-8, -12)\), \(Y\) at \((-4, -9)\), \(Z\) at \((-8, -5)\), the above steps hold. Verify the grid scale: each major grid line is 2 units? Wait, looking at the y-axis: from -14 to 14, with ticks at -14, -12, -10, ..., 12, 14. So each grid square is 1 unit. So the initial coordinates:

• \(X\): Let's check the vertical (y) and horizontal (x) positions. \(X\) is at x = -8 (8 units left of origin), y = -12 (12 units down).
• \(Y\): x = -4, y = -9.
• \(Z\): x = -8, y = -5.

Yes, that's correct. So after translation and rotation, the final coordinates are \(X(5,15)\), \(Y(8,11)\), \(Z(12,15)\).)

To graph:

1. Locate \(X(5, 15)\) (5 units right on x-axis, 15 up on y-axis).
2. Locate \(Y(8, 11)\) (8 right, 11 up).
3. Locate \(Z(12, 15)\) (12 right, 15 up).
4. Connect these three points to form the image of \(\triangle XYZ\) after the transformations.

Answer:

To solve this, we first identify the coordinates of the vertices of \(\triangle XYZ\). From the graph:

• \(X\): Let's assume the coordinates. Looking at the grid, \(X\) seems to be at \((-8, -12)\), \(Y\) at \((-4, -9)\), \(Z\) at \((-8, -5)\) (we'll confirm the grid steps: each square is 1 unit, so moving left/right and up/down by 1 unit per square).

Step 1: Translation 7 units left and 17 units up

The translation rule for a point \((x, y)\) is \((x - 7, y + 17)\).

• For \(X(-8, -12)\):

New \(X_1\): \((-8 - 7, -12 + 17) = (-15, 5)\)

• For \(Y(-4, -9)\):

New \(Y_1\): \((-4 - 7, -9 + 17) = (-11, 8)\)

• For \(Z(-8, -5)\):

New \(Z_1\): \((-8 - 7, -5 + 17) = (-15, 12)\)

Step 2: Rotation \(270^\circ\) counterclockwise around the origin

The rule for a \(270^\circ\) counterclockwise rotation around the origin is \((x, y) \to (y, -x)\) (derived from rotation matrices: \(270^\circ\) CCW is equivalent to \(90^\circ\) CW, and the transformation is \((x, y) \mapsto (y, -x)\)).

• For \(X_1(-15, 5)\):

New \(X_2\): \((5, -(-15)) = (5, 15)\)

• For \(Y_1(-11, 8)\):

New \(Y_2\): \((8, -(-11)) = (8, 11)\)

• For \(Z_1(-15, 12)\):

New \(Z_2\): \((12, -(-15)) = (12, 15)\)

Graphing the Final Image

Plot the points \(X_2(5, 15)\), \(Y_2(8, 11)\), and \(Z_2(12, 15)\) on the coordinate plane and connect them to form the transformed triangle.

(Note: If the initial coordinates were misread, adjust based on the actual grid. For example, if \(X\) is at \((-8, -12)\), \(Y\) at \((-4, -9)\), \(Z\) at \((-8, -5)\), the above steps hold. Verify the grid scale: each major grid line is 2 units? Wait, looking at the y-axis: from -14 to 14, with ticks at -14, -12, -10, ..., 12, 14. So each grid square is 1 unit. So the initial coordinates:

• \(X\): Let's check the vertical (y) and horizontal (x) positions. \(X\) is at x = -8 (8 units left of origin), y = -12 (12 units down).
• \(Y\): x = -4, y = -9.
• \(Z\): x = -8, y = -5.

Yes, that's correct. So after translation and rotation, the final coordinates are \(X(5,15)\), \(Y(8,11)\), \(Z(12,15)\).)

To graph:

1. Locate \(X(5, 15)\) (5 units right on x-axis, 15 up on y-axis).
2. Locate \(Y(8, 11)\) (8 right, 11 up).
3. Locate \(Z(12, 15)\) (12 right, 15 up).
4. Connect these three points to form the image of \(\triangle XYZ\) after the transformations.