Question

1. rotation 270° clockwise about the origin

Explanation:

To solve the rotation of a figure \(270^\circ\) clockwise about the origin, we first need to identify the coordinates of the vertices of the triangle (let's assume the vertices are \(N\), \(M\), and \(L\)). Let's find their coordinates from the grid:

Step 1: Identify Coordinates

• Let's assume the grid has each square as 1 unit.
• For point \(N\): From the origin \((0,0)\), moving left 1 unit (x = -1) and up 2 units (y = 2), so \(N(-1, 2)\).
• For point \(M\): Moving right 2 units (x = 2) and up 3 units (y = 3), so \(M(2, 3)\).
• For point \(L\): Moving right 4 units (x = 4) and up 2 units (y = 2), so \(L(4, 2)\).

Step 2: Rotation Rule for \(270^\circ\) Clockwise

The rule for rotating a point \((x, y)\) \(270^\circ\) clockwise about the origin is:
\((x, y)
ightarrow (y, -x)\)

Step 3: Apply the Rule to Each Vertex

For \(N(-1, 2)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = -1\), \(y = 2\)
New \(x\)-coordinate: \(y = 2\)
New \(y\)-coordinate: \(-x = -(-1) = 1\)
So, \(N' (2, 1)\)

For \(M(2, 3)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = 2\), \(y = 3\)
New \(x\)-coordinate: \(y = 3\)
New \(y\)-coordinate: \(-x = -2\)
So, \(M' (3, -2)\)

For \(L(4, 2)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = 4\), \(y = 2\)
New \(x\)-coordinate: \(y = 2\)
New \(y\)-coordinate: \(-x = -4\)
So, \(L' (2, -4)\)

Final Answer (Coordinates of Rotated Points)

• \(N'\): \(\boldsymbol{(2, 1)}\)
• \(M'\): \(\boldsymbol{(3, -2)}\)
• \(L'\): \(\boldsymbol{(2, -4)}\)

(Note: If you need to plot these points, they will form the rotated triangle about the origin by \(270^\circ\) clockwise.)

Answer:

To solve the rotation of a figure \(270^\circ\) clockwise about the origin, we first need to identify the coordinates of the vertices of the triangle (let's assume the vertices are \(N\), \(M\), and \(L\)). Let's find their coordinates from the grid:

Step 1: Identify Coordinates

• Let's assume the grid has each square as 1 unit.
• For point \(N\): From the origin \((0,0)\), moving left 1 unit (x = -1) and up 2 units (y = 2), so \(N(-1, 2)\).
• For point \(M\): Moving right 2 units (x = 2) and up 3 units (y = 3), so \(M(2, 3)\).
• For point \(L\): Moving right 4 units (x = 4) and up 2 units (y = 2), so \(L(4, 2)\).

Step 2: Rotation Rule for \(270^\circ\) Clockwise

The rule for rotating a point \((x, y)\) \(270^\circ\) clockwise about the origin is:
\((x, y)
ightarrow (y, -x)\)

Step 3: Apply the Rule to Each Vertex

For \(N(-1, 2)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = -1\), \(y = 2\)
New \(x\)-coordinate: \(y = 2\)
New \(y\)-coordinate: \(-x = -(-1) = 1\)
So, \(N' (2, 1)\)

For \(M(2, 3)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = 2\), \(y = 3\)
New \(x\)-coordinate: \(y = 3\)
New \(y\)-coordinate: \(-x = -2\)
So, \(M' (3, -2)\)

For \(L(4, 2)\):

Using the rule \((x, y)
ightarrow (y, -x)\):
\(x = 4\), \(y = 2\)
New \(x\)-coordinate: \(y = 2\)
New \(y\)-coordinate: \(-x = -4\)
So, \(L' (2, -4)\)

Final Answer (Coordinates of Rotated Points)

• \(N'\): \(\boldsymbol{(2, 1)}\)
• \(M'\): \(\boldsymbol{(3, -2)}\)
• \(L'\): \(\boldsymbol{(2, -4)}\)

(Note: If you need to plot these points, they will form the rotated triangle about the origin by \(270^\circ\) clockwise.)