Question

1. rotation 90° counterclockwise about the origin (x,y)→(-y,x)

Explanation:

To solve the rotation of the triangle \( UVW \) \( 90^\circ \) counterclockwise about the origin, we first need to identify the coordinates of points \( U \), \( V \), and \( W \).

Step 1: Identify Coordinates

• Let's assume the grid has each square as 1 unit.
• Point \( V \): From the graph, \( V \) is at \( (-3, 0) \) (since it's 3 units left of the origin on the x - axis and 0 on the y - axis).
• Point \( U \): Let's find the coordinates. If we count the grid, \( U \) is at \( (-4, -4) \) (4 units left on x - axis and 4 units down on y - axis).
• Point \( W \): \( W \) is at \( (1, -3) \) (1 unit right on x - axis and 3 units down on y - axis).

Step 2: Apply the Rotation Rule

The rule for a \( 90^\circ \) counterclockwise rotation about the origin is \( (x,y)\to(-y,x) \).

For Point \( V(-3,0) \):

Using the rule \( (x,y)\to(-y,x) \), substitute \( x = - 3\) and \( y=0 \).
We get \( (-0,-3)=(0, - 3) \)? Wait, no, wait. Wait, the correct rule is \( (x,y)\to(-y,x) \). So for \( V(-3,0) \), \( x=-3 \), \( y = 0 \). Then \( -y=0 \), \( x=-3 \)? Wait, no, I made a mistake. Wait, the standard rule for \( 90^\circ \) counterclockwise rotation about the origin is \( (x,y)\to(-y,x) \). Let's re - check:

If we have a point \( (x,y) \), after a \( 90^\circ \) counterclockwise rotation about the origin, the new \( x'=-y \) and \( y' = x \).

• For \( V(-3,0) \):

\( x=-3 \), \( y = 0 \)
\( x'=-y=0 \)
\( y'=x=-3 \)
So \( V'=(0, - 3) \)? Wait, no, that seems wrong. Wait, maybe I misread the coordinates of \( V \). Let's re - examine the graph. The x - axis: the origin is at the intersection of the two axes. Point \( V \) is on the x - axis, 3 units to the left of the origin, so \( V=(-3,0) \).

• For \( U(-4,-4) \):

Using the rule \( (x,y)\to(-y,x) \)
\( x=-4 \), \( y=-4 \)
\( x'=-y=-(-4) = 4 \)
\( y'=x=-4 \)
So \( U'=(4,-4) \)

• For \( W(1,-3) \):

Using the rule \( (x,y)\to(-y,x) \)
\( x = 1\), \( y=-3 \)
\( x'=-y=-(-3)=3 \)
\( y'=x = 1 \)
So \( W'=(3,1) \)

Wait, maybe I made a mistake in the initial coordinates. Let's re - identify the coordinates correctly:

Looking at the graph:

• Point \( V \): It's on the x - axis, 3 units to the left of the origin, so \( V=(-3,0) \)
• Point \( U \): Let's count the grid. From the origin, moving left 4 units (x=-4) and down 4 units (y = - 4), so \( U=(-4,-4) \)
• Point \( W \): Moving right 1 unit (x = 1) and down 3 units (y=-3), so \( W=(1,-3) \)

Now applying the \( 90^\circ \) counterclockwise rotation rule \( (x,y)\to(-y,x) \):

• For \( V(-3,0) \):

\( x=-3 \), \( y = 0 \)
New \( x=-y=0 \), new \( y=x=-3 \), so \( V'=(0,-3) \)

• For \( U(-4,-4) \):

\( x=-4 \), \( y=-4 \)
New \( x=-y = 4 \), new \( y=x=-4 \), so \( U'=(4,-4) \)

• For \( W(1,-3) \):

\( x = 1\), \( y=-3 \)
New \( x=-y = 3 \), new \( y=x = 1 \), so \( W'=(3,1) \)

Now we can plot these new points \( U'(4,-4) \), \( V'(0,-3) \), \( W'(3,1) \) to get the rotated triangle.

If we were asked to find the coordinates of the rotated points, the coordinates of the image of \( \triangle UVW \) after \( 90^\circ \) counterclockwise rotation about the origin are:

• \( U'=(4,-4) \)
• \( V'=(0,-3) \)
• \( W'=(3,1) \)

Answer:

To solve the rotation of the triangle \( UVW \) \( 90^\circ \) counterclockwise about the origin, we first need to identify the coordinates of points \( U \), \( V \), and \( W \).

Step 1: Identify Coordinates

• Let's assume the grid has each square as 1 unit.
• Point \( V \): From the graph, \( V \) is at \( (-3, 0) \) (since it's 3 units left of the origin on the x - axis and 0 on the y - axis).
• Point \( U \): Let's find the coordinates. If we count the grid, \( U \) is at \( (-4, -4) \) (4 units left on x - axis and 4 units down on y - axis).
• Point \( W \): \( W \) is at \( (1, -3) \) (1 unit right on x - axis and 3 units down on y - axis).

Step 2: Apply the Rotation Rule

The rule for a \( 90^\circ \) counterclockwise rotation about the origin is \( (x,y)\to(-y,x) \).

For Point \( V(-3,0) \):

Using the rule \( (x,y)\to(-y,x) \), substitute \( x = - 3\) and \( y=0 \).
We get \( (-0,-3)=(0, - 3) \)? Wait, no, wait. Wait, the correct rule is \( (x,y)\to(-y,x) \). So for \( V(-3,0) \), \( x=-3 \), \( y = 0 \). Then \( -y=0 \), \( x=-3 \)? Wait, no, I made a mistake. Wait, the standard rule for \( 90^\circ \) counterclockwise rotation about the origin is \( (x,y)\to(-y,x) \). Let's re - check:

If we have a point \( (x,y) \), after a \( 90^\circ \) counterclockwise rotation about the origin, the new \( x'=-y \) and \( y' = x \).

• For \( V(-3,0) \):

\( x=-3 \), \( y = 0 \)
\( x'=-y=0 \)
\( y'=x=-3 \)
So \( V'=(0, - 3) \)? Wait, no, that seems wrong. Wait, maybe I misread the coordinates of \( V \). Let's re - examine the graph. The x - axis: the origin is at the intersection of the two axes. Point \( V \) is on the x - axis, 3 units to the left of the origin, so \( V=(-3,0) \).

• For \( U(-4,-4) \):

Using the rule \( (x,y)\to(-y,x) \)
\( x=-4 \), \( y=-4 \)
\( x'=-y=-(-4) = 4 \)
\( y'=x=-4 \)
So \( U'=(4,-4) \)

• For \( W(1,-3) \):

Using the rule \( (x,y)\to(-y,x) \)
\( x = 1\), \( y=-3 \)
\( x'=-y=-(-3)=3 \)
\( y'=x = 1 \)
So \( W'=(3,1) \)

Wait, maybe I made a mistake in the initial coordinates. Let's re - identify the coordinates correctly:

Looking at the graph:

• Point \( V \): It's on the x - axis, 3 units to the left of the origin, so \( V=(-3,0) \)
• Point \( U \): Let's count the grid. From the origin, moving left 4 units (x=-4) and down 4 units (y = - 4), so \( U=(-4,-4) \)
• Point \( W \): Moving right 1 unit (x = 1) and down 3 units (y=-3), so \( W=(1,-3) \)

Now applying the \( 90^\circ \) counterclockwise rotation rule \( (x,y)\to(-y,x) \):

• For \( V(-3,0) \):

\( x=-3 \), \( y = 0 \)
New \( x=-y=0 \), new \( y=x=-3 \), so \( V'=(0,-3) \)

• For \( U(-4,-4) \):

\( x=-4 \), \( y=-4 \)
New \( x=-y = 4 \), new \( y=x=-4 \), so \( U'=(4,-4) \)

• For \( W(1,-3) \):

\( x = 1\), \( y=-3 \)
New \( x=-y = 3 \), new \( y=x = 1 \), so \( W'=(3,1) \)

Now we can plot these new points \( U'(4,-4) \), \( V'(0,-3) \), \( W'(3,1) \) to get the rotated triangle.

If we were asked to find the coordinates of the rotated points, the coordinates of the image of \( \triangle UVW \) after \( 90^\circ \) counterclockwise rotation about the origin are:

• \( U'=(4,-4) \)
• \( V'=(0,-3) \)
• \( W'=(3,1) \)