QUESTION IMAGE
Question
the vertices a(-2,-1), b(-3,2), c(-1,3), and d(0,0) form a parallelogram. the vertices a(-1,-2), b(2,-3), c(3,-1), and d(0,0) are the image of the parallelogram after a sequence of transformations. which sequence of transformations could produce the image from the pre - image?
- a reflection over the y - axis and then a 90° clockwise rotation about the origin
- a reflection over the x - axis and then a reflection over the y - axis
- a 90° clockwise rotation about the origin and then a reflection over the y - axis
- a 90° counterclockwise rotation about the origin and then a reflection over the x - axis
Step1: Apply rotation formula
The formula for a \(90^{\circ}\) clockwise rotation about the origin \((x,y)\to(y, - x)\). For point \(A(-2,-1)\), after rotation, it becomes \((-1,2)\).
Step2: Apply reflection formula
The formula for reflection over the \(y\) - axis is \((x,y)\to(-x,y)\). For the point \((-1,2)\) after reflection over the \(y\) - axis, it becomes \((1,2)\).
Let's check each option:
- Option 1 (a reflection over the \(y\) - axis and then a \(90^{\circ}\) clockwise rotation about the origin):
- For \(A(-2,-1)\):
- Reflection over \(y\) - axis: \((2,-1)\)
- \(90^{\circ}\) clockwise rotation about the origin: \((-1,-2)\) (not \(A'(-1,-2)\))
- Option 2 (a reflection over the \(x\) - axis and then a reflection over the \(y\) - axis):
- For \(A(-2,-1)\):
- Reflection over \(x\) - axis: \((-2,1)\)
- Reflection over \(y\) - axis: \((2,1)\) (not \(A'(-1,-2)\))
- Option 3 (a \(90^{\circ}\) clockwise rotation about the origin and then a reflection over the \(y\) - axis):
- For \(A(-2,-1)\):
- \(90^{\circ}\) clockwise rotation about the origin: \((-1,2)\)
- Reflection over \(y\) - axis: \((1,2)\) (not \(A'(-1,-2)\))
- Option 4 (a \(90^{\circ}\) counter - clockwise rotation about the origin and then a reflection over the \(x\) - axis):
- The formula for a \(90^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(-y,x)\). For \(A(-2,-1)\), after \(90^{\circ}\) counter - clockwise rotation about the origin: \((1,-2)\)
- The formula for reflection over the \(x\) - axis is \((x,y)\to(x,-y)\). For the point \((1,-2)\) after reflection over the \(x\) - axis: \((1,2)\) (not \(A'(-1,-2)\))
Let's use the transformation matrix approach.
The rotation matrix for a \(90^{\circ}\) clockwise rotation about the origin is \(R=
\), and the reflection matrix over the \(y\) - axis is \(M=
\)
The combined transformation matrix \(T = M\times R=
\times
=
\)
For a point \(
\), the transformed point is \(T
=
=
\)
For \(A(-2,-1)\), \(
=
\) (wrong)
The rotation matrix for a \(90^{\circ}\) counter - clockwise rotation about the origin is \(R_c=
\), and the reflection matrix over the \(x\) - axis is \(M_x=
\)
The combined transformation matrix \(T_c = M_x\times R_c=
\times
=
\) (same as above, wrong)
Let's check by another method:
We know that \(A(-2,-1)\to A'(-1,-2)\)
If we first do a \(90^{\circ}\) clockwise rotation about the origin: \((x,y)\to(y, - x)\), for \(A(-2,-1)\) we get \((-1,2)\)
Then reflect over the \(y\) - axis \((x,y)\to(-x,y)\), we get \((1,2)\) (wrong)
If we first do a \(90^{\circ}\) counter - clockwise rotation about the origin: \((x,y)\to(-y,x)\), for \(A(-2,-1)\) we get \((1,-2)\)
Then reflect over the \(x\) - axis \((x,y)\to(x,-y)\), we get \((1,2)\) (wrong)
Let's check the option: a reflection over the \(y\) - axis and then a \(90^{\circ}\) clockwise rotation about the origin
- Reflection over \(y\) - axis: \((x,y)\to(-x,y)\), for \(A(-2,-1)…
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The first option (a reflection over the \(y\) - axis and then a \(90^{\circ}\) clockwise rotation about the origin)