Question

reflect the figure over the line y = x − 5. plot all of the points of the reflected figure. you may click a plotted point to delete it.

Explanation:

To reflect a point \((x, y)\) over the line \(y = x - 5\), we use the formula for reflection over the line \(y = x + c\) (here \(c=-5\)). The reflection of a point \((x, y)\) over the line \(y = x + c\) is given by \((y - c, x + c)\). For \(y = x - 5\) (so \(c=-5\)), the reflection of \((x, y)\) is \((y - (-5), x + (-5))=(y + 5, x - 5)\).

Step 1: Identify the original points

First, we need to find the coordinates of the vertices of the original figure. From the graph (estimating the grid points):

• Let's assume the original vertices are \(A(1, -8)\), \(B(5, -3)\), \(C(6, -4)\), \(D(8, -3)\), \(E(7, -7)\) (we'll verify the reflection for these points).

Step 2: Apply the reflection formula to each point

For a point \((x, y)\), reflection over \(y = x - 5\) is \((y + 5, x - 5)\).

• For \(A(1, -8)\):

\(x = 1\), \(y = -8\)
Reflected point: \((-8 + 5, 1 - 5)=(-3, -4)\)

• For \(B(5, -3)\):

\(x = 5\), \(y = -3\)
Reflected point: \((-3 + 5, 5 - 5)=(2, 0)\)

• For \(C(6, -4)\):

\(x = 6\), \(y = -4\)
Reflected point: \((-4 + 5, 6 - 5)=(1, 1)\)

• For \(D(8, -3)\):

\(x = 8\), \(y = -3\)
Reflected point: \((-3 + 5, 8 - 5)=(2, 3)\)

• For \(E(7, -7)\):

\(x = 7\), \(y = -7\)
Reflected point: \((-7 + 5, 7 - 5)=(-2, 2)\)

Step 3: Plot the reflected points

Plot the points \((-3, -4)\), \((2, 0)\), \((1, 1)\), \((2, 3)\), and \((-2, 2)\) on the coordinate plane. Connect them to form the reflected figure.

(Note: If the original vertices were misidentified, adjust the coordinates and reapply the reflection formula. The key is using the reflection rule for \(y = x + c\): swap \(x\) and \(y\), then adjust for \(c\). For \(y = x - 5\), the reflection of \((x, y)\) is \((y + 5, x - 5)\) because solving the system of the line and the perpendicular bisector between \((x, y)\) and its reflection \((x', y')\) gives \(x' = y + 5\) and \(y' = x - 5\).)

Answer:

To reflect a point \((x, y)\) over the line \(y = x - 5\), we use the formula for reflection over the line \(y = x + c\) (here \(c=-5\)). The reflection of a point \((x, y)\) over the line \(y = x + c\) is given by \((y - c, x + c)\). For \(y = x - 5\) (so \(c=-5\)), the reflection of \((x, y)\) is \((y - (-5), x + (-5))=(y + 5, x - 5)\).

Step 1: Identify the original points

First, we need to find the coordinates of the vertices of the original figure. From the graph (estimating the grid points):

• Let's assume the original vertices are \(A(1, -8)\), \(B(5, -3)\), \(C(6, -4)\), \(D(8, -3)\), \(E(7, -7)\) (we'll verify the reflection for these points).

Step 2: Apply the reflection formula to each point

For a point \((x, y)\), reflection over \(y = x - 5\) is \((y + 5, x - 5)\).

• For \(A(1, -8)\):

\(x = 1\), \(y = -8\)
Reflected point: \((-8 + 5, 1 - 5)=(-3, -4)\)

• For \(B(5, -3)\):

\(x = 5\), \(y = -3\)
Reflected point: \((-3 + 5, 5 - 5)=(2, 0)\)

• For \(C(6, -4)\):

\(x = 6\), \(y = -4\)
Reflected point: \((-4 + 5, 6 - 5)=(1, 1)\)

• For \(D(8, -3)\):

\(x = 8\), \(y = -3\)
Reflected point: \((-3 + 5, 8 - 5)=(2, 3)\)

• For \(E(7, -7)\):

\(x = 7\), \(y = -7\)
Reflected point: \((-7 + 5, 7 - 5)=(-2, 2)\)

Step 3: Plot the reflected points

Plot the points \((-3, -4)\), \((2, 0)\), \((1, 1)\), \((2, 3)\), and \((-2, 2)\) on the coordinate plane. Connect them to form the reflected figure.

(Note: If the original vertices were misidentified, adjust the coordinates and reapply the reflection formula. The key is using the reflection rule for \(y = x + c\): swap \(x\) and \(y\), then adjust for \(c\). For \(y = x - 5\), the reflection of \((x, y)\) is \((y + 5, x - 5)\) because solving the system of the line and the perpendicular bisector between \((x, y)\) and its reflection \((x', y')\) gives \(x' = y + 5\) and \(y' = x - 5\).)