Question

graph the image of \\( \triangle def \\) after a reflection over the line \\( x = 3 \\).

Explanation:

Step1: Find the coordinates of points \(D\), \(E\), and \(F\)

• Point \(D=(0, - 8)\)
• Point \(E=(0, - 7)\)
• Point \(F=(-4, - 6)\)

Step2: Use the reflection formula over the line \(x = a\)

The formula for reflecting a point \((x,y)\) over the line \(x=a\) is \((2a - x,y)\). Here \(a = 3\).

• For point \(D=(0,-8)\):

\(x'=2\times3 - 0=6\), \(y'=-8\). So \(D'=(6,-8)\)

• For point \(E=(0,-7)\):

\(x'=2\times3 - 0 = 6\), \(y'=-7\). So \(E'=(6,-7)\)

• For point \(F=(-4,-6)\):

\(x'=2\times3-(-4)=6 + 4=10\), \(y'=-6\). So \(F'=(10,-6)\)

Step3: Plot the new points \(D'\), \(E'\), and \(F'\)

Connect the points \(D'\), \(E'\), and \(F'\) to form the reflected triangle \(\triangle D'E'F'\)

Answer:

Plot the points \(D'(6,-8)\), \(E'(6,-7)\), \(F'(10,-6)\) and connect them to get the image of \(\triangle DEF\) after reflection over the line \(x = 3\)