Question

graph the image of \\( \triangle jkl \\) after a reflection over the line \\( y = 1 \\).

Explanation:

Step1: Find the reflection formula

For a point \((x,y)\) reflected over the line \(y = a\), the formula is \((x,2a - y)\). Here \(a = 1\), so the formula is \((x,2\times1 - y)=(x,2 - y)\).

Step2: Apply the formula to each vertex

• For \(J(8,- 2)\):

Substitute \(x = 8\) and \(y=-2\) into \((x,2 - y)\).
\(2-y=2-(-2)=4\), so the image of \(J\) is \(J'(8,4)\).

• For \(K(10,-2)\):

Substitute \(x = 10\) and \(y = - 2\) into \((x,2 - y)\).
\(2-y=2-(-2)=4\), so the image of \(K\) is \(K'(10,4)\).

• For \(L(8,-7)\):

Substitute \(x = 8\) and \(y=-7\) into \((x,2 - y)\).
\(2-y=2-(-7)=9\), so the image of \(L\) is \(L'(8,9)\).

Answer:

Plot the points \(J'(8,4)\), \(K'(10,4)\) and \(L'(8,9)\) and connect them to form the reflected triangle \(\triangle J'K'L'\).