Question

graph \\( \triangle j k l \\) and its image after a reflection in the line \\( y = 1 \\).
\\( j ( 2,4 ), k ( - 4, - 2 ), l ( - 1,0 ) \\)
11

Explanation:

Step1: Find the reflection formula

The formula for reflecting a point \((x,y)\) over the line \(y = c\) is \((x,2c - y)\). Here \(c = 1\), so the formula is \((x,2\times1 - y)=(x,2 - y)\).

Step2: Apply the formula to point \(J\)

For \(J(2,4)\), substitute \(x = 2\) and \(y = 4\) into \((x,2 - y)\).
\(J'(2,2 - 4)=(2,-2)\)

Step3: Apply the formula to point \(K\)

For \(K(-4,-2)\), substitute \(x=-4\) and \(y = - 2\) into \((x,2 - y)\).
\(K'(-4,2-(-2))=(-4,4)\)

Step4: Apply the formula to point \(L\)

For \(L(-1,0)\), substitute \(x=-1\) and \(y = 0\) into \((x,2 - y)\).
\(L'(-1,2 - 0)=(-1,2)\)

Answer:

The coordinates of \(\triangle JKL\) are \(J(2,4)\), \(K(-4,-2)\), \(L(-1,0)\) and the coordinates of its image after reflection over \(y = 1\) are \(J'(2,-2)\), \(K'(-4,4)\), \(L'(-1,2)\). Plot these points on the coordinate - plane to graph \(\triangle JKL\) and its image.