Question

1. reflection across y = 3\

k(-4, 2), l(-4, 4), m(-2, 5)

Explanation:

Step1: Recall reflection over horizontal line

For a point \((x,y)\) reflected over the line \(y = k\), the formula is \((x, 2k - y)\). Here, \(k = 3\), so the formula becomes \((x, 2\times3 - y)=(x, 6 - y)\).

Step2: Reflect point K(-4,2)

Substitute \(x=-4\), \(y = 2\) into the formula: \(x=-4\), \(y=6 - 2 = 4\). So \(K'(-4,4)\).

Step3: Reflect point L(-4,4)

Substitute \(x=-4\), \(y = 4\) into the formula: \(x=-4\), \(y=6 - 4 = 2\). So \(L'(-4,2)\).

Step4: Reflect point M(-2,5)

Substitute \(x=-2\), \(y = 5\) into the formula: \(x=-2\), \(y=6 - 5 = 1\). So \(M'(-2,1)\).

Answer:

Reflected points: \(K'(-4,4)\), \(L'(-4,2)\), \(M'(-2,1)\)