Question

1. square jklm with vertices j(1, -3), k(5, 0), l(8, -4), and m(4, -7): 90° counterclockwise j’(__, ) k’(, ) l’(, ) m’(, __)

Explanation:

To rotate a point \((x, y)\) 90° counterclockwise about the origin, we use the rule \((x, y) \to (-y, x)\). We will apply this rule to each vertex of the square.

Step 1: Rotate \(J(1, -3)\)

Using the rule \((x, y) \to (-y, x)\), substitute \(x = 1\) and \(y = -3\):
\(-y = -(-3) = 3\) and \(x = 1\). So \(J' = (3, 1)\).

Step 2: Rotate \(K(5, 0)\)

Using the rule \((x, y) \to (-y, x)\), substitute \(x = 5\) and \(y = 0\):
\(-y = -0 = 0\) and \(x = 5\). So \(K' = (0, 5)\).

Step 3: Rotate \(L(8, -4)\)

Using the rule \((x, y) \to (-y, x)\), substitute \(x = 8\) and \(y = -4\):
\(-y = -(-4) = 4\) and \(x = 8\). So \(L' = (4, 8)\).

Step 4: Rotate \(M(4, -7)\)

Using the rule \((x, y) \to (-y, x)\), substitute \(x = 4\) and \(y = -7\):
\(-y = -(-7) = 7\) and \(x = 4\). So \(M' = (7, 4)\).

Answer:

\(J'(3, 1)\)
\(K'(0, 5)\)
\(L'(4, 8)\)
\(M'(7, 4)\)