Triangle \( P Q R \) has coordinates \( P(-8,3), Q(-8,6) \), and \( R(-3,6) \). If the triangle is transiated by using the rule \( (x, y) \rightarrow(x+4, y-6) \), what are the coordinates of triangle \( P^{\prime} Q^{\prime} R^{\prime} ? \) \( P^{\prime}(-12,9), Q^{\prime}(-12,12), R^{\prime}(-7,12) \) \( P^{\prime}(-4,7), Q^{\prime}(-14,0), R^{\prime}(1,10) \) \( P^{\prime}(-4,9), Q^{\prime}(-4,12), R^{\prime}(1,12) \) \( P^{\prime}(-4,-3), Q^{\prime}(-4,0), R^{\prime}(1,0) \)

Let's break down the translation of triangle \( PQR \) step by step! First, we apply the translation rule \( (x, y) \rightarrow (x + 4, y - 6) \) to each of the vertices of the triangle: - For \( P(-8, 3) \): - New \( x \): \(-8 + 4 = -4\) - New \( y \): \(3 - 6 = -3\) - So, \( P^{\prime}(-4, -3) \) - For \( Q(-8, 6) \): - New \( x \): \(-8 + 4 = -4\) - New \( y \): \(6 - 6 = 0\) - so, \( Q^{\prime}(-4, 0) \) - For \( R(-3, 6) \): - New \( x \): \(-3 + 4 = 1\) - New \( y \): \(6 - 6 = 0\) - so, \( R^{\prime}(1, 0) \) Now we have our translated points \( P^{\prime}(-4, -3), Q^{\prime}(-4, 0), R^{\prime}(1, 0) \). So, the correct answer is: \( P^{\prime}(-4,-3), Q^{\prime}(-4,0), R^{\prime}(1,0) \)