Given points \( A(-2,0), B(-5,3), C(-9,-1), P(7,6), Q(4,0) \), and \( R(-4,4) \) , which of the following proves that \( \triangle A B C \sim \triangle P Q R \) ? By the Distance Formula, \( A B=18, B C=32 \), and \( C A=50 \). Also, \( P Q=45, Q R=90 \), and \( R P=125 \). Therefore, \( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}=\frac{2}{5} \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by SAS \( \sim \). By the Distance Formula, \( A B=3 \sqrt{2}, B C=4 \sqrt{2} \), and \( C A=5 \sqrt{2} \). Also, \( P Q=3 \sqrt{5} Q R=4 \sqrt{5} \), and \( R P=5 \sqrt{5} \). Therefore, \( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5} \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by SSS \( \sim \). By the Distance Formula, \( A B=3 \sqrt{2}, B C=4 \sqrt{2} \), and \( C A=5 \sqrt{2} \). Also, \( P Q=3 \sqrt{5}, Q R=4 \sqrt{5} \), and \( R P=5 \sqrt{5} \). Therefore, \( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5} \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by SAS \( \sim \). By the Distance Formula, \( A B=50, B C=32 \), and \( C A=18 \). Also, \( P Q=125, Q R=90 \), and \( R P=45 \). Therefore, \( \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P}=\frac{2}{5} \), and therefore, \( \triangle A B C \sim \triangle P Q R \) by SSS \( \sim \).

The correct option that proves \( \triangle ABC \sim \triangle PQR \) is the **second option**. Here's why: ### **Second Option Explained:** 1. **Calculating Side Lengths Using the Distance Formula:** - For \( \triangle ABC \): - \( AB = \sqrt{(-5 - (-2))^2 + (3 - 0)^2} = \sqrt{3^2 + 3^2} = 3\sqrt{2} \) - \( BC = \sqrt{(-9 - (-5))^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = 4\sqrt{2} \) - \( CA = \sqrt{(-2 - (-9))^2 + (0 - (-1))^2} = \sqrt{7^2 + 1^2} = 5\sqrt{2} \) - For \( \triangle PQR \): - \( PQ = \sqrt{(4 - 7)^2 + (0 - 6)^2} = \sqrt{(-3)^2 + (-6)^2} = 3\sqrt{5} \) - \( QR = \sqrt{(-4 - 4)^2 + (4 - 0)^2} = \sqrt{(-8)^2 + 4^2} = 4\sqrt{5} \) - \( RP = \sqrt{(7 - (-4))^2 + (6 - 4)^2} = \sqrt{11^2 + 2^2} = 5\sqrt{5} \) 2. **Comparing the Ratios of Corresponding Sides:** - \( \frac{AB}{PQ} = \frac{3\sqrt{2}}{3\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5} \) - \( \frac{BC}{QR} = \frac{4\sqrt{2}}{4\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5} \) - \( \frac{CA}{RP} = \frac{5\sqrt{2}}{5\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5} \) All corresponding sides have the **same ratio**, which is \( \frac{\sqrt{10}}{5} \). 3. **Conclusion:** Since all corresponding sides of \( \triangle ABC \) and \( \triangle PQR \) are proportional, by the **Side-Side-Side (SSS) Similarity Criterion**, the triangles are similar. ### **Other Options:** - **First and Fourth Options:** These incorrectly calculate the side lengths and their ratios, leading to inconsistent ratios that do not support similarity. - **Third Option:** While it correctly calculates the side lengths and ratios, it incorrectly applies the **Side-Angle-Side (SAS) Similarity Criterion** instead of **SSS**. ### **Final Answer:** **By calculating all side ratios equal and applying SSS similarity, the second option is correct.**