Home › Class 10 › Mathematics › Chapter 7 › Q2 Chapter 7 · Question 2 Find the distance between (2,3)(2,3)(2,3) and (6,6)(6,6)(6,6). menu_book Back to Chapter Answer Direct Answer Distance =(6−2)2+(6−3)2=16+9=5=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{16+9}=5=(6−2)2+(6−3)2=16+9=5. Simple Explanation The points differ by 444 horizontally and 333 vertically, so distance is 555. Exam-Ready Structure Let A(2,3)A(2,3)A(2,3) and B(6,6)B(6,6)B(6,6). Then AB=(6−2)2+(6−3)2=42+32=25=5AB=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{4^2+3^2}=\sqrt{25}=5AB=(6−2)2+(6−3)2=42+32=25=5. Key Points Distance =(6−2)2+(6−3)2=16+9=5=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{16+9}=5=(6−2)2+(6−3)2=16+9=5. Related Questions Q3 How can the distance formula be used to test whether three points are collinear? arrow_forward Q4 State the section formula for internal division. arrow_forward Report a mistake Previous Q1: State the distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$. All questions Next Q3: How can the distance formula be used to test whether three points are collinear?