# Ex 10.1, 3

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex10.1, 3 Find the distance between P (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis. Since PQ is parallel to y-axis x1 = x2. Distance PQ = √("(x2 " − " x1)2 " + "(y2 " − " y1)2" ) = √("(x2 " − " x2)2 " + "(y2 " − " y1)2" ) = √("(y2 " − " y1)2" + 0) = √("(y2 " − " y1)2" ) = ± (y2 – y1) = |"y2" −" y1" | Ex10.1, 3 Find the distance between P (x1, y1) and Q (x2, y2) when: (ii) PQ is parallel to the x-axis. Since PQ is parallel to x-axis y1 = y2. Distance PQ = √("(x2 " − " x1)2 " + "(y2 " − " y1)2" ) = √("(x2 " − " x1)2 " + "(y1 " − " y1)2" ) = √("(x2 " − " x1)2" + 0) = √("(x2 " − " x1)2" ) = ± (x2 – x1) = |"x2" −" x1" |

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .