Example 18 - Find distance of (3, -5) from line 3x - 4y - 26 = 0 - Distance of a point from a line

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
Ask Download

Transcript

Example 18 Find the distance of the point (3, – 5) from the line 3x – 4y –26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |𝐴𝑥_1 + 〖𝐵𝑦〗_2 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Now, our equation is 3x – 4y –26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3 , B = − 4 , C = − 26 & we have to find the distance of the point (3, − 5) from the line So, x1 = 3 , y1 = -5 Now finding distance d = |𝐴𝑥_1 + 〖𝐵𝑦〗_2 + 𝐶|/√(𝐴^2 + 𝐵^2 ) Putting values = |3(3) + ( − 4)( − 5) − 26|/√(32 + ( − 4)2) = |9 + 20 − 26|/√(9 + 16) = |29 − 26|/√25 = |3|/√(5 × 5) = |3|/5 = 3/5 ∴ Required distance = d = 3/5 units

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail