Example 9 - Chapter 9 Class 11 Straight Lines

Last updated at May 29, 2023 by Teachoo

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

Example 18 - Chapter 10 Class 11 Straight Lines - Part 2

Example 18 - Chapter 10 Class 11 Straight Lines - Part 3

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Transcript

Example 9 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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