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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 = |𝐴π‘₯_1 + 〖𝐡𝑦〗_2 + 𝐢|/√(𝐴^2 + 𝐡^2 ) Now, our equation is 3x – 4y – 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = βˆ’4 , C = βˆ’26 & we have to find the distance of the point (3, βˆ’ 5) from the line So, x1 = 3 , y1 = βˆ’5 Now finding distance d = |𝐴π‘₯_1 + 〖𝐡𝑦〗_2 + 𝐢|/√(𝐴^2 + 𝐡^2 ) Putting values = |3(3) + (βˆ’4)( βˆ’5) βˆ’ 26|/√(32 + (βˆ’4)2) = |9 + 20 βˆ’ 26|/√(9 + 16) = |29 βˆ’ 26|/√25 = |3|/√(5 Γ— 5) = |3|/5 = 3/5 ∴ Required distance = d = πŸ‘/πŸ“ units

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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.