Check sibling questions

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 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 = |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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.