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

Transcript

Example 18 Find the distance of the point (3, โ€“5) from the line 3x โ€“ 4y โ€“26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |๐ด๐‘ฅ_1 + ใ€–๐ต๐‘ฆใ€—_2 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) Now, our equation is 3x โ€“ 4y โ€“ 26 = 0 The above equation is of the form Ax + By + C = 0 where A = 3, B = โˆ’4 , C = โˆ’26 & we have to find the distance of the point (3, โˆ’ 5) from the line So, x1 = 3 , y1 = โˆ’5 Now finding distance d = |๐ด๐‘ฅ_1 + ใ€–๐ต๐‘ฆใ€—_2 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) Putting values = |3(3) + (โˆ’4)( โˆ’5) โˆ’ 26|/โˆš(32 + (โˆ’4)2) = |9 + 20 โˆ’ 26|/โˆš(9 + 16) = |29 โˆ’ 26|/โˆš25 = |3|/โˆš(5 ร— 5) = |3|/5 = 3/5 โˆด Required distance = d = ๐Ÿ‘/๐Ÿ“ units

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.