Example 18 - Chapter 10 Class 11 Straight Lines
Last updated at Feb. 3, 2020 by Teachoo
Transcript
Example 18 Find the distance of the point (3, โ5) from the line 3x โ 4y โ26 = 0. We know that distance (d) of a point (x1, y1) from a line Ax + By + C = 0 is d = |๐ด๐ฅ_1 + ใ๐ต๐ฆใ_2 + ๐ถ|/โ(๐ด^2 + ๐ต^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 = โ4 , C = โ26 & we have to find the distance of the point (3, โ 5) from the line So, x1 = 3 , y1 = โ5 Now finding distance d = |๐ด๐ฅ_1 + ใ๐ต๐ฆใ_2 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Putting values = |3(3) + (โ4)( โ5) โ 26|/โ(32 + (โ4)2) = |9 + 20 โ 26|/โ(9 + 16) = |29 โ 26|/โ25 = |3|/โ(5 ร 5) = |3|/5 = 3/5 โด Required distance = d = ๐/๐ units
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.