Example 9 - Chapter 9 Class 11 Straight Lines
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 9 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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo