Example 18 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Sept. 6, 2021 by Teachoo
Last updated at Sept. 6, 2021 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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