Ex 10.3, 4 - Find distance of (-1, 1) from 12(x + 6) = 5(y - 2) - Distance of a point from a line

10.3,4 ii.jpg
10.3,4 iii.jpg

  1. Chapter 10 Class 11 Straight Lines
  2. Serial order wise
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Ex1 0.3, 4 Find the distance of the point (โ€“1, 1) from the line 12(x + 6) = 5(y โ€“ 2). The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is d = |๐ด๐‘ฅ_1 + ๐ต๐‘ฆ_1 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) The given line is 12(x + 6) = 5(y โ€“ 2) 12x + 12 ร— 6 = 5y โ€“ 5 ร— 2 12x + 72 = 5y โ€“ 10 12x โ€“ 5y + 82 = 0 The above equation is of the form Ax + By + C = 0 where A = 12, B = โ€“5, and C = 82 Now We have to find distance of a point (-1, 1) from a line So, x1 = โ€“ 1 & y1 = 1 Finding distance d = |๐ด๐‘ฅ_1 + ๐ต๐‘ฆ_1 + ๐ถ|/โˆš(๐ด^2 + ๐ต^2 ) Putting values d = | โˆ’ 12 โˆ’ 5 + 82|/โˆš((12)2 + ( โˆ’ 5)2) d = | โˆ’ 12 โˆ’ 5 + 82|/โˆš(144 + 25) d = 65/โˆš169 d = 65/โˆš(13 ร— 13) d = 65/13 d = 5 Thus, Required distance = 5 unit

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