Ex 10.3, 4 - Chapter 10 Class 11 Straight Lines (Term 1)
Last updated at Feb. 3, 2020 by Teachoo
Last updated at Feb. 3, 2020 by Teachoo
Transcript
Ex 10.3, 4 Find the distance of the point (โ1, 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 = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^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 = โ5, and C = 82 Now We have to find distance of a point (โ1, 1) from a line So, x1 = โ1 & y1 = 1 Finding distance d = |๐ด๐ฅ_1 + ๐ต๐ฆ_1 + ๐ถ|/โ(๐ด^2 + ๐ต^2 ) Putting values d = | โ 12 โ 5 + 82|/โ((12)2 + ( โ 5)2) d = | โ 12 โ 5 + 82|/โ(144 + 25) d = 65/โ169 d = 65/โ(13 ร 13) d = 65/13 d = 5 Thus, Required distance = 5 units
Ex 10.3
Ex 10.3, 1 (ii) Important
Ex 10.3, 1 (iii)
Ex 10.3, 2 (i)
Ex 10.3, 2 (ii)
Ex 10.3, 2 (iii) Important
Ex 10.3, 3 (i)
Ex 10.3, 3 (ii)
Ex 10.3, 3 (iii) Important
Ex 10.3, 4 You are here
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Ex 10.3, 6 (i) Important
Ex 10.3, 6 (ii)
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Ex 10.3
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