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

Ex 10.3, 4 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 4 - Chapter 10 Class 11 Straight Lines - Part 3

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Ex 10.3, 4 Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2). The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is d = |𝐴π‘₯_1 + 𝐡𝑦_1 + 𝐢|/√(𝐴^2 + 𝐡^2 ) The given line is 12(x + 6) = 5(y – 2) 12x + 12 Γ— 6 = 5y – 5 Γ— 2 12x + 72 = 5y – 10 12x – 5y + 82 = 0 The above equation is of the form Ax + By + C = 0 where A = 12, B = –5, and C = 82 Now We have to find distance of a point (βˆ’1, 1) from a line So, x1 = –1 & y1 = 1 Finding distance d = |𝐴π‘₯_1 + 𝐡𝑦_1 + 𝐢|/√(𝐴^2 + 𝐡^2 ) Putting values d = | βˆ’ 12 βˆ’ 5 + 82|/√((12)2 + ( βˆ’ 5)2) d = | βˆ’ 12 βˆ’ 5 + 82|/√(144 + 25) d = 65/√169 d = 65/√(13 Γ— 13) d = 65/13 d = 5 Thus, Required distance = 5 units

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