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Ex 10.3

Ex 10.3, 1 (i)

Ex 10.3, 1 (ii) Important

Ex 10.3, 1 (iii)

Ex 10.3, 2 (i)

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Ex 10.3, 2 (iii) Important

Ex 10.3, 3 (i) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams

Ex 10.3, 4 You are here

Ex 10.3, 5 Important

Ex 10.3, 6 (i) Important

Ex 10.3, 6 (ii)

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Last updated at Feb. 3, 2020 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 10.3, 4 Find the distance of the point (β1, 1) from the line 12(x + 6) = 5(y β 2). The distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is d = |π΄π₯_1 + π΅π¦_1 + πΆ|/β(π΄^2 + π΅^2 ) The given line is 12(x + 6) = 5(y β 2) 12x + 12 Γ 6 = 5y β 5 Γ 2 12x + 72 = 5y β 10 12x β 5y + 82 = 0 The above equation is of the form Ax + By + C = 0 where A = 12, B = β5, and C = 82 Now We have to find distance of a point (β1, 1) from a line So, x1 = β1 & y1 = 1 Finding distance d = |π΄π₯_1 + π΅π¦_1 + πΆ|/β(π΄^2 + π΅^2 ) Putting values d = | β 12 β 5 + 82|/β((12)2 + ( β 5)2) d = | β 12 β 5 + 82|/β(144 + 25) d = 65/β169 d = 65/β(13 Γ 13) d = 65/13 d = 5 Thus, Required distance = 5 units