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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) 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)
Ex 10.3, 7
Ex 10.3, 8 Important
Ex 10.3, 9 Important
Ex 10.3, 10
Ex 10.3, 11
Ex 10.3, 12 Important
Ex 10.3, 13
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important
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