Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex10.3, 15 The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c. Let the equation of line AB be y = mx + c Let OM be perpendicular to the line AB i.e. OM ⊥ AB Also, M = (−1, 2) Point M passes through the line AB So, M(–1,2) must satisfy the equation of line AB Putting values in equation y = mx + c 2 = m(–1) + c 2 = –m + c 2 = c – m c – m = 2 Also, OM is perpendicular to line AB And, we know that if two lines are perpendicular, their product of slope is -1 So, Slope of OM × Slope of AB = − 1 Now, equation of AB is y = mx + c where m is the slope So, Slope of AB = m Now, finding slope of line OM Line OM passes through the points O(0, 0) & M( − 1, 2) is Slope of OM = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 ) = (2 − 0)/( − 1 − 0) = 2/( − 1) = − 2 From (2) Slope of OM × Slope of AB = − 1 − 2 × m = − 1 m = ( − 1)/( − 2) m = 1/2 From (1) c – m = 2 Putting m = 1/2 c − 1/2 = 2 c = 2 + 1/2 c = (4 + 1)/2 c = 5/2 Thus, m = 1/2 & c = 5/2

Ex 10.3

Ex 10.3, 1
Important

Ex 10.3, 2

Ex 10.3, 3

Ex 10.3, 4

Ex 10.3, 5 Important

Ex 10.3, 6 Important

Ex 10.3, 7

Ex 10.3, 8 Important

Ex 10.3, 9

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 You are here

Ex 10.3, 16 Important

Ex 10.3, 17 Important

Ex 10.3, 18 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.