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
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 You are here
Ex 10.3, 16 Important
Ex 10.3, 17 Important
Ex 10.3, 18 Important
Ex 10.3
Last updated at Dec. 8, 2016 by Teachoo
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