Check sibling questions

Ex 10.3, 15 - Perpendicular from the origin to y = mx + c - Two lines // or/and prependicular

Ex 10.3, 15 - Chapter 10 Class 11 Straight Lines - Part 2
Ex 10.3, 15 - Chapter 10 Class 11 Straight Lines - Part 3 Ex 10.3, 15 - Chapter 10 Class 11 Straight Lines - Part 4

This video is only available for Teachoo black users


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

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.