Last updated at Feb. 3, 2020 by

Transcript

Misc 15 Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0. There are two lines Line AB 4x + 7y + 5 = 0 Line CD 2x – y = 0 Both lines meet at Q Point P(1, 2) is on line CD We need to find distance PQ. In PQ, P is (1, 2) We need to find point Q To find PQ, we must find coordinates of point Q Finding coordinate of point Q Point Q is the intersection of lines AB & CD From (1) 2x – y = 0 2x = y y = 2x Putting value of y in (2) 4x + 7(2x) + 5 = 0 4x + 14x + 5 = 0 18x = –5 x = (−5)/18 Putting x = (−5)/18 in (1) 2x – y = 0 2 ((−5)/18) – y = 0 (−5)/9 – y = 0 (−5)/9 = y y = (−5)/9 Hence coordinates of point Q is (( −𝟓)/𝟏𝟖,( −𝟓)/𝟗) Now, we have to find distance PQ We know that distance of points (x1, y1 ) & (x2, y2) is d = √((𝑥_2−𝑥_1 )2 + (𝑦_2−𝑦_1 ) ) 2 Distance PQ where P(1, 2) & Q(( −5)/18, ( −5)/9) is PQ = √((( − 5)/18−1)^2+(( − 5)/9 −2)^2 ) = √((( − 5 − 18)/18)^2+(( −5 − 18)/9)^2 ) = √(((−23)/18)^2+((−23)/9)^2 ) = √(((−23)/(2 × 9))^2+((−23)/9)^2 ) = √((( −23)/( 9))^2 (1/4+1) ) = √((( − 23)/( 9))^2 ((1 + 4)/4) ) = √((( − 23)/( 9))^2 (5/4) ) = (−23)/9 √(5/4) = (−23)/9 √(5/2^2 ) = (−23)/(9 ) √5/2 = ( −23√5)/18 But distance is always positive Hence distance of PQ = (𝟐𝟑√𝟓)/𝟏𝟖

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4

Misc 5

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13

Misc 14

Misc 15 Important You are here

Misc 16 Important

Misc 17

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23

Misc 24 Important

Chapter 10 Class 11 Straight Lines (Term 1)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.