Misc 15 - Find distance of 4x + 7y + 5 = 0 from (1, 2) along line

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Misc 14 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.