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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Misc 20 Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0. Let the given parallel lines be Line AB : 9x + 6y − 7 = 0 Line CD : 3x + 2y + 6 = 0 Multiplying equation of line CD by 3 3(3x + 2y + 6) = 3(0) 9x + 6y + 18 = 0 So equation of line CD is 9x + 6y + 18 = 0 Let the third line be PQ As PQ is equidistant from parallel lines AB & CD So, PQ parallel to the line AB & CD We need to find equation of line PQ Since Line PQ is parallel to the line AB & CD ∴ Slope of PQ = Slope of AB Finding slope of line AB 9x + 6y − 7 = 0 6y = 7 − 9x 6y = − 9x + 7 y = ( − 9𝑥 + 7)/6 y = (( − 9)/6)x + 7/9 y = (( − 3)/2)x + 7/9 The above equation of the form y = mx + c where m = slope of line ∴ Slope of AB = m = ( − 3)/2 Now, Slope of PQ = Slope of AB ∴ Slope of PQ = ( − 3)/2 Let equation of line PQ be y = mx + c where m is the slope of PQ So, m = ( − 3)/2 Putting values y = ( − 3)/2x + c 2y = -3x + c 2y + 3x − c = 0 3x + 2y − c = 0 Multiplying by 3 3(3x + 2y − c) = 3(0) 9x + 6y − 3c = 0 Also, given that Line PQ is equidistant from the line AB and line CD Distance between line AB & PQ = Distance between line CD & PQ Distance between line AB & PQ Distance between two parallel lines Ax + By + c1 = 0 & Ax + By + c2 = 0 is d = |𝑪_𝟏 − 𝑪_𝟐 |/√(𝑨^𝟐 + 𝑩^𝟐 ) Line CD : 9x + 6y + 18 = 0 The above equation is of the form Ax + By + c1 = 0 Here A = 9 , B = 6 & c1 = 18 Line PQ : 9x + 6y − 3c = 0 The above equation is of the form Ax + By + c2 = 0 Here A = 9 , B = 6 & c2 = –3c Distance between parallel line AB & PQ is d = |−3𝑐 − (−7)|/√(〖(9)〗^2 + 〖(6)〗^2 ) d = |−3𝑐 + 7|/√117 Distance between line CD & PQ d = |−3𝑐 − 18|/√117 Since, Distance between line AB & PQ = Distance between line CD & P |−3𝑐 + 7|/√117 = |−3𝑐 − 18|/√117 |−3𝑐+7| = |−3𝑐 −18| –3c + 7 = –3c – 18 or –3c + 7 = −(–3c − 18) Taking –3c + 7 = –3c − 18 –3c + 3c + 7 = −18 7 = − 18 not possible Taking –3c + 7 = − (–3c − 18) –3c + 7 = 3c + 18 7 – 18 = 3c + 3c –11 = 6c 6c = –11 c = (−11)/6 Hence, Equation of line PQ is 9x + 6y – 3c = 0 9x + 6y – 3((−11)/6) = 0 9x + 6y + 11/2 = 0 18x + 12y + 11 = 0 Which is required equation

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Davneet Singh

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.