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Misc 10 - If three lines are y = m1x + c1, y = m2x + c2 - Other Type of questions - 3 lines Concurrent

Misc 10 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 10 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 10 - Chapter 10 Class 11 Straight Lines - Part 4

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Misc 10 If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3 x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3(c1 –c2) = 0. Three lines are concurrent if they pass through a common point i.e. point of intersection of any two lines lies on the third line It is given that lines y = m1x + c1 y = m2x + c2 y = m3x + c3 are concurrent So, finding point of intersection of lines (1) & (3) Subtracting (1) from (2) y – y = (m1x + c1) − (m2x + c2 ) 0 = m1x + c1 − m2x − c2 − m1x + m2x = c1 – c2 x( − m1 + m2) = c1 – c2 x(m2 − m1) = c1 – c2 x = (𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 ) Putting value of x in equation (1) y = m1x + c1 y = m1((𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 )) + c1 y = (𝑚_1 (𝑐_1 − 𝑐_2))/(𝑚_2 〖 − 𝑚〗_1 ) + c1 So ,Point of intersection of line (1) & (2) is ((𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 ),(𝑚_1 (𝑐_1 − 𝑐_2))/(𝑚_2 〖 − 𝑚〗_1 ) " + c1" ) Since the three lines are concurrent, therefore point((𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 ),(𝑚_1 (𝑐_1 − 𝑐_2))/(𝑚_2 〖 − 𝑚〗_1 ) " + c1" ) will satisfy the equation of third line Now putting x = (𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 ) & y = (𝑚_1 (𝑐_1 − 𝑐_2))/(𝑚_2 〖 − 𝑚〗_1 ) " " + " c1" in equation (3) y = m3x + c3 (𝑚_1 (𝑐_1 − 𝑐_2))/(𝑚_2 〖 − 𝑚〗_1 ) + c1 = m3 ((𝑐_1 − 〖 𝑐〗_2)/(𝑚_2 − 𝑚_1 )) + c3 (𝑚_1 (𝑐_1 − 𝑐_2 ) + 〖 𝑐〗_1 (𝑚_2 − 𝑚_1))/(𝑚_2 〖 − 𝑚〗_1 ) = (〖𝑚_3 (𝑐〗_1 − 〖 𝑐〗_2) + 〖 𝑐〗_3 (𝑚_2 − 𝑚_1))/(𝑚_2 − 𝑚_1 ) m1(c1 − c2) + c1(m2 − m1) = m3(c1 − c2) + c3(m2 − m1) m1(c1 − c2) + c1m2 − c1m1 = m3(c1 − c2) + c3m2 − c3m1 m1(c1 − c2) + c1m2 − c1m1 − m3(c1 − c2) − c3m2 + c3m1 = 0 m1(c1 − c2) + c3m1 − c1m1 + c1m2 − c3m2 − m3(c1 − c2) = 0 m1(c1 − c2 + c3 − c1) + m2(c1 − c3) − m3(c1 − c2) = 0 m1( − c2 + c3) + m2(c1 − c3) − m3(c1 − c2) = 0 − m1(c2 − c3) − m2(c3 − c1) − m3(c1 − c2) = 0 − [m1(c2 − c3) + m2(c3 − c1) + m3(c1 − c2)] = 0 m1(c2 − c3) + m2(c3 − c1) + m3(c1 − c2) = 0 Hence proved Hence proved

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.