# Misc 9 - Chapter 9 Class 11 Straight Lines

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3

Misc 4

Misc 5 Important

Misc 6

Misc 7 Important

Misc 8 Important

Misc 9 You are here

Misc 10 Important

Misc 11 Important

Misc 12

Misc 13

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22

Misc 23 Important

Question 1 Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Misc 9 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