Q.
If three lines whose equations are y=m1x+c1, y=m2x+c2 and y=m3x+c3, are concurrent, then
I. m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0
II. m1(c1−c2)+m2(c2−c3)+m3(c3−c1)=0
Given equation of lines are y=m1x+c1.....(i) y=m2x+c2.....(ii) y=m3x+c3.....(iii)
On solving Eqs. (i) and (ii), we get m1x+c1=m2x+c2 ⇒m1x−m2x=c2−c1 ⇒x=m1−m2c2−c1
From Eq. (i), y=m1(m1−m2c2−c1)+1c1 ⇒y=m1−m2m1c2−m1c1+m1c1−m2c1 ⇒y=m1−m2m1c2−m2c1
On putting the values of x and y in Eq. (iii), we get m1−m2m1c2−m2c1=m3(m1−m2c2−c1)+1c3 ⇒m1−m2m1c2−m2c1=m1−m2m3c2−m3c1+m1c3−m2c3 ⇒m1c2−m2c1=m3c2−m3c1+m1c3−m2c3 ⇒m1c2−m2c1−m3c2+m3c1−m1c3+m2c3=0 ⇒m1(c2−c3)+m2(c3−c1)+m3(c1−c2)=0
Hence proved.