Question

If three lines whose equations are `y=m_1x+c_1,y=m_2x+c_2`and `y=m_3x+c_3`are concurrent, then show that `m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0`.

Answer 1

The given line are:
`m_(1)x-y+c_(1)=0......(i)`
`m_(2)x-y+c_(2)=0......(ii)`
`m_(3)x-y+c_(3)=0......(iii)`
On solving (i) and (ii) by cross multiplication, we get
`(x)/((-c_(2)+c_(1)))=(y)/((m_(2)c_(1)-m_(1)c_(2)))=(1)/((-m_(1)+m_(2)))`
`Rightarrow x=((c_(1)-c_(2)))/((m_(2)-m_(1)))and y=((m_(2)c_(1)-m_(1)c_(2)))/((m_(2)-m_(1)))`
Thus, the point of intersection of (i) and (ii) is P `((c_(1)-c_(2))/(m_(2)-m_(1)),(m_(2)c_(1)-m_(1)c_(2))/(m_(2)-m_(1)))`
Since the given three lines meet at a point, the point P must therefore lie on (ii) also.
`therefore m_(3).((c_(1)-c_(2)))/((m_(2)-m_(1)))-((m_(2)c_(1)-m_(1)c_(2)))/((m_(2)-m_(1)))+c_(3)=0`
`Rightarrow m_(3)(c_(1)-c_(2))-(m_(2)c_(1)-m_(1)c_(2))+c_(3)(m_(2)-m_(1))=0`
`Rightarrow m_(1)(c_(2)-c_(3))-m_(2)(c_(3)-c_(1))-m_(3)(c_(1)-c_(2))=0`