Question

If three lines whose equations are $y=m_1 x+c_1$, $y=m_2 x+c_2$ and $y=m_3 x+c_3$, are concurrent, then
I. $m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$
II. $m_1\left(c_1-c_2\right)+m_2\left(c_2-c_3\right)+m_3\left(c_3-c_1\right)=0$

• A. Both I and II are true
• B. Only l is true
• C. Only II is true
• D. Both I and II are false

Answer 1

Answer: (B)

Given equation of lines are
$ y=m_1 x+c_1 .....$(i)
$ y=m_2 x+c_2 .....$(ii)
$ y=m_3 x+c_3.....$(iii)
On solving Eqs. (i) and (ii), we get
$ m_1 x+c_1=m_2 x+c_2$
$\Rightarrow m_1 x-m_2 x=c_2-c_1$
$\Rightarrow x=\frac{c_2-c_1}{m_1-m_2}$
From Eq. (i), $ y=m_1\left(\frac{c_2-c_1}{m_1-m_2}\right)+\frac{c_1}{1}$
$\Rightarrow y=\frac{m_1 c_2-m_1 c_1+m_1 c_1-m_2 c_1}{m_1-m_2}$
$\Rightarrow y=\frac{m_1 c_2-m_2 c_1}{m_1-m_2}$
On putting the values of $x$ and $y$ in Eq. (iii), we get
$ \frac{m_1 c_2-m_2 c_1}{m_1-m_2}=m_3\left(\frac{c_2-c_1}{m_1-m_2}\right)+\frac{c_3}{1} $
$\Rightarrow \frac{m_1 c_2-m_2 c_1}{m_1-m_2}=\frac{m_3 c_2-m_3 c_1+m_1 c_3-m_2 c_3}{m_1-m_2} $
$\Rightarrow m_1 c_2-m_2 c_1=m_3 c_2-m_3 c_1+m_1 c_3-m_2 c_3 $
$\Rightarrow m_1 c_2-m_2 c_1-m_3 c_2+m_3 c_1-m_1 c_3+m_2 c_3=0 $
$\Rightarrow m_1\left(c_2-c_3\right)+m_2\left(c_3-c_1\right)+m_3\left(c_1-c_2\right)=0$
Hence proved.