Question

The lines $(p-q)x+(q-r)y+(r-p)=0,(q-r)x+(r-p)y+(p-q)=0,(r-p)x+(p-q)y+(q-r)=0$are

• A. parallel
• B. perpendicular
• C. concurrent
• D. none of these

Answer 1

The correct option is C

concurrent

Explanation for correct option

Option$C$: Concurrent

$$\begin{gathered} (p-q)x+(q-r)y+(r-p)=0.....\left(1\right) \\ (q-r)x+(r-p)y+(p-q)=0.....\left(2\right) \\ (r-p)x+(p-q)y+(q-r)=0.....\left(3\right) \end{gathered}$$

For, lines to be concurrent

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}p-q& q-r& r-p\\ q-r& r-p& p-q\\ r-p& p-q& q-r\end{array}\right|$

$$\begin{gathered} R_1=R_1+R_2+R_3 \\ \Rightarrow \left|\begin{array}{ccc}0& 0& 0\\ q-r& r-p& p-q\\ r-p& p-q& q-r\end{array}\right|=0 \end{gathered}$$ [From properties of determinants]

So, the lines are concurrent

Hence, the correct option is $C$

Explanation for incorrect options

Option $A$: Parallel

For lines to be the parallel lines, the slope of lines must be the same

$$\begin{gathered} m_1=\frac{-(p-q)}{(q-r)} \\ {m}_2=\frac{-(q-r)}{(r-p)} \\ {m}_3=\frac{-(r-p)}{(p-q)} \end{gathered}$$

Since, $m_1\ne m_2\ne m_3$

Hence, the option $A$ is incorrect.

Option $B$: Perpendicular

For lines to be perpendicular, the multiplication of slope should be equal to

$$\begin{gathered} m_1\times m_2=\frac{(p-q)}{\left(r-p\right)}\ne -1 \\ {m}_2\times m_3=\frac{\left(q-r\right)}{(p-q)}\ne -1 \\ {m}_3\times m_1=\frac{\left(r-p\right)}{(q-r)}\ne -1 \end{gathered}$$

Hence, the option $B$ is incorrect.

Hence, the only correct option is $C$, so, the given lines are concurrent.