Question

Show that the straight lines $(a-b)x+(b-c)y=c-a,(b-c)x+(c-a)y=a-b$ and $(c-a)x+(a-b)y=b-c$ are concurrent.

Answer 1

$(a-b)x+(b-c)y+a-c=0$

$(b-c)x+(c-a)y+b-a=0$

$(c-s)x+(a-b)y+c-b=0$

For these to be concurrent

$\left|\begin{array}{ccc}a-b& b-c& a-c\\ b-c& c-a& b-a\\ c-a& a-b& c-b\end{array}\right|$ should be zero

$R_1\to R_1+R_2+R_3\left|\begin{array}{ccc}0& 0& 0\\ b-c& c-a& b-a\\ c-a& a-b& c-b\end{array}\right|$ As $R_1$ is $0$

$\Rightarrow$ this det $=0$