Question

Show that the straight lines $L_1=(b+c)x+ay+1=0,L_2=(c+a)x+by+1=0$ and $L_3=(a+b)x+cy+1=0$ are concurrent.

Answer 1

The three lines are concurrent if they have the common point of intersection

$(b+c)x+ay+1=0$

$(c+a)x+by+1=0$

$(a+b)x+cy+1=0$

Solving (1) and (2)

$y=\frac{-1-(b+c)x}{a}$

Putting in (2)

$(c+a)x+b\frac{(-1-(b+c\left)x\right)}{a}+1=0$

$acx+a^{2}x+b-b^{2}x-bcx+a=0$

$x(ac+a^2-b^2-bc)=b-a$

$x(ac-bc+a^2-b^2)=b-a$

$x\left(c\right(a-b)+(a-b\left)\right(a+b\left)\right)=b-a$

$x(c+a+b)=-1\left[Cancelling\right(a-b\left)bothsides\right]$

$x=\frac{-1}{a+b+c}$

$y=\frac{-1+\frac{(b+c)(-1)}{a+b+c}}{a}=\frac{-a-b-c-b-c}{a(a+b+c)}$

Putting the value of x, y in (3) ;

$(a+b)\left(\frac{-1}{a+b+c}\right)+c\left(\frac{-a-2b-2c}{a(a+b+c)}\right)+1=0$

$-a^2-ba-ac-2bc-2c^2+a^2+ab+ac=0$

$0=0$

Hence, the lines are concurrent.