Question

If the lines ax+by+c=0, bx+cy+a=0,cx+ay+b=0,a not equal to be not equal to c are concurrent then the point of concurrency is​

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{The lines}$

$\mathsf{ax+by+c=0,bx+cy+a=0,cx+ay+b=0}\;\textsf{are concurrent}$

$\underline{\textsf{To find:}}$

$\textsf{The point of concurency}$

$\underline{\textsf{Solution:}}$

$\textsf{Since the lines are concurrent, the point of concurrency}$

$\textsf{can be obtained by solving any two of the equations}$

$\textsf{Now, we solve the equations (1) and (2) by cross multiplication rule}$

$\mathsf{ax+by+c=0}$

$\mathsf{bx+cy+a=0}$

$\textsf{By cross multiplication rule,}$

$\mathsf{\dfrac{x}{ab-c^2}=\dfrac{y}{bc-a^2}=\dfrac{1}{ac-b^2}}$

$\implies\mathsf{\dfrac{x}{ab-c^2}=\dfrac{1}{ac-b^2}}$

$\implies\mathsf{x=\dfrac{ab-c^2}{ac-b^2}}$

$\text{and}$

$\mathsf{\dfrac{y}{bc-a^2}=\dfrac{1}{ac-b^2}}$

$\mathsf{y=\dfrac{bc-a^2}{ac-b^2}}$

$\therefore\textsf{The point of concurrency is}$

$\mathsf{(\dfrac{ab-c^2}{ac-b^2},\dfrac{bc-a^2}{ac-b^2})}$