Question

The family of lines $x(a+b)+y(a-b)=2a,a,b\in R$ are concurrent at $(p,q)$ then the value of $p+q$ is

Answer 1

Given line $x(a+b)+y(a-b)=2a$
$\Rightarrow a(x+y-2)+b(x-y)=0$
$\Rightarrow (x+y-2)+\frac{b}{a}(x-y)=0$
This is a family of lines concurrent at point of intersection of lines $x-y=0$ and $x+y-2=0$.
On solving these two equations we get $(p,q)\equiv (1,1)\therefore p+q=2$