The family of straight lines \((2 a+3 b) x+(a-b) y+2 a-4 b=0\) is concurrent at the point
The family of straight lines \((2 a+3 b) x+(a-b) y+2 a-4 b=0\) is concurrent at the point
$\left(\frac{2}{5}, \frac{-14}{5}\right)$
$ \left( \frac{-2}{5}, \frac{-14}{5} \right) $
$ \left( \frac{-2}{5}, \frac{14}{5} \right) $
$\left(\frac{2}{5}, \frac{14}{5}\right)$
Solution
Rewriting the equation \((2 x+y+2) a+(3 x-y-4) b=0\) and for all \(a\), \(\mathrm{b}\) the straight lines pass through the intersection of \(2 x+y+2=0\) and \(3 x-y-4=0\) i.e., the point $\left(\frac{2}{5},-\frac{14}{5}\right)$.