88781
The family of straight lines \((2 a+3 b) x+(a-b) y+2 a-4 b=0\) is concurrent at the point
1 \(\left(\frac{2}{5}, \frac{-14}{5}\right)\)
2 \(\left(\frac{-2}{5}, \frac{-14}{5}\right)\)
3 \(\left(\frac{-2}{5}, \frac{14}{5}\right)\)
4 \(\left(\frac{2}{5}, \frac{14}{5}\right)\)
Explanation:
(A) : Given, the equation of line is,
\((2 a+3 b) x+(a-b) y+2 a-4 b=0\)
\((2 x+y+2) a+(3 x-y-4) b=0\) and for all \(a, b\) the straight lines pass through the intersection of \(2 \mathrm{x}+\mathrm{y}+2=0\) and \(3 \mathrm{x}-\mathrm{y}-4=0\)
i.e. the point \(\left(\frac{2}{5},-\frac{14}{5}\right)\).
