Question

If the straight lines $x+y-2=0,2x-y+1=0$ and $ax+by-c=0$ are concurrent, then the family of lines $2ax+3by+c=0$ ($a,b,c$ are nonzero) is concurrent at

• A. $(2,3)$
• B. $\left(\frac{1}{2},\frac{1}{3}\right)$
• C. $\left(\frac{-1}{6},\frac{-5}{9}\right)$
• D. $\left(\frac{2}{3},\frac{-7}{5}\right)$

Answer 1

Answer: (C)

$\left(\frac{-1}{6},\frac{-5}{9}\right)$

Solving lines $x+y-2=0$ and $2x-y+1=0$

We get $x=\frac{1}{3}$ $⟹y=2-x=2-\frac{1}{3}=\frac{5}{3}$
Substitute $(x,y)$ in $ax+by-c=0$

We get $\frac{a}{3}+\frac{5b}{3}=c.....\left(i\right)$
Substitute value of c in $2ax+3by+c=0$ we get $2ax+3by+\frac{a}{3}+\frac{5b}{3}=0$
$⟹a(2x+\frac{1}{3})+b(3y+\frac{5}{3})=0$ this family will always pass through $x=-\frac{1}{6},y=-\frac{5}{9}.$