Question

Show that the straight lines given by $(2+k)x+(1+k)y=5+7k$ for different value of k pass through a fixed point. Also, if that point is $(a,b)$. Find $a+b.$

Answer 1

The given straight line $(2+k)x+(1+k)y-5+7k$ can written as

$2x+y-5+k(x+y-7)=0$

This lines is of the form $L_1+K{L}_2=0$ which passes through the intersection of the lines $L_1=0$ and $L_2=0$

i.e $2x+y-5=0$ and $x+y-7=0$

On solving $2x+y=5$ and $x+y=7$ we get,

$x=-2$ and $y=9$

Hence the required fixed point is $(-2,9)=(a,b)\Rightarrow a=-2,b=9$

$\therefore a+b=-2+9=7$