Question

If $a$ and $b$ are two arbitrary constants, then the straight line $(a-2b)x+(a+3b)y+3a+4b=0$ will pass through

• A. $(-1,-2)$
• B. $(1,2)$
• C. $(-2,-3)$
• D. $(2,3)$

Answer 1

The correct option is A $(-1,-2)$

We know that family of lines $L_1+\lambda L_2=0$ passes through a fixed point which is intersection of lines $L_1$ and $L_2$

$(a-2b)x+(a+3b)y+3a+4b=0$

$=a(x+y+3)+b(-2x+3y+4)=0$

This represents a family of lines $x+y+3$ and $3y-2x+4$

$\therefore$ The given lines always passes through intersections of both the lines.

Solving lines $:$

$2x+2y+6=0$

$-2x+3y+4=0$

Adding both, we get $x=-1;y=-2$