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

$(–1,–2)$

The explanation for the correct option:

Given the equation of the straight line is

$\begin{array}{rcl}(a–2b)x+(a+3b)y+3a+4b& =& 0\dots \left(i\right)\\ ax–2bx+ay+3by+3a+4b& =& 0\\ a(x+y+3)+b(-2x+3y+4)& =& 0\\ (x+y+3)+(b/a)(-2x+3y+4)& =& 0\end{array}$

This is of the form$L_1+\lambda L_2=0$

$\begin{array}{rcl}x+y+3& =& 0\dots \left(ii\right)\\ -2x+3y+4& =& 0\dots \left(iii\right)\end{array}$

The given lines always pass through the intersection of both lines.

Solving (ii) and (iii), we get $x=-1,y=-2$

So the point of intersection is $(-1,-2).$

Hence option (A) is the correct answer.