Question

If 4a2+9b2-c2+12ab=0, then the family of straight lines ax+by+c=0 is concurrent at

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

Answer 1

Answer: (A)

(2,3)or(-2,-3)

Answer 2

The correct answer is option (A) : (2,3)or(-2,-3)
$4a^2+9b^2+12ab-c^2=0$
$\Rightarrow (2a+3b)^2-c^2=0$
$\Rightarrow (2a+3b+c)(2a+3b-c)=0$
$\Rightarrow (2a+3b+c) =0 \,\,or\,\,(2a+3b-c)=0$
$\therefore ax+by+c=0$ passes through the points (2,3) and (-2,-3)