Question

The lines 3x - y + 2 = 0 and x + 3y + 4 = 0 intersect each other in the
(A) 1st quadrant
(B) 4th quadrant
(C) 3rd quadrant
(D) 2nd quadrant​

Answer 1

Answer:

$\bold\red{(B)\:4th\: quadrant}$

Step-by-step explanation:

Let the given lines be denoted as,

L1 : 3x - y + 2 = 0 ..............(i)

and,

L2 : x + 3y +4 = 0 .................(ii)

Now, we have to find the point of intersection of these lines.

For this, multiply the eqn (i) with 3 and add to eqn (ii),

So, we get,

=> 9x - 3y + 6 + x + 3y + 4 = 0

=> 10x + 10 = 0

=> x = -1

Therefore,

=> -1 + 3y + 4 = 0

=> 3y + 3 = 0

=> y = -1

Therefore, the point of intersection is (-1,-1)

Hence, they intersect in (B) 4th quadrant.