Question

If 4a^2+9b^2-c^2+12ab=0,then the family of straight lines ax+by+c=0 is concurrent at

Answer 1

4a² + 9b² +12ab -c² =0
(2a)² + (3c)² +2(2a)(3b) -c² =0
(2a +3b)² -c² =0
(2a + 3b -c)(2a +3b +c) =0

2a + 3b =c or -c

ax + by + c =0
ax + by +2a +3b =0
a( x +2) + b( y +3) =0

hence, at x = -2 , y = -3 (-2, -3)
or ,
ax + by +(-2a -3b) =0
a(x -2) +b( y -3)=0

so, at point (2, 3)

Answer 2

Answer:

(2,3) is the answer

Step-by-step explanation:

(2,3) is the answer