Question

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

Answer 1

@Miracrown

Answer 2

Any ideas how to begin? O.o

Answer 3

Concept of pair of straight lines? o.O :-[

Answer 4

2a+3b=+-c

Answer 5

2a+3b+c=0
-2a-3b+c=0

What do I do then?

Answer 6

Are we given any restrictions on the values of a, b, and c besides the first equation?

Answer 7

No restrictions.

Answer 8

What about the 12ab term?

Answer 9

3*2*2=12

Answer 10

Dang I got the answer :P

Answer 11

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

Answer 12

ha ha, really ?

Answer 13

yeah both satisfy the equation so those are the points :|

Answer 14

(-2,-3)

Answer 15

yup!

Answer 16

@sourwing 2 points right?(2,3) or (-2,-3)

Answer 17

It seems to imply that there is some point where all lines of the form ax+by+c=0 intersect if a, b, and c satisfy the equation given.

Answer 18

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

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

either way, both equations of the the lines go through (-2,-3)