Given, 6a2−3b2−c2+7ab−ac+4bc=0 ⇒c2+c(a−4b)+3b2−6a2−7ab=0 which is the quadratic equation, from the quadratic formula,
c=
4b−a±√a2+16b2−8ab−12b2+24a2+28ab
2
=
4b−a±√25a2+4b2+20ab
2
c=
(4b−a)±(5a+2b)
2
⇒2c=4b−a+5a+2b and 2c=4b−a−5a−2b ⇒4a+6b−2c=0 and 6a−2b+2c=0 ⇒2a+3b−c=0 and 3a−b+c=0 For determining the point of concurrency with ax+by+c=0, compare both equations with ax+by+c=0
x
2
=
y
3
=
1
−1
and
x
3
+
y
(−1)
=
1
1
⇒x=−2,y=−3 and x=3,y=−1 ∴ The lines are concurrent at (−2,−3) and (3,−1).