TS EAMCET 19-July-2022 Shift 2 Solved Paper

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,
‌=‌

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).