For a, b, c {R}, if 6 a²-3 b²-c²+7 a b-a c+4 b c=0 and |a|+|b| ≠ 0, then all the lines given by a x+b y+c=0 are

Correct Answer is : concurrent at (−2,−3)(-2,-3)(−2,−3) or (3,−1)(3,-1)(3,−1)

parallel to each other a, b, c {R}

concurrent at (-2,-3) or (3,-1)

Correct Answer is : concurrent at (−2,−3)(-2,-3)(−2,−3) or (3,−1)(3,-1)(3,−1)

Test Index

TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 46 of 160

Marks: +1, -0

For

a, b, c \in \mathbb{R}

6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0

|a|+|b| \neq 0

a x+b y+c=0

are

Solution:

Given,

6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0

\Rightarrow c^2+c(a-4 b)+3 b^2-6 a^2-7 a b=0

which is the quadratic equation, from the quadratic formula,

c = \frac{4b-a \pm \sqrt{a^2+16b^2-8ab-12b^2+24a^2+28ab}}{2}

= \frac{4b-a \pm \sqrt{25a^2+4b^2+20ab}}{2}

c = \frac{(4b-a) \pm (5a+2b)}{2}

\Rightarrow 2c = 4b-a+5a+2b

and

2c = 4b-a-5a-2b

\Rightarrow 4a+6b-2c=0

and

6a-2b+2c=0

\Rightarrow 2a+3b-c=0

and

3a-b+c=0

For determining the point of concurrency with

a x+b y+c=0

, compare both equations with

a x+b y+c=0

\frac{x}{2} = \frac{y}{3} = \frac{1}{-1} \text{ and } \frac{x}{3} + \frac{y}{-1} = \frac{1}{1}

\Rightarrow x=-2, y=-3

and

x=3, y=-1

\therefore

The lines are concurrent at

(-2,-3)

and

(3,-1)

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