If 4 a^ 2 +b^ 2 +2 c^ 2 +4 a b-6 a c-3 b c=0, the family of lines ax+by+c=0 is concurrent at one or the other of the two points-

Correct answer: A. (-1,-(1)/(2)),(-2,-1) array l 4 a^ 2 +b^ 2 +2 c^ 2 +4 a b-6 a c-3 b c (2 a+b)^ 2 -3(2 a+b) c+2 c^ 2 =0 (2 a+b-2 c)(2 a+b-c)=0 c=2 a+b or c=a+(1)/(2)b array The equation of the family of lines is a(x+2)+b(y+1)=0 or a(x+1)+b( y+ (1)/(2))= 0 giving the point of concurrence (-2,-1) or (- 1,- (1)/(2))

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Mathematics
Chapter: Straight Lines

BITSAT 2013 · Mathematics · Straight Lines

If 4 a^ 2 +b^ 2 +2 c^ 2 +4 a b-6 a c-3 b c=0, the family of lines...

\quad

4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0,

the family of lines

\mathrm{ax}+\mathrm{by}+\mathrm{c}=0

is concurrent at one or the other of the two points-

Options

A $\left(-1,-\frac{1}{2}\right),(-2,-1)$
B(-1,-1), \left(- $2,$ -\frac{1}{2}\right)
C $(-1,2), \left(\frac{1}{2},-1\right)$
D $(1,2), \left(\frac{1}{2},-1\right)$

Correct answer

A. $\left(-1,-\frac{1}{2}\right),(-2,-1)$

Step-by-step solution

\begin{array}{l}4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c \\ \quad \equiv(2 a+b)^{2}-3(2 a+b) c+2 c^{2}=0 \\ \quad \Rightarrow(2 a+b-2 c)(2 a+b-c)=0 \Rightarrow c=2 a+b \\ \quad \text{ or }c=a+\frac{1}{2}b \end{array}

The equation of the family of lines is a(x+2)+b(y+1)=0

\text{ or }

a(x+1)+b\left(

y+

\frac{1}{2}\right)=

0 giving the point of concurrence (-2,-1) or

\left(-

1,-

\frac{1}{2}\right)$

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