If it is possible to draw a line which belongs to all the given family of lines y-2 x+1+λ1(2 y-x-1)=0,3 y-x-6 +λ2(y-3 x+6)=0, a x+y-2+λ3(6 x+a y-a)=0, then

Question

If it is possible to draw a line which belongs to all the given family of lines y-2 x+1+λ1(2 y-x-1)=0,3 y-x-6 +λ2(y-3 x+6)=0, a x+y-2+λ3(6 x+a y-a)=0, then (A) a=4 (B) a=3 (C) a=-2 (D) a=2

Tardigrade Admin · Accepted Answer

The first two families of lines pass through

(1, 1)

and

(3, 3)

, respectively. The point of intersection of
the lines belonging to the third family of lines will lie on line

y = x

. Hence,

a x + x - 2 = 0 and 6 x + a x - a = 0

or

\frac{2}{a + 1} = \frac{a}{6 + a}

or

a^{2} - a - 12 = 0

or

(a - 4) (a + 3) = 0

Q. If it is possible to draw a line which belongs to all the given family of lines $y - 2 x + 1 + λ_{1} (2 y - x - 1) = 0, 3 y - x - 6$ $+ λ_{2} (y - 3 x + 6) = 0, a x + y - 2 + λ_{3} (6 x + a y - a) = 0$ , then

$a = 4$

$a = 3$

$a = - 2$

$a = 2$

Q. If it is possible to draw a line which belongs to all the given family of lines y−2x+1+λ1​(2y−x−1)=0,3y−x−6 +λ2​(y−3x+6)=0,ax+y−2+λ3​(6x+ay−a)=0, then

a=4

a=3

a=−2

a=2

The first two families of lines pass through (1,1) and (3,3), respectively. The point of intersection of the lines belonging to the third family of lines will lie on line y=x. Hence, ax+x−2=0 and 6x+ax−a=0 or a+12​=6+aa​ or a2−a−12=0 or (a−4)(a+3)=0

Q. If it is possible to draw a line which belongs to all the given family of lines $y - 2 x + 1 + λ_{1} (2 y - x - 1) = 0, 3 y - x - 6$ $+ λ_{2} (y - 3 x + 6) = 0, a x + y - 2 + λ_{3} (6 x + a y - a) = 0$ , then

$a = 4$

$a = 3$

$a = - 2$

$a = 2$