Column-I Column-II
A) Lines $x - 2 y - 6 = 0, 3 x + y - 4 = 0$ and $λ x + 4 y + λ^{2} = 0$ are concurrent. The value of $λ$ is
B) The points $(λ + 1, 1), (2 λ + 1, 3)$ and $(2 λ + 2, 2 λ)$ are collinear, then the value of $λ$ is
C) If line $x + y - 1 - λ = 0$ , passing through the intersections of x - y + 1 = 0 and $3 x + y - 5 = 0$ is perpendicular
to one of them then the value of $λ$ is
D) If the line $y - x - 1 + λ = 0$ is equally inclined to axes and equidistant from the points (1, -2) and (3, 4) then $λ$ is
p) 2
q) -4
r) $- \frac{1}{2}$
s) 1

Moderate

Solution

A) $|\begin{matrix} 1 & - 2 & - 6 \\ 3 & 1 & - 4 \\ λ & 4 & λ^{2} \end{matrix}| = 0$
$\Rightarrow 1 (λ^{2} + 16) + 2 (3 λ^{2} + 4 λ) - 6 (12 - λ) = 0$
$\Rightarrow 7 λ^{2} + 14 λ - 56 = 0 \Rightarrow λ^{2} + 2 λ - 8 = 0$
$\Rightarrow λ = 2, - 4$
B) Collinear $\Rightarrow \frac{2}{λ} = 2 λ - 3 \Rightarrow λ = - \frac{1}{2}, 2$
C) Point of intersection of x-y+1=0 and 3x+y-5=0 is (1,2). (1,2) lies on- $x + y - 1 - λ = 0 \Rightarrow λ = 2$
D) Perpendicular distance form (1,-2) and (3,4) to y-x-1 $+ λ = 0$ are equal
$\Rightarrow \frac{| λ - 4 |}{\sqrt{2}} = \frac{| λ |}{\sqrt{2}} \Rightarrow λ - 4 = - λ$
$\Rightarrow λ = 2$

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