If a line L is perpendicular to the line 5x - y = 1, and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line x + by = 0 is :

Question

If a line L is perpendicular to the line 5x - y = 1, and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line x + by = 0 is : (A) (7/√5) (B) (5/√13) (C) (7/√13) (D) (5/√7)

Tardigrade Admin · Accepted Answer

Equation = of line L is x+5 y+c=0 (c2/2|1 × 5|)=5 ⇒ c=± 5 √2 ∴ Eq of line L is x+5 y ± 5 √2=0 Its distance from x+5 y=0 is (5/√13)

Q. If a line $L$ is perpendicular to the line $5 x - y = 1$ , and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ , then the distance of line $L$ from the line $x + b y = 0$ is :

$\frac{7}{5}$

$\frac{5}{13}$

$\frac{7}{13}$

$\frac{5}{7}$

Equation = of line L is
$x + 5 y + c = 0$
$\frac{c ^{2}}{2∣1 \times 5∣} = 5$
$\Rightarrow c = \pm 52$
$∴$ Eq of line $L$ is
$x + 5 y \pm 52 = 0$
Its distance from $x + 5 y = 0$ is $\frac{5}{13}$

Q. If a line L is perpendicular to the line 5x−y=1, and the area of the triangle formed by the line L and the coordinate axes is 5, then the distance of line L from the line x+by=0 is :

5​7​

13​5​

13​7​

7​5​