Q82 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
The perpendicular distance of a line from the origin is 5 units and its slope is $-1$.
The equation of the line is
- A$x+y \pm 2 \sqrt{5}=0$
- B$x-y \pm 5 \sqrt{2}=0$
- Correct answerC$x+y \pm 5 \sqrt{2}=0$
- D$x-y \pm 2 \sqrt{5}=0$
Solution
The equation of a line with slope $m = -1$ can be written in the form $y = mx + c$, which is $y = -x + c$ or $x + y - c = 0$.
The perpendicular distance $d$ of the line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \dfrac{|C|}{\sqrt{A^2 + B^2}}$.
Here, $A = 1$, $B = 1$, and $C = -c$. The distance $d = 5$.
Substituting the values into the formula: $5 = \dfrac{|-c|}{\sqrt{1^2 + 1^2}} = \dfrac{|c|}{\sqrt{2}}$.
This gives $|c| = 5\sqrt{2}$, so $c = \pm 5\sqrt{2}$.
Substituting $c$ back into the equation $x + y - c = 0$, we get $x + y - (\pm 5\sqrt{2}) = 0$, which simplifies to $x + y \pm 5\sqrt{2} = 0$.
Answer: $x+y \pm 5 \sqrt{2}=0$
The perpendicular distance $d$ of the line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \dfrac{|C|}{\sqrt{A^2 + B^2}}$.
Here, $A = 1$, $B = 1$, and $C = -c$. The distance $d = 5$.
Substituting the values into the formula: $5 = \dfrac{|-c|}{\sqrt{1^2 + 1^2}} = \dfrac{|c|}{\sqrt{2}}$.
This gives $|c| = 5\sqrt{2}$, so $c = \pm 5\sqrt{2}$.
Substituting $c$ back into the equation $x + y - c = 0$, we get $x + y - (\pm 5\sqrt{2}) = 0$, which simplifies to $x + y \pm 5\sqrt{2} = 0$.
Answer: $x+y \pm 5 \sqrt{2}=0$
Hence, the correct answer is option: 'C'.