Question

The equation of one of the straight lines passing through the point \( (0, 1) \) and is at a distance of \( \frac{3}{5} \) units from the origin is:

• A. $4x + 3y = 3$
• B. $-x + y = 1$
• C. $x + y = 1$
• D. $5x + 4y = 4$
• E. $-5x + 4y = 4$

Hint:

For lines passing through $(0, c)$, the y-intercept is already known. You can quickly test options by checking if $(0, 1)$ satisfies the equation and then calculating the distance from the origin for only the valid candidates.

Answer 1

The Correct Option is A

Concept: A line passing through $(0, 1)$ can be represented using the slope-intercept form $y = mx + 1$, or $mx - y + 1 = 0$. The perpendicular distance $d$ from the origin $(0, 0)$ to a line $Ax + By + C = 0$ is given by: \[ d = \frac{|C|}{\sqrt{A^2 + B^2}} \]

Step 1: Set up the distance formula equation.
The line equation is $mx - y + 1 = 0$. Here $A = m$, $B = -1$, and $C = 1$. The distance from $(0, 0)$ is $\frac{3}{5}$: \[ \frac{3}{5} = \frac{|1|}{\sqrt{m^2 + (-1)^2}} \]

Step 2: Solve for the slope $m$.
Square both sides: \[ \frac{9}{25} = \frac{1}{m^2 + 1} \] Cross-multiply: \[ 9(m^2 + 1) = 25 \quad \Rightarrow \quad 9m^2 + 9 = 25 \] \[ 9m^2 = 16 \quad \Rightarrow \quad m^2 = \frac{16}{9} \quad \Rightarrow \quad m = \pm \frac{4}{3} \]

Step 3: Form the equation of the line.
Using $m = -4/3$ (since options are in standard form): \[ y = -\frac{4}{3}x + 1 \quad \Rightarrow \quad 3y = -4x + 3 \quad \Rightarrow \quad 4x + 3y = 3 \] This matches option (A).