To find the equations of the lines passing through the point \( (1,0) \) and at a distance \( \frac{\sqrt{3}}{2} \) from the origin, follow these steps:
where \( A, B, C \) are the coefficients of a line equation in the form \( Ax + By + C = 0 \), and \( (x_1, y_1) \) is a point on the line. Here, the distance \( d \) from the origin to the line is given as \( \frac{\sqrt{3}}{2} \).
Substitute the point \( (1, 0) \) into the line equation \( Ax + By + C = 0 \) to find the possible forms of \( C \):
\(A \cdot 1 + B \cdot 0 + C = A + C = 0 \Rightarrow C = -A\)
Plug in the distance formula for \( d = \frac{\sqrt{3}}{2} \):
Using the condition \( C = -A \), substitute in the above equation:
\(\frac{|A|}{\sqrt{A^2 + B^2}} = \frac{\sqrt{3}}{2}\)
Squaring both sides, we get:
\(\frac{A^2}{A^2 + B^2} = \frac{3}{4}\)
Solving for the relationship between \( A \) and \( B \), we get:
\(4A^2 = 3A^2 + 3B^2 \Rightarrow A^2 = 3B^2 \Rightarrow \frac{A}{B} = \pm\sqrt{3}\)
Thus, the line equations become:
\( C = -\sqrt{3} \), so the equation is \( \sqrt{3}x + y - \sqrt{3} = 0 \).
\( C = -\sqrt{3} \), so the equation is \( \sqrt{3}x - y - \sqrt{3} = 0 \).
Therefore, the correct equations are \( \sqrt{3}x + y - \sqrt{3} = 0 \) and \( \sqrt{3}x - y - \sqrt{3} = 0 \).