Question

Select the equation for a graph that is the set of all points in the plane that are equidistant from the point f(1/2,0) and line x=-1/2

Answer 1

The equation for the graph that represents the set of all points equidistant from the point
`\( F((1)/(2), 0) \)` and the line
`\( x = -(1)/(2) \)` is
`\( (x + (1)/(2))^2 + y^2 = (1)/(4) \).`

Step-by-step explanation:

To find the equation of the graph, we consider two cases: the distance from a point and the distance from a line. The set of points equidistant from a point
`\( F((1)/(2), 0) \)` forms a circle, and the set of points equidistant from a line
`\( x = -(1)/(2) \)` forms a vertical line segment.

For the distance from a point, the general formula is
`\( √((x - x_1)^2 + (y - y_1)^2) \)`, where
`\( (x_1, y_1) \)` is the coordinates of the point. In this case, the distance from
`\( F((1)/(2), 0) \)` is
`\( \sqrt{(x - (1)/(2))^2 + y^2} \).`

For the distance from a line, the formula is the perpendicular distance, which is
`\( |ax + by + c| / √(a^2 + b^2) \),` where
`\( ax + by + c = 0 \)` is the equation of the line. Here, the line is
`\( x = -(1)/(2) \)`, so the equation becomes
`\( |x + (1)/(2)| \).`

Setting these two distances equal (as the points are equidistant from the point and the line), we get the equation
`\( \sqrt{(x - (1)/(2))^2 + y^2} = |x + (1)/(2)| \)`. Squaring both sides and simplifying yields
`\( (x + (1)/(2))^2 + y^2 = (1)/(4) \)`, which is the equation of the graph. This equation represents a circle centered at
`\( (-(1)/(2), 0) \)` with a radius of
`\( (1)/(2) \).`