Find the locus of a point whose plane is parallel to the straight line 2x-y+3=0 and the point is equidistant from the point of intersection of the line and the plane in which the point is lying. The plane is variable.

To find the locus of a point that meets the criteria you've described, we need to break down the problem step by step. We have a straight line given by the equation 2x - y + 3 = 0, and we want to determine the locus of points that are equidistant from the intersection of this line and a variable plane that is parallel to the line.

Step 1: Understanding the Line Equation

The equation of the line can be rewritten in slope-intercept form (y = mx + b) for easier analysis. Rearranging gives:

y = 2x + 3

This tells us that the slope of the line is 2, and it intersects the y-axis at (0, 3).

Step 2: Finding the Intersection Point

Next, we need to find the intersection of this line with a plane. Since the plane is variable, we can represent it in a general form. A plane can be expressed as:

Ax + By + Cz + D = 0

For our purposes, we can assume the plane is parallel to the line, which means it will have the same slope. Therefore, we can express the plane in a form that maintains this parallelism. A simple choice is:

y = 2x + k

where k is a constant that can vary.

Step 3: Finding the Intersection Point of the Line and Plane

To find the intersection of the line and the plane, we can set the equations equal to each other. The line is:

y = 2x + 3

And the plane is:

y = 2x + k

Setting these equal gives:

2x + 3 = 2x + k

This simplifies to:

3 = k

Thus, the intersection point occurs when k = 3, which means the intersection point is (x, y) = (x, 2x + 3) for any x. However, since we are looking for a specific point, we can choose x = 0, giving us the intersection point (0, 3).

Step 4: Finding the Locus of Points Equidistant from the Intersection

Now, we need to find the locus of points that are equidistant from the intersection point (0, 3). The distance from a point (x, y) to the point (0, 3) can be expressed using the distance formula:

d = √[(x - 0)² + (y - 3)²]

We want to find points that are at a constant distance from (0, 3). This describes a circle centered at (0, 3) with radius r, where r is the distance from the point to the intersection point.

Step 5: Equation of the Locus

The equation of a circle centered at (0, 3) with radius r is:

(x - 0)² + (y - 3)² = r²

This simplifies to:

x² + (y - 3)² = r²

As the radius r can vary, the locus of points that are equidistant from the intersection point (0, 3) forms a family of circles centered at (0, 3) with varying radii.

Summary

In conclusion, the locus of points that are equidistant from the intersection of the line 2x - y + 3 = 0 and a variable plane parallel to this line is a set of circles centered at the point (0, 3) with varying radii. Each circle represents a different distance from the intersection point, illustrating the relationship between the point and the line and plane configuration.