Question
What is the name of the path that a moving point follows when its distance from the lines 3x-2y=5 and 3x+2y=5 is equal?
Answers
Answered by
GPT-4o mini
To solve for the path that a moving point follows when its distance from the lines \(3x - 2y = 5\) and \(3x + 2y = 5\) is equal, we begin by determining the equations of the lines in slope-intercept form:
1. **Convert the equations into slope-intercept form:**
- For the line \(3x - 2y = 5\):
\[
-2y = -3x + 5 \implies y = \frac{3}{2}x - \frac{5}{2}
\]
- For the line \(3x + 2y = 5\):
\[
2y = -3x + 5 \implies y = -\frac{3}{2}x + \frac{5}{2}
\]
Now, the first line has a slope of \(\frac{3}{2}\) and the second line has a slope of \(-\frac{3}{2}\). This means the lines are not parallel and intersect.
2. **Find the intersection point of the two lines:**
Set the equations equal to each other:
\[
\frac{3}{2}x - \frac{5}{2} = -\frac{3}{2}x + \frac{5}{2}
\]
Multiply through by 2 to eliminate fractions:
\[
3x - 5 = -3x + 5
\]
Combine similar terms:
\[
6x = 10 \implies x = \frac{5}{3}
\]
Substitute \(x = \frac{5}{3}\) into one of the original equations to find \(y\):
\[
3\left(\frac{5}{3}\right) - 2y = 5 \implies 5 - 2y = 5 \implies 2y = 0 \implies y = 0
\]
Thus, the lines intersect at the point \(\left(\frac{5}{3}, 0\right)\).
3. **Finding the locus of points equidistant from both lines:**
The locus of points equidistant from two parallel lines is the line that is midway between them. However, since these lines have slopes that are negatives of each other, we can find the equations for the distance from these lines and solve.
The distance \(d\) from a point \((x, y)\) to a line \(Ax + By + C = 0\) is given by:
\[
d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}
\]
Applying this to both lines:
- For the line \(3x - 2y - 5 = 0\):
\[
d_1 = \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y - 5|}{\sqrt{13}}
\]
- For the line \(3x + 2y - 5 = 0\):
\[
d_2 = \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3x + 2y - 5|}{\sqrt{13}}
\]
Setting these distances equal gives us:
\[
|3x - 2y - 5| = |3x + 2y - 5|
\]
This leads to two possible cases:
1. \(3x - 2y - 5 = 3x + 2y - 5\)
- Simplifying this gives:
\[
-2y = 2y \implies 0 = 4y \implies y = 0
\]
2. \(3x - 2y - 5 = -(3x + 2y - 5)\) (removing the absolute values)
- This leads to:
\[
3x - 2y - 5 = -3x - 2y + 5 \implies 6x = 10 \implies x = \frac{5}{3}\ \text{(which produces a valid point)}
\]
The points which satisfy the equality of distances follow a specific trend. These are calculations of lines forming equal distance to both lines.
Upon evaluating the paths, we discover the midpoint or the line itself bisects the area evenly. The path, therefore, turns out to be the \(y\)-axis with a slight shift toward respective equations forming \(y=0\).
**Conclusion:**
The path that a moving point follows where its distance from both lines is equal is the line \(y = 0\), or in other terms, the x-axis (the horizontal axis).
1. **Convert the equations into slope-intercept form:**
- For the line \(3x - 2y = 5\):
\[
-2y = -3x + 5 \implies y = \frac{3}{2}x - \frac{5}{2}
\]
- For the line \(3x + 2y = 5\):
\[
2y = -3x + 5 \implies y = -\frac{3}{2}x + \frac{5}{2}
\]
Now, the first line has a slope of \(\frac{3}{2}\) and the second line has a slope of \(-\frac{3}{2}\). This means the lines are not parallel and intersect.
2. **Find the intersection point of the two lines:**
Set the equations equal to each other:
\[
\frac{3}{2}x - \frac{5}{2} = -\frac{3}{2}x + \frac{5}{2}
\]
Multiply through by 2 to eliminate fractions:
\[
3x - 5 = -3x + 5
\]
Combine similar terms:
\[
6x = 10 \implies x = \frac{5}{3}
\]
Substitute \(x = \frac{5}{3}\) into one of the original equations to find \(y\):
\[
3\left(\frac{5}{3}\right) - 2y = 5 \implies 5 - 2y = 5 \implies 2y = 0 \implies y = 0
\]
Thus, the lines intersect at the point \(\left(\frac{5}{3}, 0\right)\).
3. **Finding the locus of points equidistant from both lines:**
The locus of points equidistant from two parallel lines is the line that is midway between them. However, since these lines have slopes that are negatives of each other, we can find the equations for the distance from these lines and solve.
The distance \(d\) from a point \((x, y)\) to a line \(Ax + By + C = 0\) is given by:
\[
d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}
\]
Applying this to both lines:
- For the line \(3x - 2y - 5 = 0\):
\[
d_1 = \frac{|3x - 2y - 5|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y - 5|}{\sqrt{13}}
\]
- For the line \(3x + 2y - 5 = 0\):
\[
d_2 = \frac{|3x + 2y - 5|}{\sqrt{3^2 + 2^2}} = \frac{|3x + 2y - 5|}{\sqrt{13}}
\]
Setting these distances equal gives us:
\[
|3x - 2y - 5| = |3x + 2y - 5|
\]
This leads to two possible cases:
1. \(3x - 2y - 5 = 3x + 2y - 5\)
- Simplifying this gives:
\[
-2y = 2y \implies 0 = 4y \implies y = 0
\]
2. \(3x - 2y - 5 = -(3x + 2y - 5)\) (removing the absolute values)
- This leads to:
\[
3x - 2y - 5 = -3x - 2y + 5 \implies 6x = 10 \implies x = \frac{5}{3}\ \text{(which produces a valid point)}
\]
The points which satisfy the equality of distances follow a specific trend. These are calculations of lines forming equal distance to both lines.
Upon evaluating the paths, we discover the midpoint or the line itself bisects the area evenly. The path, therefore, turns out to be the \(y\)-axis with a slight shift toward respective equations forming \(y=0\).
**Conclusion:**
The path that a moving point follows where its distance from both lines is equal is the line \(y = 0\), or in other terms, the x-axis (the horizontal axis).