find the path of a moving point such its distance is from two lines 3x-2y=5and 3x+2y=5 are equal and state the name of path.

if the distance from two lines is equal, it must move along the angle bisector between the two lines.

There are two such paths; one is vertical and one is horizontal.

The paths are thus

x = 5/3
y = 0

To find the path of a moving point such that its distance from the lines 3x - 2y = 5 and 3x + 2y = 5 are equal, we can use the concept of perpendicular distances.

1. Start by simplifying the given equations:
Line 1: 3x - 2y = 5
Line 2: 3x + 2y = 5

2. Rewrite the equations in slope-intercept form:
Line 1: y = (3/2)x - (5/2)
Line 2: y = (-3/2)x + (5/2)

3. Let (x, y) be the coordinates of the moving point.

4. The distance from a point (x, y) to a line Ax + By + C = 0 is given by the formula:
Distance = |Ax + By + C| / sqrt(A² + B²)

5. Calculate the distance from the point (x, y) to both lines:
Distance from Line 1 = |3x - 2y - 5| / sqrt(3² + (-2)²)
Distance from Line 2 = |3x + 2y - 5| / sqrt(3² + 2²)

6. Since we want the distances to be equal, we can set them equal to each other:
|3x - 2y - 5| / sqrt(3² + (-2)²) = |3x + 2y - 5| / sqrt(3² + 2²)

7. To simplify the equation, we can square both sides:
(3x - 2y - 5)² / (3² + (-2)²) = (3x + 2y - 5)² / (3² + 2²)

8. Expand and simplify the equation:
(9x² - 12xy + 4y² - 30x + 20y + 25) / 13 = (9x² + 12xy + 4y² - 30x - 20y + 25) / 13

9. Cancel out common terms and simplify further:
12xy - 12xy + 20y + 20y = 0

10. Combine like terms:
40y = 0

11. Solve for y:
y = 0

12. Substitute y = 0 back into one of the given equations to find x:
3x - 2(0) = 5
3x = 5
x = 5/3

Therefore, the moving point lies on the line y = 0, which is the x-axis. The path followed by the moving point is thus a horizontal line called the x-axis.

To find the path of a moving point such that its distance from the lines 3x-2y=5 and 3x+2y=5 are equal, we can start by understanding the concept of distance between a point and a line.

The distance between a point (x₁, y₁) and a line Ax + By + C = 0 is given by the formula:

distance = |Ax₁ + By₁ + C| / √(A² + B²)

In this case, we have two lines:
1. 3x - 2y = 5
2. 3x + 2y = 5

Let's assume the moving point is (x, y) and its distances from the two lines are equal. We can set up the equations using the distance formula:

For line 1:
|3x - 2y - 5| / √(3² + (-2)²) = distance

For line 2:
|3x + 2y - 5| / √(3² + 2²) = distance

Since the distances are equal, we can equate the two formulas and solve for x and y to find the path of the moving point:

|3x - 2y - 5| / √(3² + (-2)²) = |3x + 2y - 5| / √(3² + 2²)

Simplifying this equation, we get:

|3x - 2y - 5| / 13 = |3x + 2y - 5| / 13

By removing the absolute value signs, we obtain:

3x - 2y - 5 = 3x + 2y - 5 (since the distance is always positive)
-2y = 2y
4y = 0
y = 0

Now we substitute y = 0 into either equation to find the value of x:

3x - 2(0) = 5
3x = 5
x = 5/3

Therefore, the path of the moving point is a horizontal line passing through the point (5/3, 0). The name of this path is a horizontal line or an x-axis.