find the equation of the locus of a point P(x,y) that is subject to the following condition: the distance of  from the line y=-5 is three-quarters its distance from the line x=2

We are asked to find the equation of the locus of a point \( P(x, y) \) such that its distance from the line \( y = -5 \) is three-quarters of its distance from the line \( x = 2 \).

### Step 1: Distance from the line \( y = -5 \)
The distance of a point \( P(x, y) \) from the line \( y = -5 \) is given by the vertical distance formula:
\[
d_1 = |y - (-5)| = |y + 5|
\]

### Step 2: Distance from the line \( x = 2 \)
The distance of a point \( P(x, y) \) from the line \( x = 2 \) is given by the horizontal distance formula:
\[
d_2 = |x - 2|
\]

### Step 3: Set up the condition
We are told that the distance from \( P(x, y) \) to the line \( y = -5 \) is three-quarters of its distance to the line \( x = 2 \). This translates mathematically into the equation:
\[
|y + 5| = \frac{3}{4} |x - 2|
\]

### Step 4: Square both sides to eliminate absolute values
To get rid of the absolute value signs, we square both sides:
\[
(y + 5)^2 = \left( \frac{3}{4}(x - 2) \right)^2
\]

### Step 5: Simplify the equation
\[
(y + 5)^2 = \frac{9}{16}(x - 2)^2
\]
Multiply both sides by 16 to eliminate the fraction:
\[
16(y + 5)^2 = 9(x - 2)^2
\]

### Step 6: Expand both sides
\[
16(y^2 + 10y + 25) = 9(x^2 - 4x + 4)
\]
Expanding both sides gives:
\[
16y^2 + 160y + 400 = 9x^2 - 36x + 36
\]

### Step 7: Rearrange into a standard form
Rearrange all terms to one side:
\[
9x^2 - 36x - 16y^2 - 160y - 364 = 0
\]

This is the equation of the locus of the point \( P(x, y) \).

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Do you have any questions or would you like further details on any step?

### Related Questions:
1. How do you derive the formula for the distance from a point to a line?
2. What happens if the ratio between the distances is different, e.g., 2/3 instead of 3/4?
3. Can the equation of a locus always be written in the form of a conic section?
4. How do you interpret the locus geometrically in this case?
5. How do you simplify equations with absolute values without squaring?

### Tip:
Squaring both sides in equations involving absolute values can introduce extraneous solutions, so it's always good to check your results against the original conditions.

Find the equation of the locus of a point P(x,y) that is subject to the following condition: the distance from the line y=-5 is three-quarters its distance from the line x=2.

Find the equation of the locus of a point P(x, y) such that its distance from the line y = -5 is three-quarters of its distance from the line x = 2. This problem involves coordinate geometry, distance formulas, and quadratic equations. Ideal for Grades 9-11.

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