If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis then the point P has

Options

A. x-coordinate =-5
B. y-coordinate =5 only
C. y-coordiante=-5 only
D. y-coordinate =5 or -5

Explanation

To solve the problem step by step, we will analyze the information given and apply the concepts of coordinate geometry.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We are given that the perpendicular distance of point P from the X-axis is 5 units. This means that the y-coordinate of point P will either be +5 or -5.

**Hint**: Remember that the distance from the X-axis is measured along the y-axis. Positive values indicate points above the X-axis, while negative values indicate points below it.

2. **Identifying the Foot of the Perpendicular**:
The foot of the perpendicular from point P to the X-axis lies on the negative direction of the X-axis. This means that the x-coordinate of the foot of the perpendicular (let's call it P') is negative.

**Hint**: The foot of the perpendicular is the point where the perpendicular line meets the X-axis, which has a y-coordinate of 0.

3. **Coordinates of Point P**:
Since the foot of the perpendicular lies on the negative X-axis, we can denote the coordinates of point P as (-a, y), where 'a' is a positive value (since it is negative on the X-axis) and y can be either +5 or -5.

**Hint**: The coordinates of any point in the Cartesian plane are given as (x, y). Here, x is negative, and y is determined by the distance from the X-axis.

4. **Using the Distance Formula**:
The distance from point P to the foot of the perpendicular P' can be calculated using the distance formula. The coordinates of P' are (-a, 0). The distance is given as 5 units.

The distance formula is:
\[ \text{Distance} = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
Plugging in the coordinates:
\[ 5 = \sqrt{(-a - (-a))^2 + (y - 0)^2} \]
This simplifies to:
\[ 5 = \sqrt{0 + y^2} \]
Thus, we have:
\[ 5 = |y| \]

**Hint**: The absolute value indicates that y can be either positive or negative.

5. **Finding Possible Values of y**:
From the equation \( |y| = 5 \), we can conclude that:
\[ y = 5 \quad \text{or} \quad y = -5 \]

**Hint**: The absolute value equation gives us two possible solutions for y.

6. **Final Coordinates of Point P**:
Therefore, the coordinates of point P can be:
- If y = 5, then P = (-a, 5)
- If y = -5, then P = (-a, -5)

Since 'a' is a positive number, we can conclude that the x-coordinate is negative.

**Hint**: The exact value of 'a' does not affect the conclusion about the possible coordinates of point P.

### Conclusion:
The point P has two possible coordinates:
1. (-a, 5) where 'a' is a positive value (indicating it's on the negative X-axis).
2. (-a, -5) where 'a' is a positive value.

Thus, the final answer is that the point P can either be at (-a, 5) or (-a, -5).