What point on the x-axis are at a distance of 4 units from the line ` 3 x - 4 y - 5 = 0 `

Options

A. `((1)/(3),0)`
B. `(0,(25)/(3))`
C. (5,1)
D. `((25)/(3),0)`

Answer

To find the points on the x-axis that are at a distance of 4 units from the line \(3x - 4y - 5 = 0\), we can follow these steps:

### Step 1: Understand the line equation
The equation of the line is given as \(3x - 4y - 5 = 0\). We can rewrite it in the standard form \(Ax + By + C = 0\) where:
- \(A = 3\)
- \(B = -4\)
- \(C = -5\)

### Step 2: Use the distance formula
The formula for the distance \(D\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
In our case, we want the distance to be 4 units, so we set \(D = 4\).

### Step 3: Identify the point on the x-axis
Since we are looking for points on the x-axis, any point on the x-axis can be represented as \((a, 0)\), where \(a\) is the x-coordinate.

### Step 4: Substitute the point into the distance formula
Substituting \(x_1 = a\) and \(y_1 = 0\) into the distance formula, we have:
\[ 4 = \frac{|3a + (-4)(0) - 5|}{\sqrt{3^2 + (-4)^2}} \]
This simplifies to:
\[ 4 = \frac{|3a - 5|}{\sqrt{9 + 16}} \]
\[ 4 = \frac{|3a - 5|}{\sqrt{25}} \]
\[ 4 = \frac{|3a - 5|}{5} \]

### Step 5: Solve the equation
Multiplying both sides by 5 gives:
\[ 20 = |3a - 5| \]
This absolute value equation leads to two cases:

**Case 1:**
\[ 3a - 5 = 20 \]
Solving for \(a\):
\[ 3a = 25 \implies a = \frac{25}{3} \]

**Case 2:**
\[ 3a - 5 = -20 \]
Solving for \(a\):
\[ 3a = -15 \implies a = -5 \]

### Step 6: Write the points
The points on the x-axis that are at a distance of 4 units from the line are:
1. \(\left(\frac{25}{3}, 0\right)\)
2. \((-5, 0)\)

### Final Answer
The points on the x-axis that are at a distance of 4 units from the line \(3x - 4y - 5 = 0\) are \(\left(\frac{25}{3}, 0\right)\) and \((-5, 0)\).

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