If the slope of a line passing through to point A(3, 2) is 3/4 then find points on the line which are 5 units away from the point A.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Write the equation of the line
Given the slope \( m = \frac{3}{4} \) and the point \( A(3, 2) \), we can use the point-slope form of the equation of a line, which is:
\[ y - y_1 = m(x - x_1) \]
Substituting the values:
\[ y - 2 = \frac{3}{4}(x - 3) \]
Now, we can rearrange this into slope-intercept form \( y = mx + c \).

### Step 2: Solve for \( c \)
Expanding the equation:
\[ y - 2 = \frac{3}{4}x - \frac{9}{4} \]
Adding 2 to both sides:
\[ y = \frac{3}{4}x - \frac{9}{4} + 2 \]
Converting 2 into quarters:
\[ y = \frac{3}{4}x - \frac{9}{4} + \frac{8}{4} \]
This simplifies to:
\[ y = \frac{3}{4}x - \frac{1}{4} \]

### Step 3: Rearranging to standard form
To convert this into standard form \( Ax + By + C = 0 \):
\[ 3x - 4y - 1 = 0 \]

### Step 4: Finding points 5 units away from point A
Let the point on the line be \( (h, k) \). Since \( k = \frac{3}{4}h - \frac{1}{4} \), we need to find the distance from point \( A(3, 2) \) to point \( (h, k) \) which is 5 units:
\[ \sqrt{(h - 3)^2 + \left(k - 2\right)^2} = 5 \]
Substituting \( k \):
\[ \sqrt{(h - 3)^2 + \left(\frac{3}{4}h - \frac{1}{4} - 2\right)^2} = 5 \]

### Step 5: Simplifying the equation
First, simplify \( k - 2 \):
\[ k - 2 = \frac{3}{4}h - \frac{1}{4} - 2 = \frac{3}{4}h - \frac{1}{4} - \frac{8}{4} = \frac{3}{4}h - \frac{9}{4} \]
Thus, the distance equation becomes:
\[ \sqrt{(h - 3)^2 + \left(\frac{3}{4}h - \frac{9}{4}\right)^2} = 5 \]
Squaring both sides:
\[ (h - 3)^2 + \left(\frac{3}{4}h - \frac{9}{4}\right)^2 = 25 \]

### Step 6: Expanding and solving the equation
Expanding both terms:
1. \( (h - 3)^2 = h^2 - 6h + 9 \)
2. \( \left(\frac{3}{4}h - \frac{9}{4}\right)^2 = \frac{9}{16}h^2 - \frac{27}{8}h + \frac{81}{16} \)

Combining these:
\[ h^2 - 6h + 9 + \frac{9}{16}h^2 - \frac{27}{8}h + \frac{81}{16} = 25 \]

### Step 7: Combine like terms and solve for \( h \)
Multiply through by 16 to eliminate fractions:
\[ 16h^2 - 96h + 144 + 9h^2 - 54h + 81 = 400 \]
Combine like terms:
\[ 25h^2 - 150h + 225 = 400 \]
Rearranging gives:
\[ 25h^2 - 150h - 175 = 0 \]
Dividing by 25:
\[ h^2 - 6h - 7 = 0 \]

### Step 8: Factor or use the quadratic formula
Factoring:
\[ (h - 7)(h + 1) = 0 \]
Thus, \( h = 7 \) or \( h = -1 \).

### Step 9: Finding corresponding \( k \) values
1. For \( h = 7 \):
\[ k = \frac{3}{4}(7) - \frac{1}{4} = \frac{21}{4} - \frac{1}{4} = \frac{20}{4} = 5 \]
So one point is \( (7, 5) \).

2. For \( h = -1 \):
\[ k = \frac{3}{4}(-1) - \frac{1}{4} = -\frac{3}{4} - \frac{1}{4} = -1 \]
So the other point is \( (-1, -1) \).

### Final Answer
The points on the line which are 5 units away from point A(3, 2) are:
\[ (7, 5) \text{ and } (-1, -1) \]