The co-ordinates of a point on the line ` y = x ` where perpendicular distance from the line `3x + 4y = 12` is 4 units, are :

Options

A. `(3/7,5/7)`
B. `(3/2,3/2)`
C. `(-8/7,-8/7)`
D. `(32/7,-32/7)`

Answer

To find the coordinates of a point on the line \( y = x \) where the perpendicular distance from the line \( 3x + 4y = 12 \) is 4 units, we can follow these steps:

### Step 1: Define the point on the line
Let the coordinates of the point on the line \( y = x \) be \( (k, k) \).

### Step 2: Use the distance formula
The formula for the distance \( d \) from a point \( (x_0, y_0) \) to a line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \( 3x + 4y - 12 = 0 \), we have:
- \( A = 3 \)
- \( B = 4 \)
- \( C = -12 \)

Substituting \( (k, k) \) into the distance formula, we get:

\[ d = \frac{|3k + 4k - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7k - 12|}{5} \]

### Step 3: Set the distance equal to 4
Since the perpendicular distance is given as 4 units, we can set up the equation:

\[ \frac{|7k - 12|}{5} = 4 \]

### Step 4: Solve for \( k \)
Multiplying both sides by 5 gives:

\[ |7k - 12| = 20 \]

This leads to two cases:

**Case 1:**
\[ 7k - 12 = 20 \]
Solving for \( k \):
\[ 7k = 32 \implies k = \frac{32}{7} \]

**Case 2:**
\[ 7k - 12 = -20 \]
Solving for \( k \):
\[ 7k = -8 \implies k = -\frac{8}{7} \]

### Step 5: Find the coordinates
Now we can find the coordinates of the points on the line \( y = x \):

1. For \( k = \frac{32}{7} \):
\[ \text{Point 1} = \left(\frac{32}{7}, \frac{32}{7}\right) \]

2. For \( k = -\frac{8}{7} \):
\[ \text{Point 2} = \left(-\frac{8}{7}, -\frac{8}{7}\right) \]

### Final Answer
The coordinates of the points on the line \( y = x \) where the perpendicular distance from the line \( 3x + 4y = 12 \) is 4 units are:
\[ \left(\frac{32}{7}, \frac{32}{7}\right) \quad \text{and} \quad \left(-\frac{8}{7}, -\frac{8}{7}\right) \]

---