Area of the triangle formed by the lines through point (6, 0) and at a perpendicular distance of 5 from point (1, 3) and line ` y = 16 ` in square units is :

Options

A. 160
B. 200
C. 240
D. 130

Answer

To find the area of the triangle formed by the given lines, we can follow these steps:

### Step 1: Find the equation of the line through the point (6, 0)
Let the slope of the line be \( m \). The equation of the line in point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (6, 0) \):
\[ y - 0 = m(x - 6) \implies y = mx - 6m \]
Rearranging gives:
\[ mx - y - 6m = 0 \quad \text{(Equation 1)} \]

### Step 2: Find the perpendicular distance from the point (1, 3) to the line
The formula for the perpendicular distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( mx - y - 6m = 0 \), we have \( A = m \), \( B = -1 \), and \( C = -6m \). The point is \( (1, 3) \).
Substituting into the distance formula:
\[ 5 = \frac{|m \cdot 1 - 1 \cdot 3 - 6m|}{\sqrt{m^2 + 1}} \]
This simplifies to:
\[ 5 = \frac{| -5m - 3 |}{\sqrt{m^2 + 1}} \]

### Step 3: Solve for \( m \)
Squaring both sides:
\[ 25(m^2 + 1) = (-5m - 3)^2 \]
Expanding both sides:
\[ 25m^2 + 25 = 25m^2 + 30m + 9 \]
This simplifies to:
\[ 25 = 30m + 9 \implies 30m = 16 \implies m = \frac{16}{30} = \frac{8}{15} \]

### Step 4: Find the second slope
The second slope is vertical, which corresponds to \( m = \infty \). This gives us the line:
\[ x = 6 \quad \text{(Equation 2)} \]

### Step 5: Find the intersection of the lines
We have two lines:
1. \( 8x - 15y - 48 = 0 \) (derived from \( m = \frac{8}{15} \))
2. \( y = 16 \)

Substituting \( y = 16 \) into the first equation:
\[ 8x - 15(16) - 48 = 0 \implies 8x - 240 - 48 = 0 \implies 8x = 288 \implies x = 36 \]
Thus, the intersection point is \( (36, 16) \).

### Step 6: Identify the vertices of the triangle
The vertices of the triangle are:
- \( A(6, 0) \)
- \( B(36, 16) \)
- \( C(6, 16) \)

### Step 7: Calculate the area of the triangle
Using the formula for the area of a triangle given by vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 6(16 - 16) + 36(16 - 0) + 6(0 - 16) \right| \]
This simplifies to:
\[ = \frac{1}{2} \left| 0 + 576 - 96 \right| = \frac{1}{2} \left| 480 \right| = 240 \]

Thus, the area of the triangle is \( \boxed{240} \) square units.