The point A(0, 0), B(1, 7), C(5, 1) are the vertices of a triangle. Find the length of the perpendicular from A to BC and hence the area of the `DeltaABC`.

Answer

To find the length of the perpendicular from point A(0, 0) to line BC and the area of triangle ABC, we will follow these steps:

### Step 1: Determine the coordinates of points A, B, and C
Given points are:
- A(0, 0)
- B(1, 7)
- C(5, 1)

### Step 2: Find the equation of line BC
To find the equation of line BC, we can use the two-point form of the line equation:
\[ \frac{y - y_2}{x - x_2} = \frac{y_3 - y_2}{x_3 - x_2} \]
Here, \( (x_2, y_2) = (1, 7) \) and \( (x_3, y_3) = (5, 1) \).

Substituting the values:
\[ \frac{y - 7}{x - 1} = \frac{1 - 7}{5 - 1} = \frac{-6}{4} = -\frac{3}{2} \]

Cross-multiplying gives:
\[ 2(y - 7) = -3(x - 1) \]
Expanding this:
\[ 2y - 14 = -3x + 3 \]
Rearranging:
\[ 3x + 2y - 17 = 0 \]

### Step 3: Identify coefficients for the perpendicular distance formula
From the equation \( 3x + 2y - 17 = 0 \):
- \( a = 3 \)
- \( b = 2 \)
- \( c = -17 \)

### Step 4: Use the formula for the length of the perpendicular from point A to line BC
The formula for the distance \( d \) from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by:
\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
Substituting \( (x_1, y_1) = (0, 0) \):
\[ d = \frac{|3(0) + 2(0) - 17|}{\sqrt{3^2 + 2^2}} = \frac{|-17|}{\sqrt{9 + 4}} = \frac{17}{\sqrt{13}} \]

### Step 5: Find the length of BC
Using the distance formula between points B and C:
\[ BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} \]
Substituting \( (x_2, y_2) = (1, 7) \) and \( (x_3, y_3) = (5, 1) \):
\[ BC = \sqrt{(5 - 1)^2 + (1 - 7)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \]

### Step 6: Calculate the area of triangle ABC
The area \( A \) of triangle ABC can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( BC \) and the height is the perpendicular distance \( d \):
\[ A = \frac{1}{2} \times (2\sqrt{13}) \times \left(\frac{17}{\sqrt{13}}\right) \]
Simplifying:
\[ A = \frac{1}{2} \times 2 \times 17 = 17 \]

### Final Results
- Length of the perpendicular from A to line BC: \( \frac{17}{\sqrt{13}} \) units
- Area of triangle ABC: \( 17 \) square units