D is a point on AC of the triangle with vertices A(2, 3), B(1, -3), C(-4, -7) and BD divides ABC into two triangles of equal area. The equation of the line drawn through B at right angles to BD is

Options

A. y - 2x + 5 = 0
B. 2y - x + 5 = 0
C. y + 2x - 5 = 0
D. 2y + x - 5 = 0

Answer

To solve the problem step by step, we will follow the process outlined in the video transcript.

### Step 1: Identify the coordinates of points A, B, and C.
Given:
- A(2, 3)
- B(1, -3)
- C(-4, -7)

### Step 2: Find the coordinates of point D, which is the midpoint of line segment AC.
The formula for the midpoint D of a line segment connecting points A(x1, y1) and C(x2, y2) is:
\[ D = \left( \frac{x1 + x2}{2}, \frac{y1 + y2}{2} \right) \]

Substituting the coordinates of A and C:
- \( x1 = 2, y1 = 3 \)
- \( x2 = -4, y2 = -7 \)

Calculating the coordinates of D:
\[ D_x = \frac{2 + (-4)}{2} = \frac{-2}{2} = -1 \]
\[ D_y = \frac{3 + (-7)}{2} = \frac{-4}{2} = -2 \]
Thus, the coordinates of D are \( D(-1, -2) \).

### Step 3: Find the slope of line BD.
Using the coordinates of points B(1, -3) and D(-1, -2), we can calculate the slope (m) of line BD using the formula:
\[ m_{BD} = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates:
\[ m_{BD} = \frac{-2 - (-3)}{-1 - 1} = \frac{-2 + 3}{-2} = \frac{1}{-2} = -\frac{1}{2} \]

### Step 4: Determine the slope of the line BE that is perpendicular to BD.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore:
\[ m_{BE} = -\frac{1}{m_{BD}} = -\frac{1}{-\frac{1}{2}} = 2 \]

### Step 5: Use the point-slope form to find the equation of line BE.
The point-slope form of a line is given by:
\[ y - y_1 = m(x - x_1) \]
Using point B(1, -3) and the slope \( m_{BE} = 2 \):
\[ y - (-3) = 2(x - 1) \]
This simplifies to:
\[ y + 3 = 2x - 2 \]
Rearranging gives:
\[ y - 2x + 5 = 0 \]

### Final Answer:
The equation of the line drawn through B at right angles to BD is:
\[ y - 2x + 5 = 0 \]