The slope of a median, drawn from the vertex A of the triangle ABC is -2. The co-ordinates of vertices B and C are respectively (-1, 3) and (3, 5). If the area of the triangle be 5 square units, then possible distance of vertex A from the origin is/are.

Options

A. (a) 6
B. (b) 4
C. (c) `2sqrt(2)`
D. (d) `3sqrt(2)`

Answer

To solve the problem step by step, we will follow the given information about the triangle and use the properties of medians and the area of triangles.

### Step 1: Find the coordinates of point D (the midpoint of BC)
Given the coordinates of points B and C:
- B(-1, 3)
- C(3, 5)

Using the midpoint formula:
\[ D\left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right) = D\left(\frac{-1 + 3}{2}, \frac{3 + 5}{2}\right) = D\left(\frac{2}{2}, \frac{8}{2}\right) = D(1, 4) \]

**Hint:** The midpoint formula is used to find the average of the x-coordinates and the y-coordinates of two points.

### Step 2: Write the equation of the median AD
The slope of median AD is given as -2. Using the point-slope form of the equation of a line, we can write the equation of line AD, which passes through point D(1, 4):
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -2\) and \(D(1, 4)\):
\[ y - 4 = -2(x - 1) \]
Simplifying this:
\[ y - 4 = -2x + 2 \implies y + 2x = 6 \]

**Hint:** The point-slope form of a line is useful when you know a point on the line and the slope.

### Step 3: Set up the area formula for triangle ABC
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Let the coordinates of A be \((x_0, y_0)\), B(-1, 3), and C(3, 5). The area is given as 5 square units:
\[ 5 = \frac{1}{2} \left| x_0(3 - 5) + (-1)(5 - y_0) + 3(y_0 - 3) \right| \]
This simplifies to:
\[ 5 = \frac{1}{2} \left| -2x_0 - 5 + y_0 + 3y_0 - 9 \right| \]
\[ 5 = \frac{1}{2} \left| -2x_0 + 4y_0 - 14 \right| \]
Multiplying both sides by 2:
\[ 10 = \left| -2x_0 + 4y_0 - 14 \right| \]

**Hint:** The area formula for a triangle based on vertices can be derived from the determinant of a matrix formed by the coordinates.

### Step 4: Solve for the coordinates of A
This gives us two equations to solve:
1. \(-2x_0 + 4y_0 - 14 = 10\)
2. \(-2x_0 + 4y_0 - 14 = -10\)

**For the first equation:**
\[ -2x_0 + 4y_0 = 24 \implies -x_0 + 2y_0 = 12 \quad \text{(Equation 1)} \]

**For the second equation:**
\[ -2x_0 + 4y_0 = 4 \implies -x_0 + 2y_0 = 2 \quad \text{(Equation 2)} \]

**Hint:** You can derive two equations from the absolute value condition, leading to two scenarios.

### Step 5: Solve the equations
**From Equation 1:**
\[ -x_0 + 2y_0 = 12 \implies y_0 = 6 + \frac{x_0}{2} \]

**From Equation 2:**
\[ -x_0 + 2y_0 = 2 \implies y_0 = 1 + \frac{x_0}{2} \]

### Step 6: Find the distance of A from the origin
For \(y_0 = 6 + \frac{x_0}{2}\):
If \(x_0 = 0\), then \(y_0 = 6\), so A(0, 6).
Distance from origin:
\[ d = \sqrt{(0-0)^2 + (6-0)^2} = 6 \]

For \(y_0 = 1 + \frac{x_0}{2}\):
If \(x_0 = 0\), then \(y_0 = 1\), so A(0, 1).
Distance from origin:
\[ d = \sqrt{(0-0)^2 + (1-0)^2} = 1 \]

### Final Answer
The possible distances of vertex A from the origin are **1 and 6**.