Points A(7, -4), B(-5, 5) and C(-3, 8) are vertices of triangle ABC, Find the length of its median through vertex A.

Answer

To find the length of the median through vertex A of triangle ABC with vertices A(7, -4), B(-5, 5), and C(-3, 8), we will follow these steps:

### Step 1: Identify the coordinates of the vertices
The coordinates of the vertices are:
- A(7, -4)
- B(-5, 5)
- C(-3, 8)

### Step 2: Find the midpoint D of side BC
To find the midpoint D of segment BC, we use the midpoint formula:
\[ D\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
where \(B(x_1, y_1) = (-5, 5)\) and \(C(x_2, y_2) = (-3, 8)\).

Calculating the coordinates of D:
\[ D\left(\frac{-5 + (-3)}{2}, \frac{5 + 8}{2}\right) = D\left(\frac{-8}{2}, \frac{13}{2}\right) = D(-4, \frac{13}{2}) \]

### Step 3: Use the distance formula to find the length of median AD
The length of the median AD can be calculated using the distance formula:
\[ AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \(A(x_1, y_1) = (7, -4)\) and \(D(x_2, y_2) = \left(-4, \frac{13}{2}\right)\).

Substituting the coordinates into the distance formula:
\[ AD = \sqrt{\left(-4 - 7\right)^2 + \left(\frac{13}{2} - (-4)\right)^2} \]
Calculating the differences:
\[ AD = \sqrt{(-11)^2 + \left(\frac{13}{2} + 4\right)^2} \]
Convert 4 to a fraction:
\[ 4 = \frac{8}{2} \quad \Rightarrow \quad AD = \sqrt{121 + \left(\frac{13 + 8}{2}\right)^2} = \sqrt{121 + \left(\frac{21}{2}\right)^2} \]
Calculating \(\left(\frac{21}{2}\right)^2\):
\[ \left(\frac{21}{2}\right)^2 = \frac{441}{4} \]
Now substituting back:
\[ AD = \sqrt{121 + \frac{441}{4}} = \sqrt{\frac{484}{4} + \frac{441}{4}} = \sqrt{\frac{925}{4}} = \frac{\sqrt{925}}{2} \]
### Step 4: Simplify \(\sqrt{925}\)
Breaking down 925:
\[ 925 = 25 \times 37 \quad \Rightarrow \quad \sqrt{925} = 5\sqrt{37} \]
Thus, we have:
\[ AD = \frac{5\sqrt{37}}{2} \]

### Final Answer
The length of the median through vertex A is:
\[ \frac{5\sqrt{37}}{2} \]

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