Determine the coordinates of the vertices of the triangle if the coordinates of its vertices are given by the formula A(x, y), B(x, y), C(x, y) and the lengths of the sides are given: AB=c, BC=a, CA=b. Write the formula for the area of the triangle.

Solution:

To determine the coordinates of the vertices of a triangle, we need more information than just the lengths of its sides. The lengths of the sides define the triangle up to congruence (meaning we can have multiple triangles with the same side lengths but different positions and orientations in the coordinate plane).

However, we can set up a system to find the coordinates if we place one vertex at the origin and one side along an axis.

Let vertex A be at the origin: \( A(0, 0) \).

Let vertex B lie on the positive x-axis. Its coordinates will be \( B(c, 0) \) since the length of side AB is \( c \).

Let vertex C have coordinates \( C(x_C, y_C) \). We know the lengths of sides BC and CA:

• The length of CA is \( b \). Using the distance formula from \( A(0, 0) \) to \( C(x_C, y_C) \): \( \sqrt{(x_C - 0)^2 + (y_C - 0)^2} = b \), which simplifies to \( x_C^2 + y_C^2 = b^2 \).
• The length of BC is \( a \). Using the distance formula from \( B(c, 0) \) to \( C(x_C, y_C) \): \( \sqrt{(x_C - c)^2 + (y_C - 0)^2} = a \), which simplifies to \( (x_C - c)^2 + y_C^2 = a^2 \).

Now we have a system of two equations with two unknowns \( x_C \) and \( y_C \):

1. \( x_C^2 + y_C^2 = b^2 \)
2. \( (x_C - c)^2 + y_C^2 = a^2 \)

From the first equation, \( y_C^2 = b^2 - x_C^2 \). Substitute this into the second equation:

\( (x_C - c)^2 + (b^2 - x_C^2) = a^2 \)

Expand \( (x_C - c)^2 \): \( x_C^2 - 2cx_C + c^2 + b^2 - x_C^2 = a^2 \)

Simplify:

\( -2cx_C + c^2 + b^2 = a^2 \)

Solve for \( x_C \):

\( 2cx_C = c^2 + b^2 - a^2 \)

\[ x_C = \frac{c^2 + b^2 - a^2}{2c} \]

Now substitute \( x_C \) back into the equation for \( y_C^2 \):

\( y_C^2 = b^2 - \left(\frac{c^2 + b^2 - a^2}{2c}\right)^2 \)

And solve for \( y_C \) (we can take the positive root for \( y_C \) if we assume the triangle is in the upper half-plane):

\[ y_C = \sqrt{b^2 - \left(\frac{c^2 + b^2 - a^2}{2c}\right)^2} \]

Thus, the coordinates of the vertices can be:

\( A(0, 0) \), \( B(c, 0) \), \( C\left(\frac{c^2 + b^2 - a^2}{2c}, \sqrt{b^2 - \left(\frac{c^2 + b^2 - a^2}{2c}\right)^2}\right) \)

Formula for the area of a triangle:

There are several formulas for the area of a triangle. Given the side lengths, we can use Heron's formula:

• First, calculate the semi-perimeter \( s \): \( s = \frac{a + b + c}{2} \)
• Then, the area \( S \) is: \[ S = \sqrt{s(s-a)(s-b)(s-c)} \]

Alternatively, using the coordinates of the vertices \( A(x_A, y_A) \), \( B(x_B, y_B) \), and \( C(x_C, y_C) \), the area can be calculated as:

\[ S = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)| \]

If we use the coordinates derived above \( A(0, 0) \), \( B(c, 0) \), \( C(x_C, y_C) \), the formula simplifies:

\[ S = \frac{1}{2} |0(0 - y_C) + c(y_C - 0) + x_C(0 - 0)| \]
\[ S = \frac{1}{2} |cy_C| \]

Since \( y_C \) is typically taken as positive for the area calculation, and \( c \) is a length (positive), this becomes:

\[ S = \frac{1}{2} c y_C \]

Substituting the expression for \( y_C \) gives the same result as Heron's formula.

Ответ: Координаты вершин можно найти, разместив одну вершину в начале координат и одну сторону вдоль оси. Например, \( A(0, 0) \), \( B(c, 0) \), \( C\left(\frac{c^2 + b^2 - a^2}{2c}, \sqrt{b^2 - \left(\frac{c^2 + b^2 - a^2}{2c}\right)^2}\right) \). Формула площади треугольника (по сторонам): \( S = \sqrt{s(s-a)(s-b)(s-c)} \), где \( s = \frac{a+b+c}{2} \).