https://file.aiquickdraw.com/mathbot/file/1727080495027zwwzrz22.jpg

Let's go through the questions in **Exercise 6.4** one by one.

### 1. Find the area of these triangles:

#### a. Triangle with vertices \( A(-2, 3) \), \( B(0, -4) \), and \( C(5, 6) \)

The area of a triangle given its three vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) 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's substitute the values for \( A(-2, 3) \), \( B(0, -4) \), and \( C(5, 6) \):

\[
\text{Area} = \frac{1}{2} \left| -2((-4) - 6) + 0(6 - 3) + 5(3 - (-4)) \right|
\]

Simplifying this:

\[
\text{Area} = \frac{1}{2} \left| -2(-10) + 5(7) \right|
= \frac{1}{2} \left| 20 + 35 \right|
= \frac{1}{2} \times 55 = 27.5
\]

So, the area of triangle \(ABC\) is **27.5 square units**.

#### b. Triangle with vertices \( P(-3, 1) \), \( Q(5, -3) \), and \( R(2, 4) \)

Using the same formula for triangle area, substitute the values for \( P(-3, 1) \), \( Q(5, -3) \), and \( R(2, 4) \):

\[
\text{Area} = \frac{1}{2} \left| -3((-3) - 4) + 5(4 - 1) + 2(1 - (-3)) \right|
\]

Simplifying this:

\[
\text{Area} = \frac{1}{2} \left| -3(-7) + 5(3) + 2(4) \right|
= \frac{1}{2} \left| 21 + 15 + 8 \right|
= \frac{1}{2} \times 44 = 22
\]

So, the area of triangle \(PQR\) is **22 square units**.

### 2. Find the area of these quadrilaterals:

The area of a quadrilateral given its four vertices can be calculated using the **shoelace theorem (Gauss's area formula)**. The formula is:

\[
\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|
\]

#### a. Quadrilateral with vertices \( A(1, 8) \), \( B(-4, 5) \), \( C(-2, -3) \), \( D(4, -2) \)

Substitute the coordinates of the vertices into the formula:

\[
\text{Area} = \frac{1}{2} \left| 1(5) + (-4)(-3) + (-2)(-2) + 4(8) - (8(-4) + 5(-2) + (-3)(4) + (-2)(1)) \right|
\]

Simplifying this:

\[
\text{Area} = \frac{1}{2} \left| 5 + 12 + 4 + 32 - (-32 + (-10) + (-12) + (-2)) \right|
= \frac{1}{2} \left| 53 - (-56) \right|
= \frac{1}{2} \times 109 = 54.5
\]

So, the area of quadrilateral \(ABCD\) is **54.5 square units**.

#### b. Quadrilateral with vertices \( P(2, 7) \), \( Q(-5, 6) \), \( R(-3, -4) \), \( S(7, 2) \)

Substitute the coordinates of the vertices into the shoelace formula:

\[
\text{Area} = \frac{1}{2} \left| 2(6) + (-5)(-4) + (-3)(2) + 7(7) - (7(-5) + 6(-3) + (-4)(7) + 2(2)) \right|
\]

Simplifying this:

\[
\text{Area} = \frac{1}{2} \left| 12 + 20 + (-6) + 49 - (-35 + (-18) + (-28) + 4) \right|
= \frac{1}{2} \left| 75 - (-77) \right|
= \frac{1}{2} \times 152 = 76
\]

So, the area of quadrilateral \(PQRS\) is **76 square units**.

---

Would you like more details on any step, or have any questions?  
Here are 5 follow-up questions you might want to explore:

1. How is the shoelace theorem derived, and what are its limitations?
2. What happens if the vertices of a triangle or quadrilateral are given in a different order?
3. Can the area formula for polygons be extended to more complex shapes?
4. How would the area change if one of the vertices were moved along the x- or y-axis?
5. What is the geometric interpretation of a negative area result?

**Tip:** When solving geometry problems involving coordinates, drawing a sketch can often clarify the problem and reduce calculation errors.

Find the area of these triangles and quadrilaterals using the given vertex coordinates.

Explore how to calculate the areas of triangles and quadrilaterals using the coordinates of their vertices with detailed step-by-step solutions. This guide covers essential concepts in coordinate geometry, including the shoelace theorem and triangle area formula.

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