Fundamental Concepts

Cartesian Plane

$Cartesian Plane$

$Quadrants$

Four quadrants labeled I to IV in counterclockwise order
Signs:
- Quadrant I: $(+,+)$
- Quadrant II: $(-,+)$
- Quadrant III: $(-,-)$
- Quadrant IV: $(+,-)$

Representation: $P(x,y)$
$Points$
Special points:
- Origin: $O(0,0)$
- On the $x$ -axis: $(a,0)$
- On the $y$ -axis: $(0,b)$

For $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ :
$Distance Between Two Points$

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Given two points in polar coordinates:

The distance $d$ between them is:

d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2 \cos(\theta_2 - \theta_1)}

Midpoint $M$ between $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ :
$Midpoint$

M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)

$Division of a Segment in a Given Ratio$

Point $P$ that divides segment $P_1P_2$ in the ratio $\frac{P_1P}{PP_2} = \frac{m}{n} = \lambda$ :

x = \frac{nx_1 + mx_2}{n + m} = \frac{x_1 + \lambda x_2}{1 + \lambda}, \quad y = \frac{ny_1 + my_2}{n + m} = \frac{y_1 + \lambda y_2}{1 + \lambda}

P\left(\frac{x_1 + \lambda x_2}{1 + \lambda}, \frac{y_1 + \lambda y_2}{1 + \lambda}\right)

$Centroid of a Triangle$

x = \frac{x_1 + x_2 + x_3}{3}

y = \frac{y_1 + y_2 + y_3}{3}

$Area of a Triangle$

Given vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ , the area is:

\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} \right|

Where the determinant evaluates to:

\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} = x_1(y_2 - y_3) - y_1(x_2 - x_3) + (x_2 y_3 - x_3 y_2)

In practice, this simplifies to:

\text{Area} = \frac{1}{2} \big| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \big|

Example

Given vertices $P_1(2, 4)$ , $P_2(5, 6)$ , and $P_3(3, 1)$ :

\begin{vmatrix} 2 & 4 & 1 \\ 5 & 6 & 1 \\ 3 & 1 & 1 \\ \end{vmatrix}

= 2(6 \cdot 1 - 1 \cdot 1) - 4(5 \cdot 1 - 3 \cdot 1) + 1(5 \cdot 1 - 3 \cdot 6)

= 2(6 - 1) - 4(5 - 3) + 1(5 - 18)

= 2(5) - 4(2) + 1(-13) = 10 - 8 - 13 = -11

\text{Area} = \frac{1}{2} |-11| = 5.5 \text{ units}^2

$Area of a Polygon$

Hint

For vertices $(x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n)$ :

\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n} \left( x_i y_{i+1} \right) - \sum_{i=1}^{n} \left( y_i x_{i+1} \right) \right|

where:

$x_{n+1} = x_1$ and $y_{n+1} = y_1$ (closes the polygon).
Vertices must be ordered clockwise or counterclockwise (no self-intersections).

Ordered list of vertices: Write coordinates in order (e.g., $P_1 \to P_2 \to P_3 \to \dots \to P_1$ ).
Sum 1 ( $\Sigma_1$ ): Multiply each $x_i$ by the $y$ of the next vertex ( $y_{i+1}$ ) and add them up.
Sum 2 ( $\Sigma_2$ ): Multiply each $y_i$ by the $x$ of the next vertex ( $x_{i+1}$ ) and add them up.
Subtract and take absolute value: Compute $|\Sigma_1 - \Sigma_2|$ and divide by 2.

Example

Vertices in order: $P_{1}(2, 4)$ , $P_{2}(5, 6)$ , $P_{3}(3, 1)$ , $P_{4}(1, 2)$ .

(2,4), (5,6), (3,1), (1,2), (2,4)

Compute $\Sigma_1$ (downward diagonals ➘):
$(2 \times 6) + (5 \times 1) + (3 \times 2) + (1 \times 4) = 12 + 5 + 6 + 4 = 27$
Compute $\Sigma_2$ (upward diagonals ➚):
$(4 \times 5) + (6 \times 3) + (1 \times 1) + (2 \times 2) = 20 + 18 + 1 + 4 = 43$
Area:

\text{Area} = \frac{1}{2} |27 - 43| = \frac{16}{2} = 8 \text{ units}^2