Co-ordinate Geometry Comprehensive Notes

Fundamentals of Co-ordinate Geometry

• Definition of Co-ordinate Geometry: The branch of mathematics which deals with the study of the position of a point in a Cartesian plane or Co-ordinate plane is called Co-ordinate Geometry.

• Abscissa: The $x$-co-ordinate of a point is also called the abscissa.

• Ordinate: The $y$-co-ordinate of a point is also called the ordinate.

Distance Formula

• Formula Definition: If $P(x_1, y_1)$ and $Q(x_2, y_2)$ are any two points in the co-ordinate plane, then the distance between points $P$ and $Q$ is given by:     $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Worked Examples:

  • Example 1: Distance between $(-5, 7)$ and $(-1, 3)$

    • Let $A(-5, 7)$ and $B(-1, 3)$ be the given points.
    • Applying the distance formula:             $AB = \sqrt{(-1 - (-5))^2 + (3 - 7)^2}$$AB = \sqrt{(-1 + 5)^2 + (3 - 7)^2}$$AB = \sqrt{(4)^2 + (-4)^2}$$AB = \sqrt{16 + 16}$$AB = \sqrt{32}$$AB = \sqrt{2 \times 2 \times 2 \times 2 \times 2}$$AB = 4\sqrt{2}\text{ units}$

  • Example 2: Distance between $(a, b)$ and $(-a, -b)$

    • Let $A(a, b)$ and $B(-a, -b)$ be the given points.
    • Applying the distance formula:             $AB = \sqrt{(-a - a)^2 + (-b - b)^2}$$AB = \sqrt{(-2a)^2 + (-2b)^2}$$AB = \sqrt{4a^2 + 4b^2}$$AB = 2\sqrt{a^2 + b^2}\text{ units}$

  • Example 3: Verification of side lengths for points $(4, 5)$, $(7, 6)$, $(4, 3)$, and $(1, 2)$

    • Let $A(4, 5)$, $B(7, 6)$, $C(4, 3)$, and $D(1, 2)$ be the given vertices.
    • Calculating side length $AB$:             $AB = \sqrt{(7 - 4)^2 + (6 - 5)^2}$$AB = \sqrt{(3)^2 + (1)^2}$$AB = \sqrt{9 + 1}$$AB = \sqrt{10}$
    • Calculating side length $BC$:             $BC = \sqrt{(4 - 7)^2 + (3 - 6)^2}$$BC = \sqrt{(-3)^2 + (-3)^2}$$BC = \sqrt{9 + 9}$$BC = \sqrt{18}$
    • Calculating side length $CD$:             $CD = \sqrt{(1 - 4)^2 + (2 - 3)^2}$$CD = \sqrt{(-3)^2 + (-1)^2}$$CD = \sqrt{9 + 1}$$CD = \sqrt{10}$
    • Calculating side length $AD$:             $AD = \sqrt{(1 - 4)^2 + (2 - 5)^2}$$AD = \sqrt{(-3)^2 + (-3)^2}$$AD = \sqrt{9 + 9}$$AD = \sqrt{18}$
    • Conclusion: All opposite sides are equal ($AB = CD = \sqrt{10}$ and $BC = AD = \sqrt{18}$).

Section Formula

• Internal Division:

  • If the point $P(x, y)$ divides the line segment joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m : n$, then by the section formula, the coordinates of point $P$ are given by:         $P(x, y) = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right)$

• Unknown Ratio Consideration:

  • If the division ratio is not explicitly given, consider the ratio as $k : 1$.

• External Division:

  • If the point $P$ divides the line segment joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ externally in the ratio $m : n$, then by the section formula, the coordinates of point $P$ are given by:         $P(x, y) = \left(\frac{m x_2 - n x_1}{m - n}, \frac{m y_2 - n y_1}{m - n}\right)$

• Sign Interpretation for Division:

  • If $k$ is positive, then the ratio indicates internal division.
  • If $k$ is negative, then the ratio indicates external division.

• Midpoint Formula:

  • If the point $P$ is the midpoint of the line segment $AB$ joining $A(x_1, y_1)$ and $B(x_2, y_2)$, then the coordinates of point $P$ are given by:         $P(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

• Worked Examples:

  • Example 1: Internal division point coordinates

    • Problem: Find the coordinates of the point which divides the join of $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$
    • Let $P(x, y)$ divide the line joining $A(-1, 7)$ and $B(4, -3)$ in the ratio $m : n = 2 : 3$
    • Applying the section formula:             $P(x, y) = \left(\frac{2(4) + 3(-1)}{2 + 3}, \frac{2(-3) + 3(7)}{2 + 3}\right)$$P(x, y) = \left(\frac{8 - 3}{5}, \frac{-6 + 21}{5}\right)$$P(x, y) = \left(\frac{5}{5}, \frac{15}{5}\right)$$P(x, y) = (1, 3)$

  • Example 2: Points of Trisection

    • Problem: Find the coordinates of points of trisection of the line segment joining $(4, -1)$ and $(-2, -3)$.

Triangle Midpoint and Area Analysis

• Setup:
  • Consider a triangle $\Delta ABC$ with vertices $A(0, -1)$, $B(2, 1)$, and $C(0, 3)$.
  • Let $D$, $E$, and $F$ be the midpoints of sides $AB$, $BC$, and $AC$ respectively.

• Calculating Midpoint Coordinates:

  • Midpoint $D$ of side $AB$ (joining $A(0, -1)$ and $B(2, 1)$):         $D = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$D = \left(\frac{0 + 2}{2}, \frac{-1 + 1}{2}\right)$$D = (1, 0)$
  • Midpoint $E$ of side $BC$ (joining $B(2, 1)$ and $C(0, 3)$):         $E = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$E = \left(\frac{2 + 0}{2}, \frac{1 + 3}{2}\right)$$E = (1, 2)$
  • Midpoint $F$ of side $AC$ (joining $A(0, -1)$ and $C(0, 3)$):         $F = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$F = \left(\frac{0 + 0}{2}, \frac{-1 + 3}{2}\right)$$F = (0, 1)$
  • Midpoint coordinates are $D(1, 0)$, $E(1, 2)$, and $F(0, 1)$.

• Area Computations:

  • Area of the midpoint triangle $\Delta DEF$ with vertices $D(1, 0)$, $E(1, 2)$, $F(0, 1)$:         $\text{ar}(\Delta DEF) = \frac{1}{2} | 1(2 - 1) + 1(1 - 0) + 0(0 - 2) |$$\text{ar}(\Delta DEF) = \frac{1}{2} | 1(1) + 1(1) + 0 |$$\text{ar}(\Delta DEF) = \frac{1}{2} | 1 + 1 + 0 |$$\text{ar}(\Delta DEF) = \frac{2}{2} = 1\text{ sq. unit}$
  • Area of the main triangle $\Delta ABC$ with vertices $A(0, -1)$, $B(2, 1)$, $C(0, 3)$:         $\text{ar}(\Delta ABC) = \frac{1}{2} | 0(1 - 3) + 2(3 - (-1)) + 0(-1 - 1) |$$\text{ar}(\Delta ABC) = \frac{1}{2} | 0 + 2(4) + 0 |$$\text{ar}(\Delta ABC) = \frac{1}{2} | 8 | = 4\text{ sq. units}$
  • Ratio of areas:         $\frac{\text{ar}(\Delta DEF)}{\text{ar}(\Delta ABC)} = \frac{1}{4}$$\text{ar}(\Delta DEF) : \text{ar}(\Delta ABC) = 1 : 4$

Area of a Quadrilateral

• Problem Statement:

  • Find the area of the quadrilateral whose vertices taken in order are $A(-4, -2)$, $B(-3, -5)$, $C(3, -2)$, and $D(2, 3)$.

• Method of Triangulation:

  • Draw diagonal $BD$ to divide quadrilateral $ABCD$ into two triangles: $\Delta ABD$ and $\Delta BDC$.

• Area Calculation of Triangle ABD:

  • Vertices: $A(-4, -2)$, $B(-3, -5)$, and $D(2, 3)$.
  • Calculation:         $\text{ar}(\Delta ABD) = \frac{1}{2} | -4(-5 - 3) - 3(3 + 2) + 2(-2 + 5) |$$\text{ar}(\Delta ABD) = \frac{1}{2} | -4(-8) - 3(8) + 2(3) |$$\text{ar}(\Delta ABD) = \frac{1}{2} | 32 - 24 + 6 |$$\text{ar}(\Delta ABD) = \frac{14}{2} = 7\text{ sq. units}$

• Area Calculation of Triangle BDC:

  • Vertices: $B(-3, -5)$, $D(2, 3)$, and $C(3, -2)$.
  • Calculation:         $\text{ar}(\Delta BDC) = \frac{1}{2} | -3(3 + 2) + 2(-2 + 5) + 3(-5 - 3) |$$\text{ar}(\Delta BDC) = \frac{1}{2} | -3(6) + 2(3) + 3(-8) |$$\text{ar}(\Delta BDC) = \frac{1}{2} | -18 + 6 - 24 |$$\text{ar}(\Delta BDC) = \frac{1}{2} | -36 | = \frac{36}{2} = 18\text{ sq. units}$

• Total Area of Quadrilateral ABCD:

  • Sum of the areas of $\Delta ABD$ and $\Delta BDC$:         $\text{ar}(ABCD) = \text{ar}(\Delta ABD) + \text{ar}(\Delta BDC)$$\text{ar}(ABCD) = 7 + \frac{36}{2} = \frac{50}{2} = 25\text{ sq. units}$