Two-Dimensional Coordinate Geometry Study Guide

Core Principles and Practical Guidance in Coordinate Geometry

• Principles of calculation in two-dimensional coordinate geometry:

  • Calculators should not be used when performing coordinate geometry calculations.

  • Gradients are frequently expressed as fractions and should not be converted into decimal approximations.

  • Equations should be kept in exact fractional form or transformed by multiplying both sides of an equation by the denominator so that all terms work with integers.

• Importance of sketching:

  • Creating a sketch of the problem on Cartesian axes helps visualize geometric relationships.

  • A clear sketch serves as an immediate sanity check to indicate when a calculated numerical answer is obviously incorrect.

Classifications of Points in Two-Dimensional Space

• Any point in the Cartesian plane falls into one of three standard categories:

  • Fixed points whose coordinates are known numerically, such as $(5, -7)$.

  • Fixed points whose coordinates are not known numerically, represented algebraically using constants such as $(a, b)$, $(x_1, y_1)$, or $(x_2, y_2)$.

  • General points that are not fixed and vary across the line or plane, represented by $(x, y)$.

Equations of Straight Lines

• Slope-intercept form of a straight line:

  • Any straight line in the Cartesian plane can be represented by the equation:     $y = m\,x + c$

  • In this equation, $m$ represents the gradient of the line, and $c$ represents the intercept on the vertical $y$-axis.

  • To find the intercept and gradient accurately from a given equation, $y$ must be made the subject.

  • Example: If the equation of a line is $2y + 3x = 8$, the gradient is not $3$; rewriting the equation with $y$ as the subject gives $y = -\frac{3}{2}\,x + 4$, showing that the gradient is $m = -\frac{3}{2}$ and the $y$-intercept is $c = 4$.

• Derivation of straight-line equations from points and gradients:

  • Given two fixed points $A(x_1, y_1)$ and $B(x_2, y_2)$ on a straight line:     

    

  • The gradient $m$ of the line $AB$ is calculated as:     $m = \frac{\text{increase in } y}{\text{increase in } x} = \frac{y_2 - y_1}{x_2 - x_1}$

  • If $P(x, y)$ is any general point on the line $AB$, then the gradient of $AP$ is also $m$:     $\frac{y - y_1}{x - x_1} = m$

  • Equating expressions for $m$ yields the point-slope form:     $y - y_1 = m(x - x_1)$

  • Alternatively, equating the gradient of $AP$ directly to the gradient of $AB$ gives:     $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

The Gradient of a Line and Collinearity

• Concept and definition of collinearity:

  • Points are collinear if they lie along the exact same straight line.

  • To determine whether three points $A$, $B$, and $C$ are collinear using gradients:

  • Calculate the gradient of line segment $AB$.

  • Calculate the gradient of line segment $BC$.

  • If the gradient of $AB$ equals the gradient of $BC$, and point $B$ is common to both segments, then points $A$, $B$, and $C$ are collinear.

The Distance Between Two Points

• Derivation of the distance formula using Pythagoras' theorem:

  • The distance $AB$ between points $A(x_1, y_1)$ and $B(x_2, y_2)$ is found by constructing a right triangle where the hypotenuse is $AB$, the horizontal base length is $(x_2 - x_1)$, and the vertical height length is $(y_2 - y_1)$.

  • By Pythagoras' theorem ($a^2 + b^2 = c^2$):     $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$

  • Taking the positive square root yields the standard distance formula:     $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$     

    

The Mid-Point of a Line Segment

• Derivation and formula for the mid-point $M$:

  • Let $M(l, n)$ be the mid-point of the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$.

  • The horizontal coordinate $l$ is calculated as:     $l = x_1 + \frac{1}{2}(x_2 - x_1) = \frac{x_1 + x_2}{2}$

  • The vertical coordinate $n$ is calculated as:     $n = y_1 + \frac{1}{2}(y_2 - y_1) = \frac{y_1 + y_2}{2}$

  • Thus, the mid-point $M$ of segment $AB$ is given by:     $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

Fully Worked Examples and Applications

• Example 8.1: Calculating line segment length

  • Objective: Find the length of the line joining $A(-2, 2)$ to $B(3, -1)$.     

    

  • Calculation:     $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$     $AB^2 = (3 - (-2))^2 + (-1 - 2)^2$     $AB^2 = (3 + 2)^2 + (-1 - 2)^2$     $AB^2 = 5^2 + (-3)^2 = 25 + 9 = 34$     $AB = \sqrt{34}$

• Example 8.2: Finding mid-point coordinates

  • Objective: Find the coordinates of the mid-point of the line joining $C(-3, -2)$ and $D(1, 3)$.

  • Calculation:     $\text{The coordinates of } M \text{ are } \left[\frac{1}{2}(x_1 + x_2), \frac{1}{2}(y_1 + y_2)\right]$     $M = \left[\frac{1}{2}(-3 + 1), \frac{1}{2}(-2 + 3)\right]$     $M = \left(-1, \frac{1}{2}\right)$

• Example 8.3: Testing collinearity of three points

  • Objective: Determine, by comparing gradients, whether the points $A(3, 1)$, $B\left(1, \frac{1}{2}\right)$, and $C(4, -4)$ are collinear.

  • Calculation:     $\text{Gradient of } AB = \frac{1 - \frac{1}{2}}{3 - 1} = -\frac{3}{2}$     $\text{Gradient of } BC = \frac{-4 - \frac{1}{2}}{4 - 1} = -\frac{3}{2}$

  • Conclusion: As the gradients of $AB$ and $BC$ are the same, and $B$ is common to both lines, points $A$, $B$, and $C$ are collinear.

• Example 8.4: Determining line equations through two points

  • Objective: Find the equation of the line passing through the points $P(1, -2)$ and $Q(3, 1)$.

  • Step 1: Find the gradient of $PQ$:     $\text{gradient of } PQ = m = \frac{1 - (-2)}{3 - 1} = \frac{3}{2}$

  • Step 2 (Method 1): Using $y - y_1 = m(x - x_1)$:     $y + 2 = \frac{3}{2}(x - 1)$     $2y + 4 = 3x - 3$     $2y = 3x - 7$

  • Step 2 (Method 2): Using $y = mx + c$:     $y = \frac{3}{2}x + c$     Substituting $x = 1$ and $y = -2$:     $-2 = \frac{3}{2}(1) + c \implies c = -2 - \frac{3}{2} = -3\frac{1}{2}$     $y = \frac{3}{2}x - 3\frac{1}{2} \quad \text{or} \quad 2y = 3x - 7$

• Example 8.5: Finding unknown endpoint from given midpoint

  • Objective: $M(2, -1)$ is the mid-point of the straight line joining point $A(-2, 4)$ to point $B$. Find the coordinates of $B$.

  • Setup: Let $B$ be the point $(p, q)$.

  • Equations:     $2 = \frac{-2 + p}{2} \implies p = 6$     $-1 = \frac{4 + q}{2} \implies q = -6$

  • Result: $B$ is the point $(6, -6)$.

• Example 8.6: Finding unknown parameter from equal segment lengths

  • Objective: The three points $A$, $B$, and $C$ have coordinates $(1, 7)$, $(-1, 3)$, and $(a, 5)$ respectively. If $AB$ and $BC$ are of equal length, calculate the two possible values of $a$.

  • Equations:     $AB^2 = BC^2$     $(-1 - 1)^2 + (3 - 7)^2 = (a + 1)^2 + (5 - 3)^2$     $(-2)^2 + (-4)^2 = (a + 1)^2 + 2^2$     $4 + 16 = (a + 1)^2 + 4$     $(a + 1)^2 = 16$

  • Solving:     $a + 1 = 4 \implies a = 3$     $a + 1 = -4 \implies a = -5$

  • Result: The two possible values of $a$ are $3$ and $-5$.

• Whiteboard Example: Step-by-step length calculation of $AB$

  • Objective: Find the distance between points $A(1, 1)$ and $B(2, 6)$.

  • Formula:     $a^2 + b^2 = c^2$     $AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$     $AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

  • Substitution:     $AB = \sqrt{(2 - 1)^2 + (6 - 1)^2}$     $AB = \sqrt{(1)^2 + (5)^2}$     $AB = \sqrt{1 + 25}$     $AB = \sqrt{26}$

Comprehensive Exercise Set (Exercise 8A)

• Problem 1: For each of the following pairs of points $A$ and $B$, find (i) the gradient of line $AB$, (ii) an equation for line $AB$, (iii) the mid-point of line $AB$, and (iv) the length of line segment $AB$:

  • (a) $A(2, 3), B(4, 7)$

  • (b) $A(1, 7), B(5, 3)$

  • (c) $A(-3, 4), B(2, -1)$

  • (d) $A(2, 5), B(-3, -2)$

  • (e) $A(2, -3), B(-1, 3)$

  • (f) $A(-2, 0), B(6, 4)$

  • (g) $A(4, 3), B(0, 0)$

  • (h) $A(2, 4), B(0, -4)$

  • (i) $A(-4, -1), B(4, -3)$

  • (j) $A(8, 3), B(5, 6)$

  • (k) $A(8, 3), B(2, 1)$

  • (l) $A(1, -3), B(-3, 5)$

• Problem 2: A triangle has vertices at $(1, 1)$, $(2, 6)$, and $(6, 2)$. Prove that the triangle is isosceles.

• Problem 3: Find the lengths of the sides of the triangle with vertices at $A(0, 1)$, $B(3, 5)$, and $C(8, -5)$. Hence prove that the triangle is right-angled, and state which angle is the right angle.

• Problem 4: Find the lengths of the sides of the triangle with vertices at $P(7, 2)$, $Q(3, 4)$, and $R(6, 10)$. Hence prove that the triangle is right-angled, and find its area.

• Problem 5: The three points $A$, $B$, and $C$ have coordinates $(2, 6)$, $(0, 2)$, and $(4, a)$ respectively. If $AB$ and $BC$ are of equal length, calculate the two possible values of a$.\n* Problem 6: The three points D$,$E$, and$F$have coordinates$(-4, 5)$,$(0, b)$, and$(-2, 1)$respectively. If$DE$and$DF$are of equal length, calculate the two possible values of$b$.

• Problem 7: The three points $G(4, -1)$, $H(c, 5)$, and $I(7, 0)$ are such that $GH$ is twice as long as $GI$. Calculate the possible values of c$.\n* Problem 8: M(6, 4)$is the mid-point of the straight line joining point$A(9, 5)$to point$B$. Find the coordinates of$B$.

• Problem 9: $L(5, 1)$ is the mid-point of the straight line joining two points.