Coordinate Geometry Study Notes

Length of a Line Segment and Midpoint

In coordinate geometry, fundamental calculations involve determining the midpoint and the length of a line segment connecting two points, $P(x_1, y_1)$ and $Q(x_2, y_2)$. The midpoint, $M$, represents the central point between the coordinates and is calculated using the average of the $x$ and $y$ values: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$. The length of the line segment $PQ$ is derived from Pythagoras' theorem and is given by the formula: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. It is a strict requirement in coordinate geometry to show appropriate calculations rather than relying on scale drawings, which are not accepted in formal assessments.

Worked Example 3.1 illustrates finding unknown variables when a midpoint is known. If the point $M(3, -11)$ is the midpoint of the line segment joining $P(-7, 4)$ and $Q(a, b)$, one can use algebraic equating or vector transitions. In Method 1 (Algebra), we equate the coordinates: $\frac{-7+a}{2} = 3$ and $\frac{4+b}{2} = -11$, resulting in $a = 10$ and $b = -26$. Method 2 (Vectors) observes the movement from $P$ to $M$ ($+10$ in $x$, $-15$ in $y$) and applies the same translation to $M$ to reach $Q$, yielding identical results.

Worked Example 3.2 applies these principles to a parallelogram $ABCD$ with vertices $A(-5, -1)$, $B(-1, -4)$, and $C(6, -2)$. To find the coordinates of vertex $D$, we utilize the property that the diagonals of a parallelogram bisect each other. The midpoint of $AC$ is $\left(\frac{-5+6}{2}, \frac{-1-2}{2}\right) = \left(\frac{1}{2}, -\frac{3}{2}\right)$. Since the midpoint of $BD$ must be the same, we set $\frac{-1+m}{2} = \frac{1}{2}$ and $\frac{-4+n}{2} = -\frac{3}{2}$, where $D$ is $(m, n)$. This process identifies $D$ as the point $(2, 1)$.

Worked Example 3.3 demonstrates finding unknown coordinates using the distance formula. Given the distance between $P(-2, a)$ and $Q(a - 2, -7)$ is $17$, we set up the equation: $\sqrt{(a - 2 - (-2))^2 + (-7 - a)^2} = 17$. Squaring both sides yields $a^2 + (-7 - a)^2 = 289$. Expanding this gives $a^2 + 49 + 14a + a^2 = 289$, which simplifies to the quadratic equation $2a^2 + 14a - 240 = 0$, or $a^2 + 7a - 120 = 0$. Factoring into $(a - 8)(a + 15) = 0$ provides two possible values for $a$: $a = 8$ or $a = -15$.

Parallel and Perpendicular Lines

The gradient of a line joining points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is defined as the ratio of vertical change to horizontal change: $\text{Gradient of } PQ = \frac{y_2 - y_1}{x_2 - x_1}$. This gradient, often denoted as $m$, determines the orientation of the line. If two lines are parallel, their gradients are equal ($m_1 = m_2$). If a line has a gradient $m$, any line perpendicular to it has a gradient that is the negative reciprocal, $-\frac{1}{m}$. This relationship can be expressed as the product of the gradients of two perpendicular lines being equal to negative one: $m_1 \times m_2 = -1$.

Worked Example 3.4 explores collinearity, where three points $A(k - 5, -15)$, $B(10, k)$, and $C(6, -k)$ lie on the same straight line. In this case, the gradient of $AB$ must equal the gradient of $BC$. Setting the ratios equal: $\frac{k - (-15)}{10 - (k - 5)} = \frac{-k - k}{6 - 10}$, which simplifies to $\frac{k + 15}{15 - k} = \frac{-2k}{-4}$. Cross-multiplying and simplifying leads to the quadratic equation $k^2 - 13k + 30 = 0$. Factoring into $(k - 3)(k - 10) = 0$ gives the possible values of $k = 3$ or $k = 10$.

Worked Example 3.5 addresses a right-angled triangle condition. For vertices $A(11, 3)$, $B(2k, k)$, and $C(-1, -11)$ where $\angle ABC = 90^{\circ}$, the product of the gradients of $AB$ and $BC$ must be $-1$. The gradient of $AB$ is $\frac{k - 3}{2k - 11}$ and the gradient of $BC$ is $\frac{-11 - k}{-1 - 2k}$. Setting their product to $-1$: $\left(\frac{k - 3}{2k - 11}\right) \times \left(\frac{-11 - k}{-1 - 2k}\right) = -1$. Simplifying the second fraction to $\frac{k + 11}{2k + 1}$ and solving the resulting equation $k^2 + 8k - 33 = -(4k^2 - 20k + 11)$ leads to $5k^2 - 12k - 44 = 0$. Factoring gives $(5k - 22)(k + 2) = 0$, resulting in $k = 4.4$ (where $B$ is $(8.8, 4.4)$) or $k = -2$ (where $B$ is $(-4, -2)$).

Equations of Straight Lines

A straight line can be represented by several forms of equations. The slope-intercept form is $y = mx + c$, where $m$ is the gradient and $c$ is the $y$-intercept. Vertical lines are represented by $x = b$, where $b$ is the $x$-intercept. An alternative, highly useful formula when a gradient $m$ and a specific point $(x_1, y_1)$ are known is the point-gradient form: $y - y_1 = m(x - x_1)$. This formula is derived from the definition of a gradient between a fixed point and a general point $P(x, y)$.

Worked Example 3.6 demonstrates this: to find the equation of a line with gradient $-2$ passing through $(4, 1)$, one substitutes into the formula: $y - 1 = -2(x - 4)$, which simplifies to $y - 1 = -2x + 8$, or $2x + y = 9$. Worked Example 3.7 shows how to find the equation given two points, $(-4, 3)$ and $(6, -2)$. First, the gradient is calculated: $m = \frac{-2 - 3}{6 - (-4)} = -\frac{5}{10} = -\frac{1}{2}$. Then, using point-gradient form with $(-4, 3)$: $y - 3 = -\frac{1}{2}(x + 4)$, leading to $2y - 6 = -x - 4$, or $x + 2y = 2$.

Worked Example 3.8 explains the process of finding the perpendicular bisector of a line segment joining $A(-5, 1)$ and $B(7, -2)$. First, the gradient of $AB$ is found: $m = \frac{-2 - 1}{7 - (-5)} = -\frac{3}{12} = -\frac{1}{4}$. The gradient of the perpendicular line is the negative reciprocal, which is $4$. Next, the midpoint of $AB$ is calculated: $M = \left(\frac{-5 + 7}{2}, \frac{1 - 2}{2}\right) = (1, -0.5)$. Finally, the line equation with gradient $4$ through $(1, -0.5)$ is found: $y - (-0.5) = 4(x - 1)$, which results in $y + 0.5 = 4x - 4$, or $2y = 8x - 9$.

The Equation of a Circle

A circle is defined geometrically as the locus of all points in a plane that are a fixed distance (the radius, $r$) from a given point (the center, $(a, b)$). By applying Pythagoras' theorem to a point $P(x, y)$ on the circumference, the completed square form of the circle equation is established: $(x - a)^2 + (y - b)^2 = r^2$. Worked Example 3.9 shows center/radius identification: for $x^2 + y^2 = 4$, the center is $(0, 0)$ and $r = 2$; for $(x - 2)^2 + (y - 4)^2 = 100$, the center is $(2, 4)$ and $r = 10$. Worked Example 3.10 constructs the equation for a circle with center $(-4, 3)$ and radius $6$: $(x + 4)^2 + (y - 3)^2 = 36$.

Worked Example 3.11 involves a circle where a diameter has endpoints $A(3, 0)$ and $B(7, -4)$. The center is the midpoint of $AB$: $\left(\frac{3+7}{2}, \frac{0-4}{2}\right) = (5, -2)$. The radius is the distance from the center to $A$: $\sqrt{(5-3)^2 + (-2-0)^2} = \sqrt{8}$. The final equation is $(x - 5)^2 + (y + 2)^2 = 8$. Expanding the circle equation yields the expanded general form: $x^2 + y^2 + 2gx + 2fy + c = 0$. In this form, the center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. It is noted that circle equations always have equal coefficients for $x^2$ and $y^2$ and contain no $xy$ term.

Worked Example 3.12 demonstrates converting from general form to completed square form using completing the square. For $x^2 + y^2 + 10x - 8y - 40 = 0$, we group terms: $(x^2 + 10x) + (y^2 - 8y) - 40 = 0$. Completed, this becomes $(x + 5)^2 - 25 + (y - 4)^2 - 16 - 40 = 0$, simplifying to $(x + 5)^2 + (y - 4)^2 = 81$. The center is $(-5, 4)$ and the radius is $9$. Several geometric facts assist in solving circle problems: the angle in a semicircle is a right angle; a perpendicular from the center to a chord bisects that chord; and the tangent at a point is perpendicular to the radius at that point.

Circle Through Three Points and Intersections

Finding the equation of a circle passing through three points, such as $P(-1, 4)$, $Q(1, 6)$, and $R(5, 4)$ in Worked Example 3.13, can be done by finding the intersection of the perpendicular bisectors of two chords (e.g., $PQ$ and $QR$). The perpendicular bisector of $PQ$ is $y = -x + 5$ and for $QR$ is $y = 2x - 1$. Solving these simultaneously gives the center at $(2, 3)$. The radius squared is found using the distance to any point: $(5 - 2)^2 + (4 - 3)^2 = 10$. The equation is $(x - 2)^2 + (y - 3)^2 = 10$. Alternatively, one can substitute the three points into $(x - a)^2 + (y - b)^2 = r^2$ to create a system of equations.

Intersections of lines and circles are analyzed using simultaneous equations and the discriminant ($b^2 - 4ac$) of the resulting quadratic. If the discriminant is greater than zero, there are two distinct intersection points. If it equals zero, the line is a tangent (one repeated root). If it is less than zero, there is no intersection. Worked Example 3.14 solves the intersection of $x = 3y + 10$ and $x^2 + y^2 = 20$, resulting in points $A(-2, -4)$ and $B(4, -2)$. The perpendicular bisector of $AB$ is $y = -3x$, which passes through the center $(0, 0)$. Finding where this bisector intersects the circle gives points $P(-\sqrt{2}, 3\sqrt{2})$ and $Q(\sqrt{2}, -3\sqrt{2})$. Worked Example 3.15 proves the line $y = x - 13$ is a tangent to the circle $x^2 + y^2 - 8x + 6y + 7 = 0$ by substituting for $y$ and showing the resulting quadratic $x^2 - 14x + 49 = 0$ has one repeated root, $x = 7$.

Questions & Discussion

Explore 3.1: This discussion focuses on whether a triangle with side lengths $2\sqrt{7}\text{ cm}$, $4\sqrt{3}\text{ cm}$, and $5\sqrt{3}\text{ cm}$ is right-angled. To verify, one must check if the sum of the squares of the two shorter sides equals the square of the longest side ($a^2 + b^2 = c^2$).

Explore 3.2: Students are prompted to use graphing software for several circle equations (e.g., $x^2 + y^2 = 25$, $(x - 2)^2 + (y - 1)^2 = 9$, $(x + 3)^2 + (y + 5)^2 = 16$) to notice the relationship between the equation's constants and the coordinates of the center and the value of the radius. The goal is to explain how these can be identified simply by looking at the equation.

Explore 3.3: This task involves using software to investigate how changing the parameters $a, b,$ and $r$ in the equation $(x - a)^2 + (y - b)^2 = r^2$ shifts the center or alters the size of the circle.

Exercise 3A, Question 7: The point $P(k, 2k)$ is equidistant from $A(8, 11)$ and $B(1, 12)$. Find the value of $k$. To solve, set the distance $PA$ equal to $PB$: $\sqrt{(k - 8)^2 + (2k - 11)^2} = \sqrt{(k - 1)^2 + (2k - 12)^2}$, square both sides, and solve the linear equation.

Exercise 3B, Question 9: The line $\frac{x}{a} + \frac{y}{b} = 1$ meets the $x$-axis at $P$ and the $y$-axis at $Q$. Given the gradient of $PQ$ is $\frac{1}{2}$ and the length is $2\sqrt{29}$, find $a$ and $b$. The coordinates are $P(a, 0)$ and $Q(0, b)$. The gradient is $\frac{b - 0}{0 - a} = -\frac{b}{a}$. Length is $\sqrt{a^2 + b^2}$. Solving these conditions allows finding the constants.

Exercise 3C, Question 16: Find two straight lines whose $x$-intercepts differ by $7$, whose $y$-intercepts differ by $5$, and whose gradients differ by $2$. This question explores the uniqueness of solutions in coordinate systems.

Exercise 3D, Question 16: A circle passes through $(2, -16)$ with a radius of $10$, and the $x$-axis is a tangent. Find the equations. Since the $x$-axis is a tangent, the $y$-coordinate of the center ($b$) must equal the radius ($\pm 10$).