M2: Coordinate Geometry

Coordinate Geometry of Straight Lines

Gradient (Slope)

The gradient $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Equations of a Line

• Gradient-intercept form: $y = mx + c$
• Point-gradient form: $y - y_1 = m(x - x_1)$
• Two-point form: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$
• General form: $Ax + By + C = 0$ where $A$, $B$, $C$ are constants

Parallel and Perpendicular Lines

• Parallel: $m_1 = m_2$
• Perpendicular: $m_1 \cdot m_2 = -1$

Distance Between Points

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

Distance from a Point to a Line

The perpendicular distance from $(x_0, y_0)$ to $Ax + By + C = 0$:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Intersection of Two Lines

To find the intersection of $A_1 x + B_1 y + C_1 = 0$ and $A_2 x + B_2 y + C_2 = 0$, solve the system of simultaneous equations.

If $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$, the lines are coincident (the same line). If $\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$, the lines are parallel (no intersection).

Angle Between Two Lines

The acute angle $\theta$ between two lines with gradients $m_1$ and $m_2$:

$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$

Coordinate Geometry of Circles

Standard Form

A circle with centre $(h, k)$ and radius $r$:

$(x - h)^2 + (y - k)^2 = r^2$

General Form

$x^2 + y^2 + Dx + Ey + F = 0$

Centre: $\left(-\frac{D}{2}, -\frac{E}{2}\right)$. Radius: $r = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}$.

For a real circle, $D^2 + E^2 - 4F > 0$.

Finding the Equation of a Circle

Given the centre and a point on the circle: Substitute into the standard form.

Given three points: Set up a system of three equations using the general form and solve for $D$, $E$, and $F$.

Given the endpoints of a diameter: The centre is the midpoint of the diameter. The radius is half the length of the diameter.

Tangent to a Circle

A tangent to a circle at point $(x_1, y_1)$ on $(x - h)^2 + (y - k)^2 = r^2$:

$(x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2$

Example: Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$.

$3x + 4y = 25$

Intersection of a Line and a Circle

Substitute the equation of the line into the equation of the circle to obtain a quadratic in $x$ (or $y$). The discriminant of this quadratic determines the nature of the intersection:

• $\Delta > 0$: The line cuts the circle at two distinct points
• $\Delta = 0$: The line is tangent to the circle
• $\Delta < 0$: The line does not meet the circle

Circle through Three Points

Given three non-collinear points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, substitute each into $x^2 + y^2 + Dx + Ey + F = 0$ to obtain a system of three linear equations in $D$, $E$, $F$.

Conic Sections: Parabola

Standard Forms

A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix).

Vertical axis (opening up or down):

$y^2 = 4ax \quad \text{(opens right)}$ $y^2 = -4ax \quad \text{(opens left)}$ $x^2 = 4ay \quad \text{(opens upward)}$ $x^2 = -4ay \quad \text{(opens downward)}$

For $y^2 = 4ax$: Focus at $(a, 0)$, directrix $x = -a$, axis of symmetry is the $x$-axis. Vertex at the origin.

Translated parabola: $(x - h)^2 = 4a(y - k)$ has vertex at $(h, k)$, focus at $(h, k + a)$, directrix $y = k - a$.

Parametric Form

For the parabola $y^2 = 4ax$, a general point is $(at^2, 2at)$ where $t$ is the parameter.

Tangent to a Parabola

For $y^2 = 4ax$, the tangent at the point $(at_1^2, 2at_1)$ is:

$ty = x + at_1^2$

Reflective Property

A ray from the focus reflects off the parabola parallel to the axis. Conversely, a ray parallel to the axis reflects through the focus. This property is used in satellite dishes, headlights, and telescopes.

Conic Sections: Ellipse

Standard Forms

An ellipse is the locus of points such that the sum of the distances from two fixed points (foci) is constant.

Horizontal major axis:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b)$

Centre: $(0, 0)$. Foci: $(\pm c, 0)$ where $c^2 = a^2 - b^2$. Major axis length: $2a$. Minor axis length: $2b$. Vertices: $(\pm a, 0)$. Co-vertices: $(0, \pm b)$. Eccentricity: $e = \frac{c}{a}$ where $0 < e < 1$.

Vertical major axis:

$\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \quad (a > b)$

Foci: $(0, \pm c)$ where $c^2 = a^2 - b^2$.

Translated ellipse: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ has centre at $(h, k)$.

Properties

• The sum of distances from any point on the ellipse to the two foci equals $2a$
• The closer $e$ is to 0, the more circular the ellipse
• The closer $e$ is to 1, the more elongated the ellipse

Tangent to an Ellipse

For $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent at $(x_1, y_1)$ is:

$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$

Conic Sections: Hyperbola

Standard Forms

A hyperbola is the locus of points such that the difference of distances from two fixed points (foci) is constant.

Horizontal transverse axis:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Centre: $(0, 0)$. Foci: $(\pm c, 0)$ where $c^2 = a^2 + b^2$. Vertices: $(\pm a, 0)$. Asymptotes: $y = \pm \frac{b}{a} x$. Eccentricity: $e = \frac{c}{a}$ where $e > 1$.

Vertical transverse axis:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Foci: $(0, \pm c)$ where $c^2 = a^2 + b^2$. Asymptotes: $y = \pm \frac{a}{b} x$.

Translated hyperbola: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ has centre at $(h, k)$.

Properties

• The difference of distances from any point on the hyperbola to the two foci equals $2a$
• Asymptotes are the lines the hyperbola approaches but never reaches
• The eccentricity $e > 1$

Tangent to a Hyperbola

For $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the tangent at $(x_1, y_1)$ is:

$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$

Comparing Conic Sections

Rectangular Hyperbola

A rectangular hyperbola has perpendicular asymptotes. Its standard equation is $xy = c^2$ or $x^2 - y^2 = a^2$.

For $xy = c^2$: Asymptotes are the coordinate axes $x = 0$ and $y = 0$.

Transformations

Translation

Replacing $x$ with $(x - h)$ and $y$ with $(y - k)$ translates the graph $h$ units right and $k$ units up.

Example: $y = (x - 2)^2 + 3$ is $y = x^2$ translated 2 units right and 3 units up.

Reflection

• $y = f(-x)$: Reflection in the $y$-axis
• $y = -f(x)$: Reflection in the $x$-axis

Scaling

• $y = af(x)$: Vertical stretch by factor $a$ (if $a > 1$) or compression (if $0 < a < 1$)
• $y = f(ax)$: Horizontal compression by factor $a$ (if $a > 1$) or stretch (if $0 < a < 1$)

Rotation of Conics

The general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ represents:

• An ellipse (or circle) if $B^2 - 4AC < 0$
• A parabola if $B^2 - 4AC = 0$
• A hyperbola if $B^2 - 4AC > 0$

Vector Methods in Proofs

Vectors in Coordinate Geometry

The position vector of point $P(x, y)$ is $\vec{OP} = \begin{pmatrix} x \\ y \end{pmatrix}$.

Vector Equation of a Line

Through point $A$ with position vector $\mathbf{a}$, in the direction of vector $\mathbf{d}$:

$\mathbf{r} = \mathbf{a} + t\mathbf{d} \quad (t \in \mathbb{R})$

In Cartesian form, if $\mathbf{d} = \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}$:

$\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2}$

Using Vectors to Prove Geometric Properties

Example: Prove that the diagonals of a parallelogram bisect each other.

Let the parallelogram have vertices $A$, $B$, $C$, $D$ with position vectors $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$.

Since $ABCD$ is a parallelogram: $\overrightarrow{AB} = \overrightarrow{DC}$, so $\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d}$, giving $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$.

The midpoint of diagonal $AC$ is $\frac{\mathbf{a} + \mathbf{c}}{2}$. The midpoint of diagonal $BD$ is $\frac{\mathbf{b} + \mathbf{d}}{2}$.

Since $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$, the midpoints coincide. Therefore, the diagonals bisect each other.

Using Dot Product for Perpendicularity

Two vectors $\mathbf{u}$ and $\mathbf{v}$ are perpendicular if and only if their dot product is zero:

$\mathbf{u} \cdot \mathbf{v} = 0$

$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$

Area Using Vectors

The area of triangle $ABC$ is:

$\text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$

In 2D, if $\overrightarrow{AB} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and $\overrightarrow{AC} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$:

$\text{Area} = \frac{1}{2}|a_1 c_2 - a_2 c_1|$

Vector Proofs for Collinearity

Three points $A$, $B$, $C$ are collinear if and only if $\overrightarrow{AB}$is parallel to $\overrightarrow{AC}$, i.e.:

$\overrightarrow{AB} = k\overrightarrow{AC} \quad \text{for some scalar } k$

Common Pitfalls

• Confusing the standard forms of the ellipse and hyperbola
• Forgetting that $c^2 = a^2 - b^2$ for the ellipse but $c^2 = a^2 + b^2$ for the hyperbola
• Misidentifying the transverse axis of a hyperbola (it is the axis containing the vertices)
• Incorrectly computing the perpendicular distance from a point to a line (sign errors in the formula)
• Forgetting to check the discriminant condition when determining intersection types
• Errors in the sign when using the translation formula for conics