M2: Coordinate Geometry | DSE - Wyatt's Notes

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

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

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

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

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

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

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).

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

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

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

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

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

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

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

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.

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 $(3, 4)$.

$3x + 4y = 25$

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

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 $D$, $E$, $F$.

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 $y = k - a$.

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

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

$ty = x + at_1^2$

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.

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 $(h, k)$.

• 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

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

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

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 $(h, k)$.

• 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 $e > 1$

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

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

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 $y = 0$.

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.

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

• $y = af(x)$: 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 $0 < a < 1$)

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 $B^2 - 4AC > 0$

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

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 $\mathbf{d} = \begin{pmatrix} d_1 \\ d_2 \end{pmatrix}$:

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

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.

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

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

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

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 $\overrightarrow{AC} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}$:

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

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

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

• 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

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