Coordinate Geometry Essentials & Perpendicular Bisectors
The Cartesian plane allows geometric shapes, lines, and intersections to be modeled using algebraic coordinates $(x, y)$.
- Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
- Midpoint Formula: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
- Gradient (Slope): $m = \frac{y_2 - y_1}{x_2 - x_1}$
- Parallel Lines: $m_1 = m_2$
- Perpendicular Lines: $m_1 \times m_2 = -1 \iff m_2 = -\frac{1}{m_1}$
The perpendicular bisector of segment $AB$ is the line that cuts $AB$ in half at $90^\circ$:
- Find the midpoint $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$.
- Find the gradient of $AB$: $m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}$.
- Compute the perpendicular gradient: $m_{\perp} = -\frac{1}{m_{AB}}$.
- Write equation using point-slope form: $y - y_M = m_{\perp}(x - x_M)$.
Rigid Transformations (Isometries): Translations, Reflections, Rotations
An isometry (rigid motion) preserves distance, angle measures, and area. The image figure is always congruent to the pre-image.
A translation slides every point by a column vector $\begin{pmatrix} a \\ b \end{pmatrix}$: $$(x, y) \to (x + a, y + b)$$
A reflection flips a figure across a mirror line:
- Reflection across $x$-axis ($y = 0$): $(x, y) \to (x, -y)$
- Reflection across $y$-axis ($x = 0$): $(x, y) \to (-x, y)$
- Reflection across $y = x$: $(x, y) \to (y, x)$
- Reflection across $y = -x$: $(x, y) \to (-y, -x)$
- $90^\circ$ counter-clockwise ($270^\circ$ clockwise): $(x, y) \to (-y, x)$
- $180^\circ$ (half-turn): $(x, y) \to (-x, -y)$
- $270^\circ$ counter-clockwise ($90^\circ$ clockwise): $(x, y) \to (y, -x)$
Dilations (Enlargements) & Negative Scale Factors
A dilation (enlargement/reduction) changes the size of a shape while preserving its angle measures. The pre-image and image are similar figures.
- If $|k| > 1$: the shape expands (enlargement).
- If $0 < |k| < 1$: the shape shrinks (reduction).
- Negative Scale Factors ($k < 0$): The image is inverted through the center of dilation. This is equivalent to a dilation of $|k|$ followed by a $180^\circ$ rotation about the center!
Geometric transformations on the Cartesian plane: reflection, inversion ($k = -1$), and dilation.
Composite Transformations & Invariant Elements
Multiple transformations applied sequentially are called composite transformations.
If transformation $A$ is applied first, followed by transformation $B$, the composition is written as $B(A(\mathbf{x}))$. Warning: Transformations are generally non-commutative: $B \circ A \neq A \circ B$. Order matters!
- Invariant Point: A point that maps onto itself: $T(P) = P$. (For example, every point on the mirror line is invariant under reflection; the center of rotation is invariant under rotation).
- Invariant Line: A line that maps onto itself (points may move along the line, but the line as a whole remains stationary).
Worked Example 1: Perpendicular Bisector Equation
Points $A(1, 3)$ and $B(7, -1)$ form a line segment $AB$. Determine the equation of the perpendicular bisector of $AB$ in standard form $ax + by + c = 0$.
Step-by-step Solution:
Step 1: Midpoint $M$ of $AB$
$M = \left(\frac{1 + 7}{2}, \frac{3 + (-1)}{2}\right) = \left(\frac{8}{2}, \frac{2}{2}\right) = (4, 1)$. [M1]
Step 2: Gradient of $AB$
$m_{AB} = \frac{-1 - 3}{7 - 1} = \frac{-4}{6} = -\frac{2}{3}$. [M1]
Step 3: Perpendicular gradient $m_{\perp}$
$m_{\perp} = -\frac{1}{m_{AB}} = -\frac{1}{-2/3} = \frac{3}{2}$. [A1]
Step 4: Equation through $(4, 1)$
$y - 1 = \frac{3}{2}(x - 4)$
$2(y - 1) = 3(x - 4) \implies 2y - 2 = 3x - 12$
$3x - 2y - 10 = 0$. [Criterion A & C: Knowledge & Communication]
Worked Example 2: Negative Scale Factor Dilation
Triangle $ABC$ has vertices $A(1, 2)$, $B(3, 2)$, and $C(1, 5)$. It is dilated by a scale factor $k = -2$ with center of dilation at $P(2, 3)$.
a) Find the coordinates of the image vertices $A'$, $B'$, and $C'$.
b) Calculate the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$.
Step-by-step Solution:
Part a: Dilation formula with center $P(a, b) = (2, 3)$ and $k = -2$:
$x' = a + k(x - a) = 2 - 2(x - 2)$
$y' = b + k(y - b) = 3 - 2(y - 3)$ [M1]
For $A(1, 2)$:
$x' = 2 - 2(1 - 2) = 2 - 2(-1) = 2 + 2 = 4$
$y' = 3 - 2(2 - 3) = 3 - 2(-1) = 3 + 2 = 5 \implies A'(4, 5)$. [A1]
For $B(3, 2)$:
$x' = 2 - 2(3 - 2) = 2 - 2(1) = 0$
$y' = 3 - 2(2 - 3) = 3 + 2 = 5 \implies B'(0, 5)$. [A1]
For $C(1, 5)$:
$x' = 2 - 2(1 - 2) = 4$
$y' = 3 - 2(5 - 3) = 3 - 2(2) = 3 - 4 = -1 \implies C'(4, -1)$. [A1]
Part b: Area ratio:
Under any dilation with scale factor $k$, the area of the image scales by $k^2$: [M1]
$\text{Area Ratio} = \frac{\text{Area}(A'B'C')}{\text{Area}(ABC)} = k^2 = (-2)^2 = 4$.
The ratio is $4 : 1$. [Criterion A: Conceptual Understanding]
Tiered Practice Exercises
10 problems advancing from basic coordinate calculations and single isometries to negative enlargements, composite transformations, and invariant geometric investigations.
Find the distance and midpoint between the points $P(-2, 7)$ and $Q(4, -1)$.
View Solution & Markscheme
Method & Marks:
Distance: $d = \sqrt{(4 - (-2))^2 + (-1 - 7)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. [M1][A1]
Midpoint: $M = \left(\frac{-2 + 4}{2}, \frac{7 + (-1)}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)$. [A1]
A point $A(3, -5)$ is translated by vector $\begin{pmatrix} -4 \\ 7 \end{pmatrix}$. State the coordinates of the image point $A'$.
View Solution & Markscheme
Method & Marks:
$A' = (3 + (-4), -5 + 7) = (-1, 2)$. [A1]
Find the image of point $B(-4, 6)$ after:
a) Reflection across the $x$-axis
b) Reflection across the line $y = x$
View Solution & Markscheme
Method & Marks:
a) Reflection across $x$-axis: $(x, -y) \implies (-4, -6)$. [A1]
b) Reflection across line $y = x$: $(y, x) \implies (6, -4)$. [A1]
Point $P(3, 1)$ is rotated $90^\circ$ counter-clockwise about the origin. Find its image $P'$. What are the coordinates if it is instead rotated $180^\circ$ about the origin?
View Solution & Markscheme
Method & Marks:
$90^\circ$ counter-clockwise rule: $(x, y) \to (-y, x)$.
For $P(3, 1)$: $P'(-1, 3)$. [A1]
$180^\circ$ rotation rule: $(x, y) \to (-x, -y)$.
Image: $P''(-3, -1)$. [A1]
Line segment $CD$ connects $C(2, 5)$ and $D(8, 7)$. Find the equation of the perpendicular bisector of $CD$ in gradient-intercept form $y = mx + c$.
View Solution & Markscheme
Method & Marks:
Midpoint $M = \left(\frac{2 + 8}{2}, \frac{5 + 7}{2}\right) = (5, 6)$. [M1]
Gradient of $CD$: $m_{CD} = \frac{7 - 5}{8 - 2} = \frac{2}{6} = \frac{1}{3}$. [M1]
Perpendicular slope: $m_{\perp} = -3$. [A1]
Equation through $(5, 6)$:
$y - 6 = -3(x - 5) \implies y - 6 = -3x + 15 \implies y = -3x + 21$. [A1]
A triangle has vertices at $A(1, 1)$, $B(4, 1)$, and $C(1, 3)$. It is enlarged with center $(0, 0)$ and scale factor $k = 3$.
a) Find the image coordinates $A', B', C'$.
b) Show that the area of $\triangle A'B'C'$ is $9$ times the area of $\triangle ABC$.
View Solution & Markscheme
Method & Marks:
a) $(x, y) \to (3x, 3y) \implies A'(3, 3), B'(9, 3), C'(3, 9)$. [A1]
b) $\triangle ABC$ is right-angled with base $b = 3$ and height $h = 2$: Area $= \frac{1}{2}(3)(2) = 3$. [M1]
$\triangle A'B'C'$ has base $b' = 9 - 3 = 6$ and height $h' = 9 - 3 = 6$: Area $= \frac{1}{2}(6)(6) = 18$.
$\frac{18}{3} = 9 = 3^2 = k^2$. $\blacksquare$ [A1][R1]
Point $M(2, 4)$ is rotated $90^\circ$ clockwise about the point $C(1, 1)$. Find the coordinates of the image point $M'$.
View Solution & Markscheme
Method & Marks:
Translate center $C(1, 1)$ to origin: [M1]
Relative vector $\mathbf{v} = M - C = (2 - 1, 4 - 1) = (1, 3)$.
Apply $90^\circ$ clockwise rotation rule $(x, y) \to (y, -x)$: [M1]
$\mathbf{v}' = (3, -1)$.
Translate back: $M' = C + \mathbf{v}' = (1 + 3, 1 + (-1)) = (4, 0)$. [A1]
Point $Q(5, -2)$ undergoes a composite transformation: first a reflection across the line $x = 2$, followed by a translation by vector $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$. Find the final image $Q''$. Does changing the order of these two transformations yield the same result? Justify algebraically.
View Solution & Markscheme
Method & Marks:
1. Reflection across $x = 2$:
Distance from $x = 2$ is $5 - 2 = 3$. Reflected $x$-coordinate is $2 - 3 = -1$. $y$ stays same: $Q'(-1, -2)$. [M1]
2. Translation:
$Q'' = (-1 - 3, -2 + 4) = (-4, 2)$. [A1]
Reversing Order:
Translate first: $Q_1 = (5 - 3, -2 + 4) = (2, 2)$.
Reflect across $x = 2$: Since $x = 2$, distance is $0$, so $Q_2(2, 2)$.
Since $(-4, 2) \neq (2, 2)$, order matters and the transformations do NOT commute. [R1]
A dilation with center $C(a, b)$ and scale factor $k = -3$ maps the point $A(1, 2)$ to $A'(-7, -6)$. Find the coordinates of the center of dilation $C$.
View Solution & Markscheme
Method & Marks:
Using the dilation formula with center $(a, b)$: [M1]
$x' = a + k(x - a) \implies -7 = a - 3(1 - a) \implies -7 = a - 3 + 3a = 4a - 3$
$4a = -4 \implies a = -1$. [A1]
$y' = b + k(y - b) \implies -6 = b - 3(2 - b) \implies -6 = b - 6 + 3b = 4b - 6$
$4b = 0 \implies b = 0$. [A1]
Therefore, the center of dilation is $C(-1, 0)$. [A1][R1]
Criterion B & D Investigation: An automated manufacturing robotic arm inspects components on a conveyor grid. A component at position $P_0(x, y)$ is rotated $180^\circ$ about point $A(2, 1)$, followed immediately by a $180^\circ$ rotation about point $B(5, 3)$.
a) Determine the algebraic mapping $(x, y) \to (x'', y'')$ for the composite transformation.
b) Prove that the composition of two $180^\circ$ rotations with different centers is always a pure translation, and state the translation vector in terms of $\vec{AB}$.
View Solution & Markscheme
Method & Marks:
Part a:
$180^\circ$ rotation about $A(a, b)$: $(x', y') = (2a - x, 2b - y) = (2(2) - x, 2(1) - y) = (4 - x, 2 - y)$. [M1]
Second $180^\circ$ rotation about $B(5, 3)$:
$x'' = 2(5) - x' = 10 - (4 - x) = x + 6$
$y'' = 2(3) - y' = 6 - (2 - y) = y + 4$. [A1]
So $(x, y) \to (x + 6, y + 4)$.
Part b: Notice that $(x'', y'') = (x, y) + \begin{pmatrix} 6 \\ 4 \end{pmatrix}$, which is a pure translation by vector $\mathbf{v} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$. [M1]
The displacement vector from $A$ to $B$ is $\vec{AB} = \begin{pmatrix} 5 - 2 \\ 3 - 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.
Therefore, the translation vector is exactly $2\vec{AB} = 2\begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \end{pmatrix}$.
In general, two $180^\circ$ rotations about centers $A$ and $B$ always produce a translation of $2\vec{AB}$. $\blacksquare$ [A1][R1: Model reflection]
Transformations & Coordinate Geometry Quick Reference
Distance Formula
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Straight-line Euclidean distance.Midpoint Formula
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Average of coordinates.Perpendicular Slope
$$m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1}$$
Negative reciprocal.Translation
$$(x, y) \to (x + a, y + b)$$
Vector shift $\begin{pmatrix} a \\ b \end{pmatrix}$.Reflection Across $y = x$
$$(x, y) \to (y, x)$$
Swap $x$ and $y$ coordinates.$90^\circ$ CCW Rotation
$$(x, y) \to (-y, x)$$
About origin $(0, 0)$.$180^\circ$ Rotation
$$(x, y) \to (-x, -y)$$
Half-turn inversion.Dilation (Origin)
$$(x, y) \to (kx, ky)$$
Area scale factor is $k^2$.Top 5 Common MYP Transformation Pitfalls
- Confusing Clockwise and Counter-Clockwise Rotations: In mathematics, the standard default direction is counter-clockwise (anticlockwise). $90^\circ$ counter-clockwise maps $(x, y) \to (-y, x)$, whereas $90^\circ$ clockwise maps $(x, y) \to (y, -x)$. Always confirm direction!
- Misunderstanding Negative Scale Factors in Dilation: A negative scale factor (e.g. $k = -2$) does NOT make dimensions negative. Lengths remain positive, scaled by $|k| = 2$. The negative sign flips the figure across the center of dilation, resulting in an upside-down (inverted) image.
- Applying Rotation Rules Directly Without Translating the Center: Standard formulas like $(x, y) \to (-y, x)$ apply ONLY when rotating about the origin $(0, 0)$. When rotating about a point $(a, b)$, first shift by $(-a, -b)$, rotate, and then shift back by $(a, b)$.
- Assuming Composite Transformations Commute: Order matters! A reflection followed by a translation is rarely equal to a translation followed by a reflection. Always perform transformations strictly in the sequence given.
- Negative Signs in Distance Formula: When subtracting negative coordinates (e.g., $x_2 - x_1$ with $x_1 = -3$), remember $(4 - (-3))^2 = (4 + 3)^2 = 7^2 = 49$. Do not accidentally write $(4 - 3)^2 = 1$.
Self-Grading Mini Test
5 quiz questions on coordinates, isometries, dilations, and invariant geometry.
What is the distance between the points $A(-1, 2)$ and $B(3, 5)$?
Explanation: $d = \sqrt{(3 - (-1))^2 + (5 - 2)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
What is the image of point $P(2, -7)$ when reflected across the line $y = x$?
Explanation: Reflection across $y = x$ swaps the coordinates: $(x, y) \to (y, x)$. Therefore, $(2, -7) \to (-7, 2)$.
Point $Q(-4, 3)$ is rotated $90^\circ$ counter-clockwise about $(0,0)$. What are its new coordinates?
Explanation: The rule for $90^\circ$ counter-clockwise rotation about origin is $(x, y) \to (-y, x)$. For $(-4, 3)$, this gives $(-3, -4)$.
Point $A(2, -3)$ is enlarged by scale factor $k = -3$ with center of enlargement at the origin $(0, 0)$. What is the image point $A'$?
Explanation: $(x, y) \to (kx, ky) = (-3 \times 2, -3 \times (-3)) = (-6, 9)$.
Line $L_1$ has equation $2x + 5y = 10$. What is the gradient of a line perpendicular to $L_1$?
Explanation: In $y = mx + c$ form: $5y = -2x + 10 \implies y = -\frac{2}{5}x + 2$, so $m_1 = -\frac{2}{5}$. The perpendicular gradient is $m_{\perp} = -\frac{1}{-2/5} = \frac{5}{2}$.