SPM Core Geometry Topic

Division of Segments, Perpendicularity, Shoelace Area & Loci

Coordinate geometry unites algebra and spatial analysis. Master internal segment ratios $m:n$, perpendicular bisectors ($m_1 m_2 = -1$), shoelace polygon formulas, and circular distance-ratio loci $PA:PB = m:n$ in maritime and radar tracking applications.

1. Core Coordinate Geometry Formulas

Division of Line Segment ($m:n$)

$\left(\frac{nx_1 + mx_2}{m + n}, \frac{ny_1 + my_2}{m + n}\right)$

Cross-multiply: $n$ multiplies $(x_1, y_1)$ and $m$ multiplies $(x_2, y_2)$!

Parallel & Perpendicular Lines

$\text{Parallel: } m_1 = m_2$
$\text{Perpendicular: } m_1 m_2 = -1$

Perpendicular bisector passes through the midpoint with slope $-\frac{1}{m}$.

Shoelace Polygon Area

$\text{Area} = \frac{1}{2} \left| \begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix} \right|$

Collinear points $\iff \text{Area} = 0$. Arrange vertices anti-clockwise.

Equation of Locus: For a moving point $P(x, y)$ maintaining a constant ratio of distances from two fixed points $A$ and $B$: $\frac{PA}{PB} = \frac{m}{n} \implies n^2 PA^2 = m^2 PB^2 \implies n^2[(x - x_A)^2 + (y - y_A)^2] = m^2[(x - x_B)^2 + (y - y_B)^2]$. This simplifies into a general circle equation: $x^2 + y^2 + 2gx + 2fy + c = 0$.

2. SPM Examiner Pitfalls in Coordinate Geometry

Trap 1: Cross-Multiplication Error in Ratio Formula:

Students often write $\frac{mx_1 + nx_2}{m+n}$. Remember the cross rule: the ratio segment $AP = m$ multiplies point $B(x_2, y_2)$, and $PB = n$ multiplies point $A(x_1, y_1)$!

Trap 2: Forgetting to Repeat the First Coordinate in Shoelace Area:

The Shoelace matrix MUST close the polygon by repeating the starting coordinate at the end: $\begin{matrix} x_1 & x_2 & x_3 & x_1 \\ y_1 & y_2 & y_3 & y_1 \end{matrix}$. Omitting the closing column yields a wildly incorrect area.

Trap 3: Forgetting to Square Ratio in Locus:

$\frac{PA}{PB} = \frac{1}{2} \implies 2PA = PB \implies 4 PA^2 = PB^2$. Squaring 2 to become 4 is frequently missed, leaving a factor of 2 that derails the entire locus circle equation.

3. Progressive Worked Examples with SPM Marking Rubrics

Example 1 • Perpendicular Bisector & Area [6 Marks]
Given points $A(-2, 5)$ and $B(4, 1)$.

(a) Find the equation of the perpendicular bisector of line segment $AB$. [3 marks]

(b) The perpendicular bisector intersects the $x$-axis at point $C$. Calculate the area of triangle $ABC$. [3 marks]

(a) Midpoint $M = \left(\frac{-2 + 4}{2}, \frac{5 + 1}{2}\right) = (1, 3)$. [1m: K1: Compute midpoint]
Gradient $m_{AB} = \frac{1 - 5}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3} \implies m_{\perp} = -\frac{1}{-2/3} = \frac{3}{2}$. [1m: K1: Perpendicular gradient $m_1 m_2 = -1$]
Equation: $y - 3 = \frac{3}{2}(x - 1) \implies 2y - 6 = 3x - 3 \implies 3x - 2y + 3 = 0$. [1m: N1: Correct equation]
(b) Point $C$ on $x$-axis: $y = 0 \implies 3x + 3 = 0 \implies x = -1 \implies C(-1, 0)$. [1m: K1: Find coordinates of C]
$\text{Area} = \frac{1}{2} \left| \begin{matrix} -2 & 4 & -1 & -2 \\ 5 & 1 & 0 & 5 \end{matrix} \right| = \frac{1}{2} |[(-2)(1) + 4(0) + (-1)(5)] - [5(4) + 1(-1) + 0(-2)]|$. [1m: K1: Shoelace formula substitution]
$= \frac{1}{2} |(-2 + 0 - 5) - (20 - 1 + 0)| = \frac{1}{2} |-7 - 19| = \frac{1}{2} |-26| = 13\text{ unit}^2$. [1m: N1: Correct area]
Example 2 • Circular Distance Ratio Locus [5 Marks]
Point $P(x, y)$ moves such that its distance from point $A(1, 2)$ is twice its distance from point $B(4, -1)$.

(a) Find the equation of the locus of $P$. [3 marks]

(b) Hence, show that the locus of $P$ is a circle and state its center and radius. [2 marks]

(a) $PA = 2PB \implies PA^2 = 4PB^2$. [1m: P1: Square ratio $PA^2 = 4PB^2$]
$(x - 1)^2 + (y - 2)^2 = 4[(x - 4)^2 + (y + 1)^2]$. [1m: K1: Distance formula substitution]
$x^2 - 2x + 1 + y^2 - 4y + 4 = 4[x^2 - 8x + 16 + y^2 + 2y + 1] = 4x^2 - 32x + 4y^2 + 8y + 68$. [1m: K1: Expand accurately]
$3x^2 + 3y^2 - 30x + 12y + 63 = 0 \implies x^2 + y^2 - 10x + 4y + 21 = 0$. [1m: N1: Locus equation in simplest form]
(b) Completing the square: $(x - 5)^2 - 25 + (y + 2)^2 - 4 + 21 = 0 \implies (x - 5)^2 + (y + 2)^2 = 8$. Circle with center $(5, -2)$ and radius $r = \sqrt{8} = 2\sqrt{2}\text{ units}$. [1m: N1: Center and radius stated]
Example 3 • KBAT Maritime Radar Geofence Tracking [5 Marks]
A coastal lighthouse is located at origin $L(0, 0)$ and a radio communication tower is located at $T(6, 8)$ (units in kilometers). A fishing trawler $S(x, y)$ navigates within coastal waters such that its distance from tower $T$ is always $5\text{ km}$.

(a) Find the Cartesian equation of the circular boundary locus of the trawler $S$. [2 marks]

(b) Determine the closest possible distance between the trawler and the lighthouse $L$. [3 marks]

(a) $ST = 5 \implies (x - 6)^2 + (y - 8)^2 = 5^2 \implies (x - 6)^2 + (y - 8)^2 = 25$. [1m: K1: Distance to center]
$x^2 - 12x + 36 + y^2 - 16y + 64 = 25 \implies x^2 + y^2 - 12x - 16y + 75 = 0$. [1m: N1: Locus equation]
(b) Distance from lighthouse $L(0,0)$ to tower center $T(6,8)$: $LT = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ km}$. [1m: K1: Distance between centers]
Closest distance to circle perimeter $= LT - \text{radius} = 10 - 5 = 5\text{ km}$. [2m: K1, N1: Closest distance $= 5\text{ km}$]
Open Form 4 Chapter 7 Worksheet