Additional Mathematics notes, chapter 7

Coordinate Geometry

Textbook Additional Mathematics Form 4 KSSM (DLP), pages 174–209

What this chapter covers

{'Relate the position of a point on a line segment to the ratio m': 'n, derive the formula for the point that divides a segment, and use it to find coordinates and ratios (7.1.1, 7.1.2)'}
Solve problems involving the divisor of a line segment (7.1.3)
Make and verify conjectures about the gradients of parallel lines (m1 = m2) and perpendicular lines (m1m2 = −1), and use them to find equations and unknown values (7.2.1, 7.2.2)
Derive and use the formula for the area of a triangle, quadrilateral and polygon from the coordinates of its vertices, including collinear points (7.3.1, 7.3.2, 7.3.3, 7.3.4)
Solve problems involving areas of polygons (7.3.5)
Find the equation of a locus that is a fixed distance from a point, in a fixed ratio of distances from two points, or equidistant from two points (7.4.1)
Solve problems involving equations of loci (7.4.2)

7.1Divisor of a line segment

Internal point: a point P on segment AB that divides it in the ratio m : n, written AP : PB = m : n. The segment is cut into (m + n) equal parts, with m parts from A to P and n parts from P to B. If m = n, P is the midpoint.

The position of P decides the ratio, and the ratio decides the position. Read the ratio in the order of the letters: AP : PB.

AP : PB = 2 : 3 means P is 2/5 of the way from A to B. So AP : AB = 2 : 5.

The formula on a Cartesian plane

If A(x₁, y₁) and B(x₂, y₂) and P divides AB in the ratio m : n, then

P(x, y) = ( (nx₁ + mx₂) / (m + n) , (ny₁ + my₂) / (m + n) )
For the midpoint (m = n): M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )

The ratio number m goes with the far point B and n goes with the near point A. Swapping them gives the point on the wrong side of the midpoint.

Cartesian plane with A(2, −4) and B(12, 6) joined by a line; P(6, 0) lies on it with AP : PB = 2 : 3 — P divides AB in the ratio 2 : 3. Textbook method, p. 178.

Example: find the point (based on textbook p. 179)

A(−3, 5) and B(9, −1). P divides AB in the ratio 1 : 2. Here m = 1, n = 2.
x = (2(−3) + 1(9)) / 3 = 3/3 = 1
y = (2(5) + 1(−1)) / 3 = 9/3 = 3
P = (1, 3)

Example: find the ratio (based on textbook p. 180)

P(3, 2) divides A(1, 0) and B(9, 8). Let AP : PB = m : n.
x: (n(1) + m(9)) / (m + n) = 3, so n + 9m = 3m + 3n, so 6m = 2n, so m : n = 1 : 3.
Check with y: (n(0) + m(8)) / (m + n) = 8/4 = 2.

A(0, 0) and B(10, 5). Find the point P that divides AB in the ratio 3 : 2.

Show answerHide answer

x = (2(0) + 3(10))/5 = 6, y = (2(0) + 3(5))/5 = 3. P = (6, 3).

Problems with moving objects (textbook p. 181): if something moves at a steady speed, the ratio of distances equals the ratio of times. After 18 s of a 60 s journey, EP : EG = 18 : 60 = 3 : 10, so EP : PG = 3 : 7.

7.2Parallel and perpendicular lines

Gradient: m = (y₂ − y₁)/(x₂ − x₁) = tan θ, where θ is the angle the line makes with the positive x-axis (0° ≤ θ < 180°).

Lines	Condition	Why
Parallel	m₁ = m₂	Corresponding angles are equal, so tan θ₁ = tan θ₂
Perpendicular	m₁m₂ = −1	θ₂ = 90° + θ₁, so tan θ₂ = −1/tan θ₁

Three lines: two parallel with gradient 1/2 and one perpendicular to them with gradient −2 — L₁ and L₂ are parallel (same gradient). L₃ is perpendicular to both because ½ × (−2) = −1.

Put every equation in the form y = mx + c first. For ax + by = c the gradient is −a/b. For x/a + y/b = 1 (intercept form) the gradient is −b/a.

Perpendicular does not mean the gradients are negatives of each other. For m = 3 the perpendicular gradient is −1/3, not −3.

Example: parallel or perpendicular? (based on textbook p. 186-187)

(a) 2x + 3y = 9 and 4x + 6y = 7: both have gradient −2/3, so they are parallel.
(b) 3y = x + 5 and y = −3x + 1: gradients 1/3 and −3. Product = −1, so they are perpendicular.

Example: find an unknown (based on textbook p. 187)

A(2, 3), B(5, 9) and C(11, c) with ∠ABC = 90°. Gradient AB = (9 − 3)/(5 − 2) = 2. So gradient BC = −1/2.
(c − 9)/(11 − 5) = −1/2, so c − 9 = −3, so c = 6.

The line 2y = 6 − kx is perpendicular to y = 2x + 1. Find k.

Show answerHide answer

2y = 6 − kx gives y = 3 − (k/2)x, so m = −k/2. m × 2 = −1 gives −k = −1, so k = 1.

Equation of a line through (x₁, y₁) with gradient m: y − y₁ = m(x − x₁). For the perpendicular bisector of AB use the midpoint of AB and the gradient −1/m_AB. To find where two lines meet, solve their equations simultaneously (textbook p. 188-189).

7.3Areas of polygons

Shoelace formula: for vertices (x₁, y₁), (x₂, y₂), …, (x_n, y_n) taken in order round the polygon,
Area = ½ | (x₁y₂ + x₂y₃ + … + x_ny₁) − (x₂y₁ + x₃y₂ + … + x₁y_n) |

Write the vertices in a row, in order, and repeat the first at the end.
Multiply downwards to the right and add (first sum).
Multiply downwards to the left and add (second sum).
Subtract, take the absolute value and halve.

If the vertices are not in order round the shape, the formula gives the wrong area. Sketch the points first. The absolute value signs make sure the area is positive.

Triangle with vertices A(−2, 1), B(4, −1) and C(1, 6) — Triangle ABC, area 18 units².

Example: area of a triangle (based on textbook p. 193)

A(−2, 1), B(4, −1), C(1, 6).
First sum: (−2)(−1) + 4(6) + 1(1) = 2 + 24 + 1 = 27
Second sum: 4(1) + 1(−1) + (−2)(6) = 4 − 1 − 12 = −9
Area = ½ |27 − (−9)| = ½ × 36 = 18 units²

Collinear points have area 0. To show three points lie on one straight line, show the shoelace area is 0 (textbook p. 194). If an area is given and one coordinate is unknown, write ± the area: this gives two possible values (Example 9, p. 194).

Find the area of the triangle with vertices (0, 0), (6, 0) and (0, 4).

Show answerHide answer

First sum 0 + 6(4) + 0 = 24. Second sum 0. Area = ½ × 24 = 12 units². (Check: ½ × 6 × 4 = 12.)

The same method works for a quadrilateral, pentagon or any polygon, as long as the vertices are listed in order (textbook p. 195-197). For a parallelogram, the diagonals share a midpoint, so use the midpoint formula to find a missing vertex.

7.4Equations of loci

Locus: the path of a moving point P(x, y) that obeys a rule. Use the distance formula d = √((x − x₁)² + (y − y₁)²), then square both sides to remove the root.

Rule for P	Shape	Equation
Fixed distance r from A(x₁, y₁)	Circle	(x − x₁)² + (y − y₁)² = r²
PA : PB = m : n	Circle	n²[(x − x₁)² + (y − y₁)²] = m²[(x − x₂)² + (y − y₂)²]
PA = PB	Straight line (perpendicular bisector of AB)	(x − x₁)² + (y − y₁)² = (x − x₂)² + (y − y₂)²

Circle with centre A(1, −2) and radius 3, with a moving point P(x, y) on it — Locus of P: distance 3 from A(1, −2).

Example: fixed distance (based on textbook p. 201)

P is 3 units from A(1, −2).
(x − 1)² + (y + 2)² = 9
x² − 2x + 1 + y² + 4y + 4 = 9
x² + y² − 2x + 4y − 4 = 0

Square the ratio too. PA : PB = 1 : 2 means PA² : PB² = 1 : 4, so 4PA² = PB². Do not write 2PA² = PB².

Find the locus of P that is equidistant from O(0, 0) and A(6, 0).

Show answerHide answer

x² + y² = (x − 6)² + y². So 0 = −12x + 36, x = 3. It is the perpendicular bisector of OA.

When the question gives a diagram, check your final equation by substituting a point you can see on the locus (textbook p. 204).