Problems on points and vectors – Collinear Points and Orthogonal Vectors in 2D (Senior 3 Mathematics) | Ganzaa

Collinear Points and Orthogonal Vectors in 2D

Lessons

Senior 3 · Mathematics

Collinear Points and Orthogonal Vectors in 2D

3 lessons

Lesson 3 of 3

Problems on points and vectors

25 min

Checking a plot without a tape measure

A land surveyor records the corners of a plot as coordinates, not as a drawing. From four pairs of numbers you can decide whether the plot is a square, a rectangle, a rhombus or just a parallelogram, and whether a proposed path meets a boundary at a right angle. Everything needed is in the last two lessons; this lesson puts it to work.

Important

Quadrilateral $A B C D$ from vectors (vertices in order round the shape):

Shape	Vector test
Parallelogram	$\overrightarrow{AB} = \overrightarrow{DC}$
Rectangle	parallelogram and $\overrightarrow{AB} \cdot \overrightarrow{AD} = 0$
Rhombus	parallelogram and $\|\overrightarrow{AB}\| = \|\overrightarrow{AD}\|$ (or diagonals orthogonal: $\overrightarrow{AC} \cdot \overrightarrow{BD} = 0$ )
Square	rectangle and rhombus
Trapezium	$\overrightarrow{AB} = k\overrightarrow{DC}$ with $k \neq 1$ (one pair of parallel sides)

Worked example 1: is it a square?

Show that $A (- 2, 1)$ , $B (2, - 1)$ , $C (4, 3)$ , $D (0, 5)$ are the vertices of a square.

Solution $\overrightarrow{AB} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ and $\overrightarrow{DC} = \begin{pmatrix} 4 - 0 \\ 3 - 5 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ : equal, so $A B C D$ is a parallelogram.

$\overrightarrow{AD} = \begin{pmatrix} 2 \\ 4 \end{pmatrix}$ : $\overrightarrow{AB} \cdot \overrightarrow{AD} = 8 - 8 = 0$ , so the angle at $A$ is a right angle: a rectangle.

$|\overrightarrow{AB}| = \sqrt{16 + 4} = \sqrt{20}$ and $|\overrightarrow{AD}| = \sqrt{4 + 16} = \sqrt{20}$ : adjacent sides are equal, so it is a square.

Worked example 2: a rhombus that is not a square

Classify the quadrilateral $A (0, 0)$ , $B (5, 0)$ , $C (8, 4)$ , $D (3, 4)$ .

Solution $\overrightarrow{AB} = \begin{pmatrix} 5 \\ 0 \end{pmatrix} = \overrightarrow{DC}$ : a parallelogram. $|\overrightarrow{AB}| = 5$ and $|\overrightarrow{AD}| = \left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = 5$ : a rhombus. But $\overrightarrow{AB} \cdot \overrightarrow{AD} = 15 \neq 0$ , so it is not a square.

As a rhombus should, its diagonals are orthogonal: $\overrightarrow{AC} \cdot \overrightarrow{BD} = \begin{pmatrix} 8 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 4 \end{pmatrix} = -16 + 16 = 0$ .

Worked example 3: a right angle from a point on an axis

$A (1, 3)$ and $B (6, 2)$ are given. Find every point $P$ on the $x$ -axis with $\angle APB = 90^\circ$ .

Solution Let $P (x, 0)$ . Then $\overrightarrow{PA} = \begin{pmatrix} 1 - x \\ 3 \end{pmatrix}$ and $\overrightarrow{PB} = \begin{pmatrix} 6 - x \\ 2 \end{pmatrix}$ . A right angle at $P$ means

(1 - x)(6 - x) + 3 \times 2 = 0 \iff x^2 - 7x + 12 = 0 \iff (x - 3)(x - 4) = 0 .

So $P (3, 0)$ or $P (4, 0)$ . Both lie on the circle with diameter $A B$ , exactly as the angle-in-a-semicircle theorem predicts.

Worked example 4: a perpendicular path

A straight boundary runs through $A (1, 2)$ and $B (5, 4)$ . A path from $A$ must meet it at a right angle and end on the $y$ -axis at $D$ . Find $D$ .

Solution Let $D (0, y)$ : $\overrightarrow{AD} = \begin{pmatrix} -1 \\ y - 2 \end{pmatrix}$ and $\overrightarrow{AB} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ . Orthogonal: $- 4 + 2 (y - 2) = 0$ , so $y = 4$ and $D (0, 4)$ .

Worked example 5: both tests at once, the foot of a perpendicular

Find the point $H$ on the line through $O (0, 0)$ and $B (4, 2)$ that is closest to $C (5, 5)$ , and the distance $C H$ .

Solution $H$ is on line $O B$ (collinear), so $\overrightarrow{OH} = t\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and $H (2 t, t)$ . The closest point is the foot of the perpendicular, so $\overrightarrow{CH} \cdot \begin{pmatrix} 2 \\ 1 \end{pmatrix} = 0$ :

(2t - 5) \times 2 + (t - 5) \times 1 = 0 \iff 5t - 15 = 0 \iff t = 3 .

So $H (6, 3)$ , $\overrightarrow{CH} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $CH = \sqrt{1 + 4} = \sqrt{5} \approx 2.24$ units.

Warning

Name the vertices in order. For $A B C D$ , compare $\overrightarrow{AB}$ with $\overrightarrow{DC}$ , not with $\overrightarrow{CD}$ : $\overrightarrow{CD} = -\overrightarrow{DC}$ , and the test would wrongly fail.

A rectangle needs the parallelogram test first. One right angle alone does not make a rectangle.

Tip

In a classification question, go in this order and write one line for each: parallelogram? (equal vectors) → right angle? (dot product) → equal sides? (lengths). Then name the shape.

The GCSE Maths Tutor: Vectors and vector proofs. Watch how every proof ends by quoting a multiple (for parallel or collinear) or a shared point, the same habits as this unit.

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Practice

The last two are harder.

$A (1, 1)$ , $B (4, 2)$ , $C (6, 5)$ , $D (3, 4)$ . Show that $A B C D$ is a parallelogram. Is it a rectangle? A rhombus?
Show that $K (0, 0)$ , $L (3, 1)$ , $M (2, 4)$ , $N (- 1, 3)$ is a square.
$A (1, 2)$ , $B (5, 4)$ and $C (3, 8)$ are three vertices of a rectangle $A B C D$ . Check the right angle at $B$ and find $D$ .
$A (0, 0)$ , $B (6, 0)$ , $C (4, 3)$ , $D (1, 3)$ . Show that $A B C D$ is a trapezium and compare $A B$ with $D C$ .
$A (- 2, 1)$ and $B (4, 3)$ . Find the points $P$ on the $y$ -axis with $\angle APB = 90^\circ$ .
Find the foot $H$ of the perpendicular from $C (5, 0)$ to the line through $O (0, 0)$ and $B (3, 4)$ , and the distance $C H$ .
$A (2, 1)$ and $B (6, 4)$ . Find the two points $C$ such that triangle $A B C$ has a right angle at $B$ and $B C = A B$ .

Show answersHide answers

$\overrightarrow{AB} = \overrightarrow{DC} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ : parallelogram. $\overrightarrow{AD} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ : $\overrightarrow{AB} \cdot \overrightarrow{AD} = 9 \neq 0$ , not a rectangle; $\sqrt{10} \neq \sqrt{13}$ , not a rhombus.
$\overrightarrow{KL} = \overrightarrow{NM} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ ; $\overrightarrow{KN} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ : $\overrightarrow{KL} \cdot \overrightarrow{KN} = -3 + 3 = 0$ and both lengths are $\sqrt{10}$ . A square.
$\overrightarrow{BA} \cdot \overrightarrow{BC} = \begin{pmatrix} -4 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 4 \end{pmatrix} = 8 - 8 = 0$ . $\overrightarrow{AD} = \overrightarrow{BC}$ , so $D (- 1, 6)$ .
$\overrightarrow{AB} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}$ and $\overrightarrow{DC} = \begin{pmatrix} 3 \\ 0 \end{pmatrix}$ , so $\overrightarrow{AB} = 2\overrightarrow{DC}$ : parallel but not equal, a trapezium with $A B = 2 D C$ .
$P (0, y)$ : $(- 2) (4) + (1 - y) (3 - y) = 0$ , $y^{2} - 4 y - 5 = 0$ , so $y = 5$ or $y = - 1$ : $P (0, 5)$ or $P (0, - 1)$ .
$H (3 t, 4 t)$ : $(3t - 5) \times 3 + 4t \times 4 = 0$ , $25 t = 15$ , $t = 0.6$ . $H (1.8, 2.4)$ and $CH = \sqrt{3.2^2 + 2.4^2} = 4$ .
$\overrightarrow{AB} = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$ ; $\overrightarrow{BC}$ must be orthogonal with length 5: $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$ or $\begin{pmatrix} 3 \\ -4 \end{pmatrix}$ . So $C (3, 8)$ or $C (9, 0)$ .

1Checking a plot without a tape measure

Checking a plot without a tape measure