Chapter 1: Orienting Yourself: The Use of Coordinates (Coordinate Geometry)

The door starts at D₁, which is at 8 on the x-axis. The distance of a point from the y-axis is its x-coordinate, so the door is 8 ft from the left wall.

The door lies on the x-axis itself, so its distance from the x-axis is 0.

The door is 8 ft from the left wall (y-axis) and 0 ft from the x-axis (it lies on the x-axis).

D₁ is 8 units to the right of O and lies on the x-axis, so its y-coordinate is 0.

D₁ = (8, 0)

Both D₁ and R₁ lie on the x-axis, so the width of the door is the difference of their x-coordinates:

$$\text{Width}=11.5-8=3.5\text{ft}$$

A usual room door is about 3 ft wide, so 3.5 ft is comfortable. A standard wheelchair is about 2–2.5 ft wide and needs a clear opening of roughly 3 ft, so a wheelchair user can pass through this door easily.

The door is 3.5 ft wide. This is a comfortable width, and a person in a wheelchair can enter easily.

Both points lie on the y-axis, so the width is the difference of their y-coordinates:

$$\text{Width of bathroom door}=4-1.5=2.5\text{ft}$$

Since $2.5<3.5$, the bathroom door is narrower, by $3.5-2.5=1$ ft.

The bathroom door (2.5 ft) is narrower than the room door (3.5 ft) by 1 ft.

The table is a rectangle, so its sides are parallel to the axes. The feet (8, 9) and (11, 9) share y = 9; the feet (11, 9) and (11, 7) share x = 11. The fourth foot must have the x-coordinate of (8, 9) and the y-coordinate of (11, 7).

The fourth foot is at (8, 7).

The table occupies x = 8 to 11 and y = 7 to 9. The bed ends at x = 6.5, so there is a gap of $8-6.5=1.5$ ft between the bed and the table, and the table does not block the room door (at x = 8 to 11.5 on y = 0) or the wardrobe. The only thing to adjust is the plant in the top-right corner, which may touch the table; it can be shifted a little.

Yes, it is a reasonable spot: it is clear of the bed, the wardrobe and both doors (the corner plant may need to be shifted slightly).

$$\text{Length}=11-8=3\text{ft},\text{Width}=9-7=2\text{ft}$$

The plan shows only the floor (two dimensions: length and breadth), so it gives no information about height.

Length = 3 ft, width = 2 ft. The height cannot be found from a 2-D floor plan.

The door B₁B₂ is $4-1.5=2.5$ ft wide. When it swings open into the bedroom about the hinge B₁, its free end moves on a quarter circle of radius 2.5 ft with centre B₁ (0, 1.5).

The corner of the wardrobe nearest to B₁ is W₄ (3, 2). Its distance from B₁ is

$$B_1{W}_4=\sqrt{(3-0{)}^2+(2-1.5{)}^2}=\sqrt{9+0.25}=\sqrt{9.25}\approx 3.04\text{ft}$$

Since $3.04>2.5$, the door's free end never reaches the wardrobe. It also stays below the bed, which starts at y = 5.

If the door is made wider than about 3 ft, it will strike the wardrobe corner W₄, and a door wider than 3.5 ft (reaching y = 5) would also run into the bed. Suggestions for a wider door: make it open into the bathroom instead of the bedroom, use a sliding door, or put the hinge at B₂ so it swings towards the empty wall area instead of the wardrobe.

No, the 2.5 ft door will not hit the wardrobe (nearest corner W₄ is about 3.04 ft from B₁). For a wider door, make it open inwards (into the bathroom) or use a sliding door.

O is the origin; F is on the y-axis, 9 units up; R is 6 units to the left of F; P is 6 units to the left of O.

O (0, 0), F (0, 9), R (−6, 9), P (−6, 0)

The corners are S (−6, 6), H (−3, 6), W (−2, 9) and R (−6, 9). SH lies on the line y = 6 and RW lies on the line y = 9, so SH ∥ RW. The other two sides, HW and SR, are not parallel (SR is vertical, HW is slanted). A quadrilateral with exactly one pair of parallel sides is a trapezium. Also SR is perpendicular to both SH and RW, so it is a right trapezium.

$$SH=-3-(-6)=3\text{ft},RW=-2-(-6)=4\text{ft},SR=9-6=3\text{ft}$$

SHWR is a (right) trapezium with S (−6, 6), H (−3, 6), W (−2, 9), R (−6, 9).

The free part of the bathroom is the rectangle x = −6 to 0, y = 0 to 6 (below the showering area). Both fittings should go against the left wall PS, leaving space near the door B₁B₂ on the right. One possible arrangement (answers may differ):

• Washbasin (3 ft along the wall, 2 ft deep): corners (−6, 0.5), (−4, 0.5), (−4, 3.5), (−6, 3.5).
• Toilet (2 ft along the wall, 3 ft deep): corners (−6, 3.8), (−3, 3.8), (−3, 5.8), (−6, 5.8).

Check: neither space overlaps the showering area (which starts at y = 6) and both are inside the bathroom.

Washbasin: (−6, 0.5), (−4, 0.5), (−4, 3.5), (−6, 3.5). Toilet: (−6, 3.8), (−3, 3.8), (−3, 5.8), (−6, 5.8). (Other correct layouts are possible.)

P is (−6, 0) and A is (12, 0), so PA $=12-(-6)=18$ ft, which is the length. The room door D₁R₁ is on the x-axis and opens from the dining room, so the dining room lies below the x-axis. Its width is 15 ft, so the other two corners are 15 units below P and A.

Corners of the dining room: (−6, 0), (12, 0), (12, −15) and (−6, −15).

The centre of the room is midway between the corners:

$$x=\frac{-6+12}{2}=3,y=\frac{0+(-15)}{2}=-7.5$$

So the centre is (3, −7.5). Keep the 5 ft side along the length of the room. The table then extends $\frac{5}{2}=2.5$ ft on each side of x = 3 and $\frac{3}{2}=1.5$ ft on each side of y = −7.5:

$$x=3\pm 2.5=0.5\text{ or }5.5,y=-7.5\pm 1.5=-6\text{ or }-9$$

Feet of the table: (0.5, −6), (5.5, −6), (5.5, −9) and (0.5, −9).

The axes meet at the origin O. It is at zero distance from both axes.

x-coordinate = 0 and y-coordinate = 0, i.e. the origin O (0, 0).

Every point on a line parallel to the y-axis is at the same distance from the y-axis, so all such points have the same x-coordinate. The line through W is the line x = −5. Hence H = (−5, y) for some number y.

• If y > 0, H is in Quadrant II (x negative, y positive).
• If y < 0, H is in Quadrant III (both negative).
• If y = 0, H = (−5, 0) lies on the x-axis (in no quadrant).

H = (−5, y). H can lie in Quadrant II or Quadrant III (or on the x-axis at (−5, 0)).

Prediction from the coordinates:

• A (0, −2) and M (−5, −2) have the same y-coordinate, so AM is horizontal (parallel to the x-axis).
• M (−5, −2) and P (−5, 2) have the same x-coordinate, so MP is vertical (parallel to the y-axis).
• A horizontal line and a vertical line are perpendicular, so AM ⊥ MP.
• M (−5, −2) and P (−5, 2) differ only in the sign of the y-coordinate, so they are mirror images in the x-axis.

The plot confirms: AM runs along y = −2, MP runs along x = −5, they meet at a right angle at M, and P is the reflection of M in the x-axis.

(i) AM ⊥ MP. (ii) AM ∥ x-axis (also MP ∥ y-axis). (iii) M (−5, −2) and P (−5, 2) are mirror images in the x-axis.

The easiest right triangle uses lines parallel to the axes. Take N at the origin (0, 0) and I (5, 0), directly above Z on the x-axis. Then IZ is vertical and NI is horizontal, so ∠NIZ = 90°.

$$NI=5-0=5\text{units},IZ=0-(-6)=6\text{units}$$

$$NZ=\sqrt{5^2+6^2}=\sqrt{25+36}=\sqrt{61}\approx 7.81\text{units}$$

With N (0, 0) and I (5, 0): NI = 5 units, IZ = 6 units, NZ = √61 ≈ 7.81 units (right angle at I).

Without negative numbers, both coordinates would be 0 or positive. We could only describe points to the right of the y-axis and above the x-axis, that is, Quadrant I and the positive parts of the two axes. Points to the left of O or below O (Quadrants II, III and IV) would have no coordinates.

Such a system can still work for a bounded region if we put the origin at its bottom-left corner, as computer screens do (see Q15), but it cannot locate every point of the whole, endless plane.

It would cover only Quadrant I (and the positive axes), so it would not locate all the points of the plane; points in Quadrants II, III and IV would be left out.

Method: three points lie on one straight line exactly when the sum of the two shorter distances equals the longest distance (if they made a triangle, the sum of two sides would be greater than the third). We use the distance formula.

$$MA=\sqrt{(0+3{)}^2+(0+4{)}^2}=\sqrt{9+16}=5$$

$$AG=\sqrt{(6-0{)}^2+(8-0{)}^2}=\sqrt{36+64}=10$$

$$MG=\sqrt{(6+3{)}^2+(8+4{)}^2}=\sqrt{81+144}=\sqrt{225}=15$$

$MA+AG=5+10=15=MG$, so A lies on the segment MG.

Another check: moving from M to A we go 3 right and 4 up; from A to G we go 6 right and 8 up. The rise per unit run is $\frac{4}{3}$ in both cases, so the direction does not change.

Yes, M, A and G are on the same straight line, because MA + AG = 5 + 10 = 15 = MG.

$$RB=\sqrt{(-2+5{)}^2+(-5+1{)}^2}=\sqrt{9+16}=5$$

$$BC=\sqrt{(4+2{)}^2+(-12+5{)}^2}=\sqrt{36+49}=\sqrt{85}\approx 9.220$$

$$RC=\sqrt{(4+5{)}^2+(-12+1{)}^2}=\sqrt{81+121}=\sqrt{202}\approx 14.213$$

$RB+BC\approx 5+9.220=14.220$, which is not equal to $RC\approx 14.213$. To be sure this is not a rounding error, square both: $(5+\sqrt{85}{)}^2=110+10\sqrt{85}\approx 202.2$, while $R{C}^2=202$. They are different.

Rise per unit run: R to B is $\frac{-4}{3}\approx -1.33$, but B to C is $\frac{-7}{6}\approx -1.17$. The direction changes at B.

On a graph R, B, C look almost in line, which is exactly why the calculation is the better test.

No. RB + BC ≈ 14.22 ≠ RC ≈ 14.21, so R, B and C are not on one straight line (they are very nearly collinear). M, A, G (Problem 6) are collinear.

(i) Take O (0, 0), P (4, 0) and Q (0, 4). OP = OQ = 4 and the axes are perpendicular, so ∠POQ = 90°. Also $PQ=\sqrt{4^2+4^2}=4\sqrt{2}$.

(ii) Take O (0, 0), A (−3, −4) in Quadrant III and B (3, −4) in Quadrant IV. A and B are mirror images in the y-axis, so

$$OA=\sqrt{3^2+4^2}=5=OB,AB=3-(-3)=6$$

(i) O (0, 0), P (4, 0), Q (0, 4). (ii) O (0, 0), A (−3, −4), B (3, −4) with OA = OB = 5. (Other answers are possible.)

The midpoint of ST is halfway along in both directions, so its coordinates are the averages of the coordinates of S and T. We compare that with M.

Connection: when M is the midpoint, each coordinate of M is the average of the corresponding coordinates of S and T. For S $(x_1,y_1)$ and T $(x_2,y_2)$:

$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$

Yes, Yes, No, No. M is the midpoint of ST exactly when $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

$$\frac{3+x}{2}=-7\Rightarrow 3+x=-14\Rightarrow x=-17$$

$$\frac{-4+y}{2}=1\Rightarrow -4+y=2\Rightarrow y=6$$

Check: $\left(\frac{3-17}{2},\frac{-4+6}{2}\right)=(-7,1)$ ✓

B = (−17, 6)

P and Q divide AB into three equal parts: AP = PQ = QB. So P is the midpoint of AQ, and Q is the midpoint of PB. Moving from A to B, the total change is

$$\Delta x=16-4=12,\Delta y=-2-7=-9$$

Each third of the journey changes x by $12÷3=4$ and y by $-9÷3=-3$.

$$P=(4+4,7-3)=(8,4),Q=(8+4,4-3)=(12,1)$$

Check with the midpoint idea: midpoint of A (4, 7) and Q (12, 1) is $\left(\frac{16}{2},\frac{8}{2}\right)=(8,4)=P$ ✓; midpoint of P (8, 4) and B (16, −2) is $(12,1)=Q$ ✓. Also, the midpoint of AB, (10, 2.5), is the midpoint of PQ ✓.

In general, for A $(x_1,y_1)$ and B $(x_2,y_2)$: $P=\left(\frac{2x_1+x_2}{3},\frac{2y_1+y_2}{3}\right)$ and $Q=\left(\frac{x_1+2x_2}{3},\frac{y_1+2y_2}{3}\right)$.

P = (8, 4) and Q = (12, 1)

(i) A point lies on a circle with centre O if its distance from O equals the radius. The distance of $(x,y)$ from O is $\sqrt{x^2+y^2}$.

$$OA=\sqrt{1^2+(-8{)}^2}=\sqrt{65},OB=\sqrt{(-4{)}^2+7^2}=\sqrt{65},OC=\sqrt{(-7{)}^2+(-4{)}^2}=\sqrt{65}$$

All three are at the same distance √65 from O, so they lie on a circle K with centre O and radius √65 ≈ 8.06 units.

(ii) Compare each distance with √65:

$$OD=\sqrt{(-5{)}^2+6^2}=\sqrt{61}<\sqrt{65}\Rightarrow D\text{ is inside }K$$

$$OE=\sqrt{0^2+9^2}=9=\sqrt{81}>\sqrt{65}\Rightarrow E\text{ is outside }K$$

(i) OA = OB = OC = √65, so A, B, C lie on circle K of radius √65 ≈ 8.06 units. (ii) D is inside K (OD = √61); E is outside K (OE = 9).

Take D as the midpoint of BC, E of CA and F of AB (the usual convention). Let A $(x_1,y_1)$, B $(x_2,y_2)$, C $(x_3,y_3)$. Using the midpoint formula and doubling:

$$x_2+x_3=10,x_3+x_1=12,x_1+x_2=0$$

$$y_2+y_3=2,y_3+y_1=10,y_1+y_2=6$$

Adding the three x-equations: $2(x_1+x_2+x_3)=22$, so $x_1+x_2+x_3=11$. Subtract each equation from 11:

$$x_1=11-10=1,x_2=11-12=-1,x_3=11-0=11$$

Similarly $y_1+y_2+y_3=9$, so

$$y_1=9-2=7,y_2=9-10=-1,y_3=9-6=3$$

Check: midpoint of BC = $\left(\frac{-1+11}{2},\frac{-1+3}{2}\right)=(5,1)=D$ ✓, midpoint of CA = $(6,5)=E$ ✓, midpoint of AB = $(0,3)=F$ ✓.

A (1, 7), B (−1, −1), C (11, 3)

(i) Draw 10 vertical lines (N-S streets) and 10 horizontal lines (E-W streets), each 1 cm apart, since 200 m is shown as 1 cm. Number the streets 1 to 10 from west to east and from south to north. The two main roads are among these lines and cross near the centre.

(ii) The 4th N-S street and the 3rd E-W street are two straight lines that cross at exactly one point. Similarly the 3rd N-S street and the 4th E-W street meet at exactly one point. These two crossings are different points, because the order of the numbers matters, just as (4, 3) and (3, 4) are different points on a graph.

(a) Only one intersection is (4, 3). (b) Only one intersection is (3, 4), and it is a different crossing from (4, 3).

The screen covers $0\le x\le 800$ and $0\le y\le 600$.

(i) A circle stays on screen if its leftmost, rightmost, lowest and highest points are inside.

• Circle A: x from $100-80=20$ to $100+80=180$; y from $150-80=70$ to $150+80=230$. All within the screen.
• Circle B: x from $250-100=150$ to $350$; y from $230-100=130$ to $330$. All within the screen.

(ii) Distance between the centres:

$$AB=\sqrt{(250-100{)}^2+(230-150{)}^2}=\sqrt{150^2+80^2}=\sqrt{22500+6400}=\sqrt{28900}=170$$

Sum of radii $=80+100=180$ and difference $=100-80=20$. Since $20<170<180$, the circles overlap and cross each other at two points (if AB were more than 180 they would be apart; if it were exactly 180 they would just touch).

(i) No part of either circle lies outside the screen. (ii) Yes, the circles intersect (at two points), since 170 < 80 + 100 = 180.

Sides:

$$AB=\sqrt{(-1-2{)}^2+(2-1{)}^2}=\sqrt{9+1}=\sqrt{10},BC=\sqrt{(-2+1{)}^2+(-1-2{)}^2}=\sqrt{1+9}=\sqrt{10}$$

$$CD=\sqrt{(1+2{)}^2+(-2+1{)}^2}=\sqrt{10},DA=\sqrt{(2-1{)}^2+(1+2{)}^2}=\sqrt{10}$$

Diagonals:

$$AC=\sqrt{(-2-2{)}^2+(-1-1{)}^2}=\sqrt{16+4}=\sqrt{20},BD=\sqrt{(1+1{)}^2+(-2-2{)}^2}=\sqrt{4+16}=\sqrt{20}$$

All four sides are equal (so ABCD is a rhombus) and the diagonals are also equal, so every angle is 90°. A rhombus with equal diagonals is a square. (Check one angle: $A{B}^2+B{C}^2=10+10=20=A{C}^2$, so ∠B = 90° by the converse of the Baudhayana-Pythagoras theorem.)

$$\text{Area}=(\text{side}{)}^2=(\sqrt{10}{)}^2=10\text{square units}$$

Yes, ABCD is a square: all sides are √10 and the diagonals are both √20. Area = 10 square units.