Exercise Set 1.1
(i) From the figure, $D_1$ lies on the x-axis at $x = 8$.
Distance from left wall (y-axis) = 8 ft
Distance from x-axis = 0 ft
(ii)
$$D_1 = (8, 0)$$
(iii)
$$\text{Width of door} = 11.5 – 8 = 3.5 \text{ ft}$$
A standard room door is about 3 ft wide, so 3.5 ft is comfortable. A wheelchair user can enter easily.
(iv)
$$\text{Bathroom door width} = 4 – 1.5 = 2.5 \text{ ft}$$
Since $2.5 < 3.5$, the bathroom door is narrower than the room door.
Exercise Set 1.2
Question 1 — Study table with feet at $(8,9)$, $(11,9)$, $(11,7)$
(i) The table is a rectangle. The fourth foot completes the rectangle:
$$\text{Fourth foot} = (8, 7)$$
(ii) The point $(8,7)$ is in the bedroom, clear of the wardrobe. It is a reasonable spot.
(iii)
$$\text{Width} = 11 – 8 = 3 \text{ ft}$$
$$\text{Length} = 9 – 7 = 2 \text{ ft}$$
Height cannot be found from a 2D floor map.
Question 2 — Bathroom door
The door is 2.5 ft wide. When it swings 90 degrees into the bedroom, the tip reaches up to $x = 2.5$. The wardrobe left edge is at $x = 3$. Since $2.5 < 3$, the door will not hit the wardrobe. If the door is made wider, a sliding door or outward opening door should be used.
Question 3 — Reiaan’s bathroom
(i) Four corners of the bathroom:
$$O = (0, 0), \quad F = (0, 10), \quad R = (-6, 10), \quad P = (-6, 0)$$
(ii) The showering area SHWR is a rectangle. Coordinates:
$$S = (-6, 6), \quad H = (-6, 9), \quad W = (-3, 9), \quad R = (-3, 6)$$
(iii) Washbasin $3 \times 2$ ft, one possible placement:
$$(-3, 8), \quad (-3, 10), \quad (0, 10), \quad (0, 8)$$
Toilet $2 \times 3$ ft, one possible placement:
$$(-6, 0), \quad (-6, 3), \quad (-4, 3), \quad (-4, 0)$$
Question 4 — Other rooms
(i) Dining room corners:
$$P(-6, 0), \quad A(12, 0), \quad (12, -15), \quad (-6, -15)$$
(ii) Centre of dining room:
$$x_c = \frac{-6 + 12}{2} = 3$$
$$y_c = \frac{0 + (-15)}{2} = -7.5$$
Dining table $5 \times 3$ ft centred at $(3, -7.5)$, feet at:
$$(0.5, -6), \quad (5.5, -6), \quad (5.5, -9), \quad (0.5, -9)$$
Section 1.4 — Distance Formula Worked Examples
For $A(3,4)$, $D(7,1)$, $M(9,6)$:
$$AD = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units}$$
$$DM = \sqrt{(9-7)^2 + (6-1)^2} = \sqrt{4 + 25} = \sqrt{29} \text{ units}$$
$$MA = \sqrt{(3-9)^2 + (4-6)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units}$$
End of Chapter Exercises
Question 1
The axes intersect at the origin:
$$x\text{-coordinate} = 0, \quad y\text{-coordinate} = 0$$
Question 2
Point $H$ on the vertical line through $W(-5, y)$:
$$H = (-5, y) \text{ for any value of } y$$
If $y > 0$, H is in Quadrant II. If $y < 0$, H is in Quadrant III. If $y = 0$, H is on the x-axis.
Question 3
Points $R(3,0)$, $A(0,-2)$, $M(-5,-2)$, $P(-5,2)$
(i) Side $AM$ is horizontal. Side $MP$ is vertical. Therefore:
$$AM \perp MP$$
(ii) Side $AM$ from $(0,-2)$ to $(-5,-2)$ is parallel to the x-axis.
(iii) $M(-5,-2)$ and $P(-5,2)$ have the same x-coordinate and opposite y-values. They are mirror images in the x-axis.
Question 4
Triangle with $I(0,0)$, $N(5,0)$, $Z(5,-6)$:
$$IZ = \sqrt{(5-0)^2 + (-6-0)^2} = \sqrt{25 + 36} = \sqrt{61} \text{ units}$$
$$ZN = \sqrt{(5-5)^2 + (0+6)^2} = \sqrt{0 + 36} = 6 \text{ units}$$
$$IN = \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{25} = 5 \text{ units}$$
Verification using Pythagoras:
$$IN^2 + ZN^2 = 25 + 36 = 61 = IZ^2$$
Question 5
Without negative numbers, only Quadrant I (where $x \geq 0$ and $y \geq 0$) can be used. Points in Quadrants II, III and IV cannot be located. So the system cannot locate all points on a 2D plane.
Question 6
Check if $M(-3,-4)$, $A(0,0)$, $G(6,8)$ are collinear by comparing slopes:
$$\text{Slope of } MA = \frac{0-(-4)}{0-(-3)} = \frac{4}{3}$$
$$\text{Slope of } AG = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$$
Both slopes are equal. The three points are collinear.
Question 7
Check if $R(-5,-1)$, $B(-2,-5)$, $C(4,-12)$ are collinear:
$$\text{Slope of } RB = \frac{-5-(-1)}{-2-(-5)} = \frac{-4}{3}$$
$$\text{Slope of } BC = \frac{-12-(-5)}{4-(-2)} = \frac{-7}{6}$$
$$\frac{-4}{3} \neq \frac{-7}{6}$$
The points are not collinear.
Question 8
(i) Right-angled isosceles triangle using $O(0,0)$, $A(4,0)$, $B(0,4)$:
$$OA = 4, \quad OB = 4, \quad AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}$$
Right angle is at $O$ and $OA = OB$, so this is a right-angled isosceles triangle.
(ii) Using $O(0,0)$, $C(-3,-4)$ in Quadrant III, $D(3,-4)$ in Quadrant IV:
$$OC = \sqrt{9 + 16} = 5, \quad OD = \sqrt{9 + 16} = 5$$
Since $OC = OD$, triangle $OCD$ is isosceles.
Question 9 — Midpoint Table
For $S(-3,0)$, $M(0,0)$, $T(3,0)$:
$$\frac{-3+3}{2} = 0 \quad \text{and} \quad \frac{0+0}{2} = 0$$
M is the midpoint. Yes.
For $S(2,3)$, $M(3,4)$, $T(4,5)$:
$$\frac{2+4}{2} = 3 \quad \text{and} \quad \frac{3+5}{2} = 4$$
M is the midpoint. Yes.
For $S(0,0)$, $M(0,5)$, $T(0,-10)$:
$$\frac{0+(-10)}{2} = -5 \neq 5$$
M is not the midpoint. No.
For $S(-8,7)$, $M(0,-2)$, $T(6,-3)$:
$$\frac{-8+6}{2} = -1 \neq 0$$
M is not the midpoint. No.
The midpoint formula is:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
Question 10
$M(-7,1)$ is midpoint of $A(3,-4)$ and $B(x,y)$:
$$\frac{3 + x}{2} = -7 \implies 3 + x = -14 \implies x = -17$$
$$\frac{-4 + y}{2} = 1 \implies -4 + y = 2 \implies y = 6$$
$$B = (-17, 6)$$
Question 11
$A(4,7)$ and $B(16,-2)$. Point $P$ divides $AB$ in ratio $1:2$ from $A$:
$$P = \left(\frac{2 \times 4 + 1 \times 16}{3}, \frac{2 \times 7 + 1 \times (-2)}{3}\right) = \left(\frac{24}{3}, \frac{12}{3}\right) = (8, 4)$$
Point $Q$ divides $AB$ in ratio $2:1$ from $A$:
$$Q = \left(\frac{1 \times 4 + 2 \times 16}{3}, \frac{1 \times 7 + 2 \times (-2)}{3}\right) = \left(\frac{36}{3}, \frac{3}{3}\right) = (12, 1)$$
$$P = (8, 4), \quad Q = (12, 1)$$
Question 12
(i) Distance from origin to each point:
$$OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$$
$$OB = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$$
$$OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$$
All three equal $\sqrt{65}$, so A, B, C lie on circle $K$ with centre $O$ and:
$$\text{Radius} = \sqrt{65} \approx 8.06 \text{ units}$$
(ii)
$$OD = \sqrt{(-5)^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$$
Since $\sqrt{61} < \sqrt{65}$, point $D$ lies inside circle $K$.
$$OE = \sqrt{0^2 + 9^2} = 9$$
Since $9 > \sqrt{65} \approx 8.06$, point $E$ lies outside circle $K$.
Question 13
Midpoints $D(5,1)$, $E(6,5)$, $F(0,3)$ of sides $BC$, $CA$, $AB$:
$$A = E + F – D = (6+0-5, 5+3-1) = (1, 7)$$
$$B = D + F – E = (5+0-6, 1+3-5) = (-1, -1)$$
$$C = D + E – F = (5+6-0, 1+5-3) = (11, 3)$$
$$A = (1, 7), \quad B = (-1, -1), \quad C = (11, 3)$$
Question 14
(i) Draw a $10 \times 10$ grid with scale $1 \text{ cm} = 200 \text{ m}$. The two main roads are the axes. Streets are numbered 1 to 5 on each side in both directions.
(ii)
(a) Street intersection $(4,3)$ — exactly 1 such intersection.
(b) Street intersection $(3,4)$ — exactly 1 such intersection.
Note: $(4,3)$ and $(3,4)$ are different intersections, just as coordinates are ordered pairs.
Question 15
(i) Check Circle A, centre $A(100,150)$, radius $= 80$:
$$100 – 80 = 20 > 0, \quad 100 + 80 = 180 < 800$$
$$150 – 80 = 70 > 0, \quad 150 + 80 = 230 < 600$$
Circle A lies entirely inside the screen.
Check Circle B, centre $B(250,230)$, radius $= 100$:
$$250 – 100 = 150 > 0, \quad 250 + 100 = 350 < 800$$
$$230 – 100 = 130 > 0, \quad 230 + 100 = 330 < 600$$
Circle B lies entirely inside the screen. Neither circle lies outside.
(ii) Distance between centres:
$$AB = \sqrt{(250-100)^2 + (230-150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170 \text{ px}$$
Sum of radii $= 80 + 100 = 180$ px
Difference of radii $= 100 – 80 = 20$ px
Since $20 < 170 < 180$, that is $|r_1 – r_2| < AB < r_1 + r_2$, the two circles intersect at two points.
Question 16
Points $A(2,1)$, $B(-1,2)$, $C(-2,-1)$, $D(1,-2)$
Step 1 — Side lengths:
$$AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9+1} = \sqrt{10}$$
$$BC = \sqrt{(-2+1)^2 + (-1-2)^2} = \sqrt{1+9} = \sqrt{10}$$
$$CD = \sqrt{(1+2)^2 + (-2+1)^2} = \sqrt{9+1} = \sqrt{10}$$
$$DA = \sqrt{(2-1)^2 + (1+2)^2} = \sqrt{1+9} = \sqrt{10}$$
All four sides are equal.
Step 2 — Diagonals:
$$AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16+4} = \sqrt{20}$$
$$BD = \sqrt{(1+1)^2 + (-2-2)^2} = \sqrt{4+16} = \sqrt{20}$$
Both diagonals are equal.
Step 3 — Midpoints of diagonals:
$$\text{Midpoint of } AC = \left(\frac{2+(-2)}{2}, \frac{1+(-1)}{2}\right) = (0, 0)$$
$$\text{Midpoint of } BD = \left(\frac{-1+1}{2}, \frac{2+(-2)}{2}\right) = (0, 0)$$
The diagonals bisect each other at the origin. Since all sides are equal, diagonals are equal, and they bisect each other, ABCD is a square.
$$\text{Area} = (\sqrt{10})^2 = 10 \text{ square units}$$