Orienting Yourself: The Use of Coordinates

Solution

From Fig. 1.3, we can read that point $D_1$ is located at approximately $(8, 0)$ on the x-axis (the bottom wall of the room).

Distance from the left wall (y-axis): The x-coordinate of $D_1$ tells us how far it is from the y-axis. Reading from the figure, $D_1$ appears at x = 8. So the door is 8 feet from the left wall.

Distance from the x-axis: Since the door is along the bottom wall of the room, which lies on the x-axis itself, the door is at y = 0. The distance from the x-axis is 0 feet (the door is on the x-axis).

Solution

From the figure, $D_1$ lies on the x-axis. Reading the x-value: $D_1$ is at x = 8 (approximately, reading the grid carefully from the figure where $W_2$ is at x = 8 and $D_1$ is just to its right near x = 8).

Since $D_1$ is on the x-axis, its y-coordinate = 0.

Solution

Both $D_1$ and $R_1$ lie on the x-axis (y = 0). Width of the door = difference in x-coordinates:

Standard room doors are about 2.5 to 3 feet (30–36 inches) wide. A 3.5-foot door is wider than standard, which is quite comfortable.

Wheelchair access: A standard wheelchair requires at least 32–36 inches (about 2.7–3 feet) of clear width. Since the door is 3.5 feet wide, yes, a person in a wheelchair should be able to enter comfortably.

Solution

Both $B_1$ and $B_2$ lie on the y-axis (x = 0). The bathroom door runs vertically (along the left wall).

Room door width = 3.5 feet (from Q iii). Since $2.5 \lt 3.5$, the bathroom door is narrower.

Solution

Let the three given feet be $P_1(8,9)$, $P_2(11,9)$, $P_3(11,7)$. Since this is a rectangle, opposite sides are parallel and equal.

(i) Fourth foot: The fourth vertex of a rectangle is found by noting that diagonals bisect each other, OR by observing the pattern: $P_1$ and $P_2$ share y = 9 (horizontal side). $P_2$ and $P_3$ share x = 11 (vertical side). So the fourth foot must share y = 7 with $P_1$‘s row and x = 8 with $P_1$‘s column.

(ii) Is this a good spot? The table occupies x from 8 to 11 and y from 7 to 9. The bedroom is 12 ft × 10 ft (from O to A and C). The table is well within the room, away from the bed and wardrobe area. It’s near the right wall, which is fine for a study corner. Yes, it is a reasonable spot.

(iii) Dimensions: Width = $|11 - 8| = 3$ ft. Length = $|9 - 7| = 2$ ft. Height cannot be determined from a 2-D floor map — the map only shows the floor view, not the vertical dimension.

Solution

$B_1 = (0, 1.5)$ and $B_2 = (0, 4)$. The door is 2.5 ft wide, hinged at $B_1(0, 1.5)$ and swinging into the bedroom (positive x direction). When fully open, the free end of the door sweeps through a quarter-circle of radius 2.5 ft centred at $B_1$.

The wardrobe ($W_1W_2W_3W_4$) has its closest corner at approximately $W_1(3, 0)$ and $W_4(3, 2)$. The door’s arc, starting at $B_2(0, 4)$ and swinging right, would pass through points like $(2.5, 1.5)$ when horizontal. This is close to the wardrobe but may just avoid it at its current width.

If the door is made wider (say, 3 ft), the arc extends further into the room, and the risk of hitting the wardrobe increases significantly. Suggestions: (a) Make the door open outward into the bathroom, (b) Use a sliding door, or (c) Move the wardrobe further from the bathroom wall.

Solution

From Fig. 1.5, the bathroom occupies a region to the left of the y-axis and above the x-axis. Reading the coordinates:

(i) Bathroom corners:
$O = (0, 0)$ (origin, bottom-right of bathroom)
$F = (0, 10)$ (top-right of bathroom, on y-axis)
$R = (-6, 10)$ (top-left of bathroom)
$P = (-6, 0)$ (bottom-left of bathroom)
The bathroom is 6 ft wide and 10 ft tall — matching the “6 ft × 9 ft” label (approximately).

(ii) Showering area SHWR: This appears as a square/rectangular region in the upper portion of the bathroom. From the figure, approximate corners are: $S(-6, 6)$, $H(-6, 10)$, $W(-3, 10)$, $R(-3, 6)$. Shape: Rectangle (4 ft wide, 4 ft tall approximately).

(iii) Suggested placement:
Washbasin (3 ft × 2 ft): corners at $(-6, 0)$, $(-3, 0)$, $(-3, 2)$, $(-6, 2)$ — along the bottom-left wall.
Toilet (2 ft × 3 ft): corners at $(-3, 0)$, $(-1, 0)$, $(-1, 3)$, $(-3, 3)$ — next to washbasin.

Solution

From Fig. 1.5, $P = (-6, 0)$ and $A = (12, 0)$. The length $PA = 12 - (-6) = 18$ ft ✓. The width of 15 ft extends downward (in the negative y direction) since the bedroom occupies the positive-y region.

(i) Dining room corners:
$P = (-6, 0)$, $A = (12, 0)$, and two corners directly below:
$A' = (12, -15)$, $P' = (-6, -15)$

(ii) Dining table at the centre: Centre of the dining room = midpoint of the diagonal = $\left(\frac{-6+12}{2}, \frac{0+(-15)}{2}\right) = (3, -7.5)$

The table is 5 ft × 3 ft. Placing it centred at (3, –7.5), its feet are at: $\left(3 \pm 2.5,\ -7.5 \pm 1.5\right)$, i.e., at $(0.5, -6)$, $(5.5, -6)$, $(5.5, -9)$, $(0.5, -9)$.

Solution

The two axes (x-axis and y-axis) intersect at the origin O. The origin is the reference point for both axes — it is zero distance from both axes.

Solution

A line parallel to the y-axis is a vertical line. Any point on the vertical line through W has the same x-coordinate as W, i.e., x = –5. The y-coordinate of H can be any real number.

So H = (–5, y) for any value of y. Depending on 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 (no quadrant).

Solution

Let us think before plotting. Shape RAMP has vertices R(3,0) → A(0,–2) → M(–5,–2) → P(–5,2) → back to R.

(i) Perpendicular sides: Side MP goes from M(–5,–2) to P(–5,2) — this is a vertical segment (x is constant). Side AM goes from A(0,–2) to M(–5,–2) — this is a horizontal segment (y is constant). A vertical and a horizontal line are perpendicular. So MP ⊥ AM.

(ii) Side parallel to an axis: AM is horizontal (both points have y = –2), so AM is parallel to the x-axis. MP is vertical (both have x = –5), so MP is parallel to the y-axis.

(iii) Mirror images: M(–5, –2) and P(–5, 2): same x-coordinate, y-values are negatives of each other. They are mirror images in the x-axis.

Solution

One natural approach: Use Z(5,–6), and choose I and N such that angle IZN = 90°. A simple choice: I = (5, 0) directly above Z on the y = 0 line, and N = (0, –6) to the left of Z on the same horizontal.

Note: Answers will vary depending on which I and N are chosen — the question says “Answers may differ from person to person.”

Solution

Without negative numbers, we could only use non-negative coordinates. The coordinate system would only cover the region where both x ≥ 0 and y ≥ 0 — that is, only Quadrant I (and the positive parts of the axes). This covers only one-quarter of the entire 2-D plane.

Points in Quadrants II, III, and IV would be completely unreachable. So no, such a system cannot locate all points in a 2-D plane. This is exactly why Brahmagupta’s formalisation of negative numbers was so crucial — it enabled the four-quadrant Cartesian plane we use today.

Solution

Method (using distance): Three points are collinear if the sum of two distances equals the third. Calculate MA, AG, and MG.

Check: $MA + AG = 5 + 10 = 15 = MG$ ✓. Since the distances add up perfectly, A lies exactly between M and G on a straight line. The three points are collinear.

Alternative method: Check if the slope (rise ÷ run) between each pair is the same. Slope MA = (0–(–4))/(0–(–3)) = 4/3. Slope AG = (8–0)/(6–0) = 8/6 = 4/3. Same slope ✓.

Solution

Use the slope method. If all three slopes are equal, the points are collinear.

$\dfrac{-4}{3} \ne \dfrac{-7}{6}$, so the slopes are not equal. The three points are not collinear — they do not lie on the same straight line.

Solution

M is the midpoint of ST if and only if M = $\left(\frac{x_S+x_T}{2}, \frac{y_S+y_T}{2}\right)$

Solution

Using the midpoint formula:

Solution

If P and Q trisect AB, then AP = PQ = QB = one-third of AB. So P is the midpoint of A and the midpoint of AB, and Q is the midpoint of P and B.

Step 1: Find the midpoint M of AB:

Step 2: P = midpoint of A and M:

Step 3: Q = midpoint of M and B:

Solution

A point lies on a circle centred at O if its distance from O equals the radius. Compute OA, OB, OC:

All three distances equal $\sqrt{65}$, so A, B, C all lie on circle K. Radius = $\sqrt{65} \approx 8.06$ units.

(ii)

$\sqrt{61} \lt \sqrt{65}$ → D is inside K.   $9 \approx 9 \gt \sqrt{65} \approx 8.06$ → E is outside K.

Solution

With 10 streets in each direction, the N–S streets are numbered 1 through 10, and E–W streets are numbered 1 through 10. A street intersection (a, b) refers to the crossing of the a-th N–S street and the b-th E–W street.

(a) How many intersections can be called (4, 3)? There is exactly one 4th N–S street and one 3rd E–W street. They cross at exactly one point. Answer: 1 intersection.

(b) How many intersections can be called (3, 4)? Similarly, there is one 3rd N–S street and one 4th E–W street. Their crossing is exactly one point — a different intersection from (4,3). Answer: 1 intersection.

This illustrates that ordered pairs matter: (4,3) ≠ (3,4) just as coordinates are ordered.

Solution

(i) Circle at A(100,150), radius 80: The circle extends from x = 100–80 = 20 to x = 100+80 = 180, and y = 150–80 = 70 to y = 150+80 = 230. All within [0,800] × [0,600]. Circle A stays within the screen.

Circle at B(250,230), radius 100: x: 250–100=150 to 250+100=350 ✓. y: 230–100=130 to 230+100=330 ✓. All within bounds. Circle B also stays within the screen.

(ii) Do circles intersect? Find the distance between centres A and B:

Sum of radii = 80 + 100 = 180. Difference of radii = 100 – 80 = 20. Since $20 \lt 170 \lt 180$ (distance between centres is between the difference and sum of radii), the circles intersect at two points.

Solution

Find all four side lengths:

All four sides are equal ($\sqrt{10}$). Now check a diagonal:

For a square with side $\sqrt{10}$, diagonal = $\sqrt{10} \times \sqrt{2} = \sqrt{20}$ ✓. So all sides are equal and diagonals confirm right angles. ABCD is a square.

Area = side² = $(\sqrt{10})^2 = \mathbf{10}$ square units.