Orienting Yourself: The Use of Coordinates

(i) The door D₁R₁ lies on the x-axis (the bottom wall), so its distance from the x-axis is 0. Its distance from the left wall (y-axis) equals the x-coordinate of D₁ — read this value off Fig. 1.3.

(ii) Since D₁ lies on the x-axis, its coordinates are of the form (x, 0), where x is the distance you read from the figure.

(iii) Width of the door = difference of the x-coordinates:

A standard room door is about 3 feet (36 inches) wide, which is comfortable for daily use. For a wheelchair user, a clear width of at least 32–36 inches (≈ 2.7–3 ft) is recommended — so a 3-ft door allows a wheelchair to enter, though wider is easier.

(iv) The bathroom door runs along the y-axis from B₁ (0, 1.5) to B₂ (0, 4):

Since 2.5 ft is less than the room-door width (≈ 3 ft), the bathroom door is narrower than the room door.

(i) For a rectangle, opposite sides are parallel to the axes here. Three feet are at (8, 9), (11, 9) and (11, 7). The fourth foot must share x = 8 with the first point and y = 7 with the third point:

(ii) The table occupies the region from x = 8 to 11 and y = 7 to 9 — the top-right corner of the room, against the walls. This is a good spot: it leaves the middle of the room free to move around and does not block the door. (Answers may vary with reasoning.)

(iii) Length = 11 − 8 = 3 ft; width = 9 − 7 = 2 ft. The height cannot be determined — Fig. 1.5 is a map of the floor (2-D), and height is the third dimension, which a floor map does not show.

The door is hinged at B₁ (0, 1.5) and its free end is at B₂ (0, 4), so its length is 2.5 ft. When it opens into the bedroom, the free end sweeps a quarter circle of radius 2.5 ft centred at B₁ — covering all points within 2.5 ft of B₁ on the bedroom side.

Method: Check whether any corner of the wardrobe (from Fig. 1.5) lies within distance 2.5 of B₁ (0, 1.5), using the distance formula \(d = \sqrt{(x-0)^2 + (y-1.5)^2}\). If d < 2.5 for some part of the wardrobe, the door will hit it.

If the door is made wider, its sweep radius increases, so the wardrobe (or other furniture) may need to be shifted farther from B₁, or the door could be hinged at B₂ instead / made to open into the bathroom / replaced by a sliding door.

(i) Read the four corners of the bathroom rectangle directly from Fig. 1.5. Remember: a corner’s coordinates are (distance from y-axis, distance from x-axis). One corner is the origin O (0, 0).

(ii) The showering area SHWR is a rectangle (or square, if adjacent sides read equal in the figure). Write its corners in order, e.g. S, H, W, R, taking each pair of coordinates from the figure.

(iii) This part is open-ended — any placement works as long as the washbasin rectangle spans 3 units along one wall and 2 units along the perpendicular direction, the toilet spans 2 × 3, and neither overlaps the shower or the door’s swing. Example: a washbasin along the bottom wall from (2, 0) to (5, 0) has corners (2, 0), (5, 0), (5, 2), (2, 2). (Answers may vary.)

(i) Locate P and A in Fig. 1.5; the segment PA (18 ft) is one side of the dining room, and the room extends 15 ft perpendicular to PA (below the x-axis in the figure, which is why the graph sheet extends to (0, −15)). Each corner is then fixed by moving 15 units perpendicular to PA from its endpoints.

(ii) First find the centre of the dining room — the point where its diagonals meet, i.e. the midpoint of any diagonal, using \(\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)\). The 5 ft × 3 ft table centred there has its feet at centre ± 2.5 in the length direction and centre ± 1.5 in the width direction. For example, if the centre works out to (c₁, c₂) with the table’s length parallel to the x-axis, the feet are (c₁ − 2.5, c₂ − 1.5), (c₁ + 2.5, c₂ − 1.5), (c₁ + 2.5, c₂ + 1.5), (c₁ − 2.5, c₂ + 1.5).

The point of intersection of the x-axis and the y-axis is the origin O. Its x-coordinate is 0 and its y-coordinate is 0, i.e. O = (0, 0).

The line through W parallel to the y-axis is the vertical line x = −5. Every point on it has x-coordinate −5, so:

Since the x-coordinate is negative, H lies in Quadrant II when y > 0 and in Quadrant III when y < 0. (When y = 0, H is the point (−5, 0) on the x-axis itself.)

Joining in order gives sides RA, AM, MP and PR.

(i) A (0, −2) and M (−5, −2) have the same y-coordinate, so AM is horizontal (along y = −2). M (−5, −2) and P (−5, 2) have the same x-coordinate, so MP is vertical (along x = −5). Hence AM ⊥ MP.

(ii) AM is parallel to the x-axis (equally, MP is parallel to the y-axis).

(iii) M (−5, −2) and P (−5, 2) have equal x-coordinates and opposite y-coordinates, so they are mirror images of each other in the x-axis.

Plotting the four points confirms all three predictions.

Answers may vary. One example: take I (0, −6) and N (5, 0), with the right angle at Z (5, −6):

• ZI is horizontal: ZI = |5 − 0| = 5 units
• ZN is vertical: ZN = |0 − (−6)| = 6 units
• Hypotenuse IN, by the Baudhāyana–Pythagoras theorem:

Any triangle with one horizontal and one vertical side through Z is a valid construction.

Without negative numbers, both coordinates of every point would have to be positive (or zero). The coordinate system would then consist of only the first quadrant, together with the positive x-axis and positive y-axis.

No — such a system cannot locate all points of the 2-D plane. Any point to the left of the y-axis or below the x-axis (Quadrants II, III and IV) needs a negative coordinate to describe its position relative to the origin. This is precisely why Brahmagupta’s formalisation of zero and negative numbers was essential for the full four-quadrant Cartesian plane.

Method (without plotting): compute the ratio \(\frac{\text{change in } y}{\text{change in } x}\) (the slope) between pairs of points. If the slopes between M–A and A–G are equal, the three points are collinear.

The ratios are equal and both segments pass through A, so M, A and G are collinear.

Distance check (alternative): MA = \(\sqrt{3^2+4^2}\) = 5, AG = \(\sqrt{6^2+8^2}\) = 10, MG = \(\sqrt{9^2+12^2}\) = 15. Since MA + AG = MG (5 + 10 = 15), the points lie on one straight line. ✓

Compare slopes:

Since \(-\dfrac{4}{3} \neq -\dfrac{7}{6}\), the points R, B and C are not collinear — they do not lie on the same straight line.

Plotting confirms it: the point C lies slightly off the line through R and B.

Answers may vary. Example constructions:

(i) O (0, 0), A (4, 0), B (0, 4). Sides OA = OB = 4 (isosceles) and OA is horizontal while OB is vertical, so the angle at O is 90° — a right-angled isosceles triangle. Hypotenuse AB = \(\sqrt{4^2+4^2} = 4\sqrt{2}\).

(ii) O (0, 0), P (−3, −4) in Quadrant III, Q (3, −4) in Quadrant IV. Then:

So triangle OPQ is isosceles with the required vertices. ✓

M is the midpoint of ST exactly when \(M = \left(\dfrac{x_S + x_T}{2},\ \dfrac{y_S + y_T}{2}\right)\). Checking each row:

Connection: each coordinate of the midpoint is the average of the corresponding coordinates of the endpoints:

Using the midpoint relations:

Therefore B = (−17, 6).

Check: midpoint of (3, −4) and (−17, 6) = \(\left(\dfrac{3-17}{2}, \dfrac{-4+6}{2}\right)\) = (−7, 1) = M ✓

Idea: P and Q divide AB into three equal parts, so P is one-third of the way from A to B and Q is two-thirds of the way. Equivalently: P is the midpoint of A and Q, and Q is the midpoint of P and B.

Moving one-third of the change in each coordinate:

Check: midpoint of A (4, 7) and Q (12, 1) = (8, 4) = P ✓; midpoint of P (8, 4) and B (16, −2) = (12, 1) = Q ✓

(i) A point lies on a circle centred at O exactly when its distance from O equals the radius. Compute each distance:

All three points are at the same distance from O, so they lie on the circle K with centre O and radius \(\sqrt{65} \approx 8.06\) units.

(ii) Compare each point’s distance from O with \(\sqrt{65}\):

Take D as the midpoint of BC, E as the midpoint of CA, and F as the midpoint of AB. Adding the midpoint equations shows that each vertex equals the sum of the two adjacent midpoints minus the opposite midpoint:

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

(i) Draw two perpendicular main roads through the centre, then 10 streets parallel to each: since streets lie on both sides of each main road, 5 run on each side, spaced 1 cm (200 m) apart.

(ii) The label “4th N–S street” does not say which side of the N–S main road the street is on (east or west), and similarly the “3rd E–W street” can be north or south of the E–W main road. So there are 2 choices for the N–S street and 2 for the E–W street:

Note that (4, 3) and (3, 4) refer to different intersections — the order of the numbers matters. The ambiguity about the side of the main road is exactly what signed coordinates (positive and negative directions) resolve in the Cartesian system.

(i) A circle lies fully on the screen if its centre is at least one radius away from every edge (x = 0, x = 800, y = 0, y = 600).

• Circle A (centre (100, 150), r = 80): distances to the edges are 100, 700, 150 and 450 — all ≥ 80. Fully on screen.
• Circle B (centre (250, 230), r = 100): distances to the edges are 250, 550, 230 and 370 — all ≥ 100. Fully on screen.

So no part of either circle lies outside the screen.

(ii) Distance between the centres:

Sum of radii = 80 + 100 = 180; difference of radii = 20. Since

the distance between centres is less than the sum of the radii (and more than their difference), so the two circles intersect each other at two points.

Sides (by the distance formula):

All four sides are equal, so ABCD is at least a rhombus.

Diagonals:

A rhombus whose diagonals are equal is a square — so ABCD is a square. ✓

Area: