Chapter 1: Orienting Yourself: The Use of Coordinates

Standard Door Widths:

• The standard width for a room door (interior door) is typically 0.9 metres (90 cm) in India.
• For main entrance doors, the standard width can be 1.0 to 1.2 metres.
• Bathroom doors are usually narrower, around 0.7 to 0.8 metres (70–80 cm).

Activity: Measure the width of doors in your home and school using a measuring tape and compare them with these standard values.

You will likely find that most doors in homes and schools are about 0.9 m wide.

Wheelchair Accessibility of School Doors:

For a door to be accessible to a wheelchair user:

• The minimum recommended door width is 0.9 metres (90 cm).
• There should also be enough space on either side to manoeuvre the wheelchair.

Checking school doors:

• Most standard wheelchair widths are about 0.6–0.7 metres.
• A door of at least 0.9 m allows a wheelchair to pass through comfortably.

Conclusion: Students should measure the doors in their school. If the doors are at least 0.9 m wide and have no steps or barriers in front of them, they are suitable for wheelchair users. Many older school buildings may have narrow doors (less than 0.9 m) and steps, which make them difficult for wheelchair users.

(This is an activity-based reflective question. Answers may vary based on the student's school.)

Understanding the y-axis:

The y-axis is the vertical axis. Any point on the y-axis has no horizontal distance from itself — it is directly on the vertical line.

Since the x-coordinate measures the perpendicular distance of a point from the y-axis:

• A point ON the y-axis has zero horizontal distance from the y-axis.

Therefore, the x-coordinate of any point on the y-axis is 0.

For example: Points like (0, 2), (0, –3), (0, 5) all lie on the y-axis and all have x-coordinate = 0.

Understanding the x-axis:

Yes! There is a similar generalisation for points on the x-axis.

The y-coordinate measures the perpendicular distance of a point from the x-axis.

• A point lying ON the x-axis has zero vertical distance from the x-axis.

Therefore, the y-coordinate of any point on the x-axis is 0.

For example: Points like (3, 0), (–4, 0), (7, 0) all lie on the x-axis and all have y-coordinate = 0.

Summary:

• Point on y-axis → x-coordinate = 0, written as (0, y)
• Point on x-axis → y-coordinate = 0, written as (x, 0)

When do P(x, y) and Q(y, x) coincide?

Two points coincide (are the same point) only when their coordinates are exactly equal.

For P(x, y) and Q(y, x) to be the same point:

• x-coordinate of P = x-coordinate of Q → x = y
• y-coordinate of P = y-coordinate of Q → y = x

Both conditions give the same requirement: x = y

Conclusion: Point Q(y, x) coincides with point P(x, y) if and only if x = y.

Example:

• If x = 3, y = 3: P = (3, 3) and Q = (3, 3) → they coincide ✓
• If x = 2, y = 5: P = (2, 5) and Q = (5, 2) → they do NOT coincide ✗

So, yes, Q(y, x) can coincide with P(x, y), but only when x equals y.

Verifying the claim:

The claim states two things:

1. If x ≠ y, then (x, y) ≠ (y, x)
2. (x, y) = (y, x) if and only if x = y

Proof of Claim 1: If x ≠ y

• The point (x, y) has x-coordinate = x and y-coordinate = y.
• The point (y, x) has x-coordinate = y and y-coordinate = x.
• Since x ≠ y, the x-coordinates are different (x ≠ y), so the points are different.
• Therefore, (x, y) ≠ (y, x) when x ≠ y. ✓

Proof of Claim 2: (x, y) = (y, x) only if x = y

• For (x, y) = (y, x), we need:
  • x-coordinates equal: x = y
  • y-coordinates equal: y = x
• Both give the same condition: x = y
• So (x, y) = (y, x) ↔ x = y. ✓

Example:

• x = 4, y = 7: (4, 7) ≠ (7, 4) since 4 ≠ 7 ✓
• x = 5, y = 5: (5, 5) = (5, 5) since 5 = 5 ✓

The claim is TRUE.

Understanding the figure:

In Fig. 1.3, the room has corners O, A, B, C with O at the origin. The y-axis represents the left wall and the x-axis represents the bottom wall.

From the figure, point D1 (the starting point of the door) is located at coordinates (9, 0).

Distance from the left wall (y-axis): The x-coordinate of D1 gives its horizontal distance from the y-axis.

• Distance from y-axis = x-coordinate of D1 = 9 units (i.e., 9 metres, assuming each unit = 1 metre)

Distance from the x-axis (bottom wall): The door D1R1 lies along the x-axis (the bottom wall), so:

• Distance from x-axis = y-coordinate of D1 = 0 units

The door is 9 metres from the left wall and lies right on the x-axis (bottom wall), so its distance from the x-axis is 0.

Finding the coordinates of D1:

From the figure (Fig. 1.3), the door D1R1 is located along the x-axis (the bottom wall of the room).

• The x-coordinate of D1 = 9 (its horizontal distance from the y-axis)
• The y-coordinate of D1 = 0 (since it lies on the x-axis)

Therefore, the coordinates of D1 are (9, 0).

Calculating the width of the door:

The door goes from D1 to R1 along the x-axis.

• Coordinates of D1 = (9, 0)
• Coordinates of R1 = (11.5, 0)

Width of the door = x-coordinate of R1 − x-coordinate of D1 = 11.5 − 9 = 2.5 units (i.e., 2.5 metres)

Is this a comfortable width?

A standard room door is typically about 0.9 metres (90 cm) wide. A width of 2.5 metres is much wider than a standard door — this seems unusually wide for a regular room door. (Note: It is possible the figure uses a scale where 1 unit ≠ 1 metre. If the scale is, say, 1 unit = 0.4 m, then 2.5 units = 1 metre, which is a reasonable door width.)

Assuming the scale gives a realistic door width of about 0.9–1.0 metres, this would be a comfortable width.

For a wheelchair user: The recommended minimum door width for wheelchair access is about 0.9 metres (90 cm). A door of approximately 0.9 m or wider would allow a wheelchair user to enter comfortably. If the actual width comes out to about 0.9 m or more (depending on the scale used), then yes, a wheelchair user can enter easily.

Calculating the width of the bathroom door:

The bathroom door goes from B1 to B2 along the y-axis.

• Coordinates of B1 = (0, 1.5)
• Coordinates of B2 = (0, 4)

Width of the bathroom door = y-coordinate of B2 − y-coordinate of B1 = 4 − 1.5 = 2.5 units

Width of the room door (from question iii): = 11.5 − 9 = 2.5 units

Comparison: Both the bathroom door and the room door have a width of 2.5 units.

Therefore, the bathroom door and the room door are equal in width — neither is narrower nor wider than the other.

(Note: Typically in real life, bathroom doors are narrower than room doors. Students can observe and compare with actual measurements in their homes.)

Understanding the problem: Three corners of a rectangle are given. We need to find the fourth corner.

Given three corners: A = (8, 9), B = (11, 9), C = (11, 7)

Part (i): Finding the fourth foot

Step 1: Observe the pattern of coordinates.

• A = (8, 9) and B = (11, 9) share the same y-coordinate → AB is a horizontal side.
• B = (11, 9) and C = (11, 7) share the same x-coordinate → BC is a vertical side.

Step 2: For a rectangle, the fourth corner D must complete the shape.

• D must have the same x-coordinate as A (x = 8) and the same y-coordinate as C (y = 7).
• Therefore, D = (8, 7)

Part (ii): Is this a good spot?

Looking at the floor plan (Fig. 1.5), the region around coordinates (8–11, 7–9) falls in Reiaan's room. This appears to be an open area in the room, so yes, it is a good spot for the study table — it does not overlap with other furniture like the bed or wardrobe.

Part (iii): Width, Length, and Height

Step 1: Find the length (longer side).

• Horizontal distance = x-coordinate of B − x-coordinate of A = 11 − 8 = 3 ft

Step 2: Find the width (shorter side).

• Vertical distance = y-coordinate of A − y-coordinate of C = 9 − 7 = 2 ft

So, the table is 3 ft long and 2 ft wide.

Step 3: Height of the table.

• The coordinates are only 2-dimensional (x and y), so the height cannot be determined from the floor plan. Height is a third dimension not represented on this graph.

Understanding the problem: When a door opens, it sweeps an arc. The hinge is fixed at B1, and the door rotates around it. We need to check if this arc reaches the wardrobe.

Step 1: Identify the hinge position B1 and the door width from Fig. 1.5. From the figure, B1 is located at approximately (−1, 0) (on the boundary between the bathroom and bedroom). The door swings into the bedroom.

Step 2: Determine the arc swept by the door. The door swings in a quarter-circle (90°) arc. The radius of this arc equals the width of the door (approximately 2–3 ft as shown in the figure).

Step 3: Check if the arc overlaps with the wardrobe. The wardrobe in the figure is placed near the wall in the bedroom. If the door swings fully open, depending on the door width, the arc may reach the wardrobe's position.

Conclusion:

• If the door width is about 2 ft, it may just clear the wardrobe.
• If the door is made wider, the arc increases and it will likely hit the wardrobe.

Suggested changes if the door is made wider:

1. Move the wardrobe further away from the door's swing path.
2. Change the door to open outward (into the bathroom/corridor) instead of into the bedroom.
3. Install a sliding door instead of a hinged door to avoid any collision.
4. Rehang the door with the hinge on the other side so it swings in the opposite direction away from the wardrobe.

Reading coordinates from Fig. 1.5:

Using the graph (scale: 1 cm = 1 unit = 1 ft), locate the four corners of the bathroom.

From the figure, the bathroom appears to occupy the region in the lower-left portion of the floor plan:

(Note: Exact coordinates depend on the actual figure provided in Fig. 1.5. Students should read these directly from their graph sheet.)

Step 1: Locate corner O where two walls of the bathroom meet near the origin area. Step 2: Move along the x-axis and y-axis to find the other three corners F, R, P. Step 3: Record the (x, y) coordinate pair for each corner by reading the gridlines.

Result: The bathroom is a rectangle with the four corners at the coordinates read from the graph.

Step 1: Identify the showering area SHWR on the floor plan. From Fig. 1.5, the showering area SHWR is a region within the bathroom marked with the labels S, H, W, R at its four corners.

Step 2: Determine the shape. Since it is labeled with four corners and appears to have equal sides or right angles on the plan, the showering area SHWR is a square (or rectangle).

Step 3: Read coordinates from the graph. Using the floor plan grid:

(Students must read exact values from their own copy of Fig. 1.5)

Step 4: Verify the shape.

• SH: horizontal segment (same y-coordinate)
• HW: vertical segment (same x-coordinate)
• WR: horizontal segment (same y-coordinate)
• RS: vertical segment (same x-coordinate) → All angles are 90°, confirming it is a rectangle (possibly a square if length = width).

Conclusion: The showering area SHWR is a rectangle (or square) shape.

Step 1: Find available space in the bathroom. The bathroom has corners approximately at O(−1, 0), F(−1, −6), R(−7, −6), P(−7, 0). The showering area already occupies part of this space.

Placing the Washbasin (3 ft × 2 ft): A suitable location is near the wall along the top (y = 0 to y = −2) and to the right side of the bathroom.

Coordinates of washbasin corners (example placement):

• (−1, 0), (−4, 0), (−4, −2), (−1, −2)

This gives:

• Width along x-axis: −1 to −4 = 3 ft ✓
• Width along y-axis: 0 to −2 = 2 ft ✓

Placing the Toilet (2 ft × 3 ft): A suitable location is adjacent to the washbasin or along another wall.

Coordinates of toilet corners (example placement):

• (−4, 0), (−6, 0), (−6, −3), (−4, −3)

This gives:

• Width along x-axis: −4 to −6 = 2 ft ✓
• Width along y-axis: 0 to −3 = 3 ft ✓

Note: Students should ensure these spaces do not overlap with each other or with the showering area SHWR on their own copy of Fig. 1.5. Exact coordinates may vary based on where they choose to place the fixtures.

Step 1: Identify point P from the floor plan. From Fig. 1.5, point P is a corner shared with the bathroom/bedroom area. Based on the figure, P appears to be at approximately (−7, 0).

Step 2: Determine the direction of the length. The length of the dining room (18 ft) extends from P to A along the x-axis direction (horizontally).

• P = (−7, 0)
• A = P + 18 units along x-axis = (−7 + 18, 0) = (11, 0)

Step 3: Determine the width (15 ft). The width extends downward (in the negative y-direction) from the PA line:

• Width = 15 ft downward → y goes from 0 to −15

Step 4: Find all four corners.

Step 5: Sketch on graph sheet. Draw the rectangle with these four corners. Label each corner with its coordinates.

Verification:

• PA = 11 − (−7) = 18 ft ✓ (length)
• PQ = 0 − (−15) = 15 ft ✓ (width)

Step 1: Find the centre of the dining room.

Dining room corners: P(−7, 0), A(11, 0), B(11, −15), Q(−7, −15)

Centre x-coordinate = (−7 + 11) ÷ 2 = 4 ÷ 2 = 2 Centre y-coordinate = (0 + (−15)) ÷ 2 = −15 ÷ 2 = −7.5

So, centre of dining room = (2, −7.5)

Step 2: Place the 5 ft × 3 ft table centred at (2, −7.5).

The table extends:

• 5/2 = 2.5 ft on each side along the length (x-direction)
• 3/2 = 1.5 ft on each side along the width (y-direction)

Step 3: Calculate the four corner coordinates.

Step 4: Verify dimensions.

• Length: 4.5 − (−0.5) = 5 ft ✓
• Width: −6 − (−9) = 3 ft ✓
• Centre: ((−0.5 + 4.5)/2, (−6 + (−9))/2) = (2, −7.5) ✓

Result: The four feet of the dining table are at (−0.5, −6), (4.5, −6), (4.5, −9), and (−0.5, −9).

The two axes (x-axis and y-axis) meet at a special point called the Origin.

• The x-coordinate of the origin = 0
• The y-coordinate of the origin = 0

So the coordinates of the point of intersection are (0, 0).

Using the midpoint formula:

If M is the midpoint of AB, then: $M_x = \frac{A_x + B_x}{2} \quad \text{and} \quad M_y = \frac{A_y + B_y}{2}$

Given: A = (3, –4), M = (–7, 1), B = (x, y)

Finding x (x-coordinate of B): $-7 = \frac{3 + x}{2}$ $-14 = 3 + x$ $x = -14 - 3 = -17$

Finding y (y-coordinate of B): $1 = \frac{-4 + y}{2}$ $2 = -4 + y$ $y = 2 + 4 = 6$

∴ B = (–17, 6)

Verification: Midpoint of A(3,–4) and B(–17,6) = ((3–17)/2, (–4+6)/2) = (–14/2, 2/2) = (–7, 1) = M ✓

Understanding the trisection:

P and Q divide AB into three equal parts: $A \xrightarrow{1/3} P \xrightarrow{1/3} Q \xrightarrow{1/3} B$

Key insight: Q is the midpoint of PB, and P is the midpoint of AQ.

Alternative approach: P divides AB in ratio 1:2 and Q divides AB in ratio 2:1.

Step-by-step using midpoints:

Step 1: Find the midpoint M of AB: $M = \left(\frac{4+16}{2}, \frac{7+(-2)}{2}\right) = \left(\frac{20}{2}, \frac{5}{2}\right) = \left(10, 2.5\right)$

Step 2: Find P = midpoint of A and Q.

But first, note that Q is the midpoint of segment from P to B, and P is 1/3 of the way from A to B.

Using section formula approach:

Since P is 1/3 of the way from A to B: $P_x = A_x + \frac{1}{3}(B_x - A_x) = 4 + \frac{1}{3}(16-4) = 4 + \frac{12}{3} = 4 + 4 = 8$ $P_y = A_y + \frac{1}{3}(B_y - A_y) = 7 + \frac{1}{3}(-2-7) = 7 + \frac{-9}{3} = 7 - 3 = 4$ $\therefore P = (8, 4)$

Since Q is 2/3 of the way from A to B: $Q_x = A_x + \frac{2}{3}(B_x - A_x) = 4 + \frac{2}{3}(12) = 4 + 8 = 12$ $Q_y = A_y + \frac{2}{3}(B_y - A_y) = 7 + \frac{2}{3}(-9) = 7 - 6 = 1$ $\therefore Q = (12, 1)$

Verification using midpoints:

• Midpoint of AP = midpoint of (4,7) and (8,4) = (6, 5.5)
• Midpoint of PQ = midpoint of (8,4) and (12,1) = (10, 2.5)
• Midpoint of QB = midpoint of (12,1) and (16,–2) = (14, –0.5)
• AP = √((8–4)²+(4–7)²) = √(16+9) = √25 = 5
• PQ = √((12–8)²+(1–4)²) = √(16+9) = √25 = 5
• QB = √((16–12)²+(–2–1)²) = √(16+9) = √25 = 5 ✓

All three segments are equal, confirming P(8,4) and Q(12,1) are the trisection points.

Concept: A point lies on a circle with centre O and radius r if its distance from O equals r. It lies inside if distance < r, and outside if distance > r.

Distance formula from origin: distance = √(x² + y²)

(i) Finding radius and checking A, B, C:

Distance OA (from origin to A(1, –8)): $OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$

Distance OB (from origin to B(–4, 7)): $OB = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$

Distance OC (from origin to C(–7, –4)): $OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$

Since OA = OB = OC = √65, all three points are equidistant from the origin.

∴ A, B, and C all lie on circle K with centre O(0,0) and radius r = √65 ≈ 8.06 units.

(ii) Checking D(–5, 6) and E(0, 9):

Distance OD (from origin to D(–5, 6)): $OD = \sqrt{(-5)^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$

Since √61 < √65, we have OD < r. ∴ D(–5, 6) lies INSIDE circle K.

Distance OE (from origin to E(0, 9)): $OE = \sqrt{0^2 + 9^2} = \sqrt{81} = 9$

Now, r = √65 ≈ 8.06

Since 9 > √65, we have OE > r. ∴ E(0, 9) lies OUTSIDE circle K.

Setting up the problem:

Let A = (x₁, y₁), B = (x₂, y₂), C = (x₃, y₃)

The midpoints of the sides are:

• D = midpoint of BC → D(5, 1)
• E = midpoint of AC → E(6, 5)
• F = midpoint of AB → F(0, 3)

Setting up equations using midpoint formula:

From D = midpoint of BC: $\frac{x_2 + x_3}{2} = 5 \Rightarrow x_2 + x_3 = 10 \quad ...(1)$ $\frac{y_2 + y_3}{2} = 1 \Rightarrow y_2 + y_3 = 2 \quad ...(2)$

From E = midpoint of AC: $\frac{x_1 + x_3}{2} = 6 \Rightarrow x_1 + x_3 = 12 \quad ...(3)$ $\frac{y_1 + y_3}{2} = 5 \Rightarrow y_1 + y_3 = 10 \quad ...(4)$

From F = midpoint of AB: $\frac{x_1 + x_2}{2} = 0 \Rightarrow x_1 + x_2 = 0 \quad ...(5)$ $\frac{y_1 + y_2}{2} = 3 \Rightarrow y_1 + y_2 = 6 \quad ...(6)$

Solving for x-coordinates:

Adding equations (1), (3), (5): $(x_2 + x_3) + (x_1 + x_3) + (x_1 + x_2) = 10 + 12 + 0$ $2(x_1 + x_2 + x_3) = 22$ $x_1 + x_2 + x_3 = 11 \quad ...(7)$

From (7) – (1): x₁ = 11 – 10 = 1 From (7) – (3): x₂ = 11 – 12 = –1 From (7) – (5): x₃ = 11 – 0 = 11

Solving for y-coordinates:

Adding equations (2), (4), (6): $(y_2 + y_3) + (y_1 + y_3) + (y_1 + y_2) = 2 + 10 + 6$ $2(y_1 + y_2 + y_3) = 18$ $y_1 + y_2 + y_3 = 9 \quad ...(8)$

From (8) – (2): y₁ = 9 – 2 = 7 From (8) – (4): y₂ = 9 – 10 = –1 From (8) – (6): y₃ = 9 – 6 = 3

∴ A = (1, 7), B = (–1, –1), C = (11, 3)

Verification:

• Midpoint of BC = ((–1+11)/2, (–1+3)/2) = (5, 1) = D ✓
• Midpoint of AC = ((1+11)/2, (7+3)/2) = (6, 5) = E ✓
• Midpoint of AB = ((1–1)/2, (7–1)/2) = (0, 3) = F ✓

(i) Drawing the model:

• There are 10 streets running N-S and 10 streets running E-W.
• Streets are 200 m apart; using 1 cm = 200 m, they are 1 cm apart on paper.
• Draw a grid of 10 × 10 lines, giving a 9 cm × 9 cm grid (10 lines with 9 gaps between them in each direction).
• The two main roads are along the centre. Label N-S streets 1 to 10 from left to right, and E-W streets 1 to 10 from bottom to top (or vice versa).

(ii) Street intersections:

In the convention given, (m, n) means the intersection of the mth N-S street and the nth E-W street.

Important note: The numbers used are street numbers (1st, 2nd, ..., 10th), not coordinates with positive and negative values. So each street has a unique number from 1 to 10.

(a) How many intersections can be called (4, 3)?

(4, 3) means: 4th N-S street meets 3rd E-W street.

• There is exactly one 4th N-S street and exactly one 3rd E-W street.
• They meet at exactly one point.

∴ There is exactly 1 intersection called (4, 3).

(b) How many intersections can be called (3, 4)?

(3, 4) means: 3rd N-S street meets 4th E-W street.

• Again, there is exactly one 3rd N-S street and exactly one 4th E-W street.
• They meet at exactly one point.

∴ There is exactly 1 intersection called (3, 4).

Key observation: (4, 3) and (3, 4) refer to different intersections (just like (4,3) ≠ (3,4) as coordinates), showing that order matters in this notation.

Setting up the coordinate system:

• Bottom-left corner = origin (0, 0)
• Bottom-right corner = (800, 0)
• Top-left corner = (0, 600)
• Top-right corner = (800, 600)

For a circle to be fully inside the screen, every point on it must satisfy: 0 ≤ x ≤ 800 and 0 ≤ y ≤ 600.

(i) Does any part of either circle lie outside the screen?

Circle 1: Centre A(100, 150), Radius = 80

Check all four boundaries:

• Left boundary (x = 0): Distance from A to left edge = 100. Since 100 > 80, circle doesn't cross left edge. ✓
• Right boundary (x = 800): Distance from A to right edge = 800 – 100 = 700. Since 700 > 80, ✓
• Bottom boundary (y = 0): Distance from A to bottom = 150. Since 150 > 80, ✓
• Top boundary (y = 600): Distance from A to top = 600 – 150 = 450. Since 450 > 80, ✓

∴ Circle 1 lies completely INSIDE the screen.

Circle 2: Centre B(250, 230), Radius = 100

• Left boundary: Distance = 250. Since 250 > 100, ✓
• Right boundary: Distance = 800 – 250 = 550. Since 550 > 100, ✓
• Bottom boundary: Distance = 230. Since 230 > 100, ✓
• Top boundary: Distance = 600 – 230 = 370. Since 370 > 100, ✓

∴ Circle 2 also lies completely INSIDE the screen.

Neither circle lies outside the screen.

(ii) Do the two circles intersect each other?

Two circles intersect if the distance between their centres is less than the sum of their radii and greater than the absolute difference of their radii.

Distance between centres A(100, 150) and B(250, 230): $AB = \sqrt{(250-100)^2 + (230-150)^2}$ $= \sqrt{150^2 + 80^2}$ $= \sqrt{22500 + 6400}$ $= \sqrt{28900}$ $= 170 \text{ pixels}$

Sum of radii = 80 + 100 = 180 pixels

Difference of radii = |100 – 80| = 20 pixels

Check: 20 < 170 < 180

Since the distance between centres (170) is less than the sum of radii (180) and greater than the difference of radii (20):

∴ The two circles DO intersect each other (they overlap).

Step 1: Calculate all side lengths using the distance formula:

AB (from A(2,1) to B(–1,2)): $AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9+1} = \sqrt{10}$

BC (from B(–1,2) to C(–2,–1)): $BC = \sqrt{(-2-(-1))^2 + (-1-2)^2} = \sqrt{1+9} = \sqrt{10}$

CD (from C(–2,–1) to D(1,–2)): $CD = \sqrt{(1-(-2))^2 + (-2-(-1))^2} = \sqrt{9+1} = \sqrt{10}$

DA (from D(1,–2) to A(2,1)): $DA = \sqrt{(2-1)^2 + (1-(-2))^2} = \sqrt{1+9} = \sqrt{10}$

All four sides are equal: AB = BC = CD = DA = √10 ✓

Step 2: Check the diagonals (for a square, diagonals must be equal and perpendicular):

Diagonal AC (from A(2,1) to C(–2,–1)): $AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$

Diagonal BD (from B(–1,2) to D(1,–2)): $BD = \sqrt{(1-(-1))^2 + (-2-2)^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$

AC = BD = 2√5 ✓ (Diagonals are equal)

Step 3: Check if sides are perpendicular:

Slope of AB = (2–1)/(–1–2) = 1/(–3) = –1/3

Slope of BC = (–1–2)/(–2–(–1)) = –3/(–1) = 3

Product of slopes = (–1/3) × 3 = –1 ✓

Since the product of slopes of adjacent sides = –1, they are perpendicular.

∴ ABCD IS a square — all sides equal AND adjacent sides perpendicular.

Step 4: Find the area:

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

Finding coordinates of H:

A line through W that is parallel to the y-axis is a vertical line. Every point on a vertical line has the same x-coordinate.

Since W has x-coordinate = –5, every point on this vertical line also has x-coordinate = –5.

The y-coordinate of H can be any real number (it is not fixed).

So the coordinates of H are (–5, y) where y can be any real number.

Which quadrants can H lie in?

Since the x-coordinate is –5 (negative):

• If y > 0: H is in Quadrant II (negative x, positive y)
• If y < 0: H is in Quadrant III (negative x, negative y)
• If y = 0: H lies on the x-axis (not in any quadrant)

So H can lie in Quadrant II or Quadrant III (or on the x-axis, but not in Quadrants I or IV).

Setting up the quadrilateral RAMP:

The vertices in order are:

• R = (3, 0)
• A = (0, –2)
• M = (–5, –2)
• P = (–5, 2)

Sides: RA, AM, MP, PR

(i) Two sides perpendicular to each other:

Let us check side MP and side PM direction:

• Side AM: from A(0, –2) to M(–5, –2). Both y-coordinates are –2, so AM is horizontal.
• Side MP: from M(–5, –2) to P(–5, 2). Both x-coordinates are –5, so MP is vertical.

A horizontal line and a vertical line are always perpendicular.

∴ Sides AM and MP are perpendicular to each other.

(ii) One side parallel to one of the axes:

• Side AM goes from (0, –2) to (–5, –2): y-coordinate is the same (–2) → parallel to the x-axis.
• Side MP goes from (–5, –2) to (–5, 2): x-coordinate is the same (–5) → parallel to the y-axis.

So AM is parallel to the x-axis and MP is parallel to the y-axis.

(iii) Two points that are mirror images in one axis:

Compare M(–5, –2) and P(–5, 2):

• Same x-coordinate: –5
• y-coordinates are opposite in sign: –2 and +2

This means M and P are reflections of each other in the x-axis.

∴ M(–5, –2) and P(–5, 2) are mirror images of each other in the x-axis.

Verification by plotting: Plot all four points on graph paper and join them in order R→A→M→P→R. You will observe:

• AM is horizontal (parallel to x-axis)
• MP is vertical (parallel to y-axis)
• AM ⊥ MP
• M and P are symmetric about the x-axis

One convenient way to construct a right-angled triangle using Z(5, –6):

Choose the right angle at a point that shares a coordinate with Z.

Let us choose:

• Z = (5, –6) — one vertex
• I = (5, 0) — on the y-axis, directly above Z (same x-coordinate)
• N = (0, 0) — the origin

This forms triangle IZN with a right angle at I (since IZ is vertical and IN is horizontal... wait, let us check).

Actually, let us choose:

• I = (0, –6) — same y-coordinate as Z
• N = (0, 0) — origin
• Z = (5, –6)

Right angle at I = (0, –6), because:

• IZ is horizontal (both have y = –6)
• IN is vertical (both have x = 0)
• IZ ⊥ IN ✓

Calculating the three side lengths:

Side IZ (horizontal, from I(0,–6) to Z(5,–6)): $IZ = |5 - 0| = 5 \text{ units}$

Side IN (vertical, from I(0,–6) to N(0,0)): $IN = |0 - (–6)| = 6 \text{ units}$

Side ZN (hypotenuse, from Z(5,–6) to N(0,0)): $ZN = \sqrt{(5-0)^2 + (-6-0)^2} = \sqrt{25 + 36} = \sqrt{61} \approx 7.81 \text{ units}$

Verification using Pythagoras Theorem: $IZ^2 + IN^2 = 25 + 36 = 61 = ZN^2 ✓$

Note: Different students may choose different points for I and N, giving different triangles — all are valid as long as a right angle is formed.

What the system would look like without negative numbers:

If we only had non-negative numbers (0, 1, 2, 3, ...), the coordinate system would only have one quadrant — Quadrant I (where both x ≥ 0 and y ≥ 0).

The x-axis would extend only to the right of the origin, and the y-axis would extend only upward from the origin.

Would it locate all points on a 2D plane?

No, such a system would NOT be able to locate all points on a 2D plane.

Reason:

• A full 2D plane has four quadrants.
• Without negative numbers, we can only represent points in Quadrant I (where x ≥ 0 and y ≥ 0).
• Points in Quadrant II (x < 0, y > 0), Quadrant III (x < 0, y < 0), and Quadrant IV (x > 0, y < 0) cannot be represented.
• Also, points on the negative x-axis or negative y-axis cannot be located.

Conclusion: A coordinate system without negative numbers can only cover one-quarter of the plane. To locate ALL points on a 2D plane, we need negative numbers.

Method to check without plotting — using the slope:

Three points are collinear (lie on the same straight line) if the slope between any two pairs of points is equal.

Slope formula: slope = (y₂ – y₁)/(x₂ – x₁)

Slope of MA (from M(–3,–4) to A(0,0)): $\text{slope} = \frac{0-(-4)}{0-(-3)} = \frac{4}{3}$

Slope of AG (from A(0,0) to G(6,8)): $\text{slope} = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$

Both slopes are equal (4/3).

Since A is a common point and slopes MA and AG are equal, all three points lie on the same straight line.

∴ Yes, M(–3, –4), A(0, 0), and G(6, 8) are collinear.

Alternative method using ratios: Check if (y_A – y_M)/(x_A – x_M) = (y_G – y_A)/(x_G – x_A). Both give 4/3, confirming collinearity.

Applying the slope method:

Slope of RB (from R(–5,–1) to B(–2,–5)): $\text{slope} = \frac{-5-(-1)}{-2-(-5)} = \frac{-4}{3}$

Slope of BC (from B(–2,–5) to C(4,–12)): $\text{slope} = \frac{-12-(-5)}{4-(-2)} = \frac{-7}{6}$

Comparing slopes: $\frac{-4}{3} = \frac{-8}{6} \neq \frac{-7}{6}$

The slopes are not equal.

∴ R(–5, –1), B(–2, –5), and C(4, –12) are NOT collinear — they do NOT lie on the same straight line.

Verification by plotting: When you plot R(–5,–1), B(–2,–5), and C(4,–12) on graph paper and try to join them with a single straight line, you will find that all three points do not fall on the same line, confirming our calculation.

Contrast with Problem 6: M, A, G were collinear (slope = 4/3 throughout), but R, B, C are not collinear (slopes –4/3 and –7/6 are different).

(i) Right-angled isosceles triangle with origin as one vertex:

For a right-angled isosceles triangle, we need:

• The right angle at one vertex
• Two equal legs

Choose:

• O = (0, 0) — origin (right angle here)
• A = (4, 0) — on x-axis
• B = (0, 4) — on y-axis

Verification:

• OA = 4 units (along x-axis)
• OB = 4 units (along y-axis)
• OA = OB ✓ (isosceles)
• Angle AOB = 90° (axes are perpendicular) ✓ (right-angled)
• AB = √(4² + 4²) = √32 = 4√2 units (hypotenuse)

This triangle O(0,0), A(4,0), B(0,4) is a right-angled isosceles triangle.

(ii) Isosceles triangle with origin as one vertex, one vertex in Quadrant III and another in Quadrant IV:

Choose:

• O = (0, 0) — origin (third vertex)
• P = (–3, –4) — in Quadrant III
• Q = (3, –4) — in Quadrant IV

Verification:

• OP = √(9 + 16) = √25 = 5 units
• OQ = √(9 + 16) = √25 = 5 units
• OP = OQ ✓ (isosceles)
• PQ = |3 – (–3)| = 6 units (horizontal base)

Triangle O(0,0), P(–3,–4), Q(3,–4) is an isosceles triangle with one vertex in Quadrant III and another in Quadrant IV.

Note: These are suggested answers. Other valid choices of points are also acceptable.

Rule for midpoint: M is the midpoint of ST if and only if: $M_x = \frac{S_x + T_x}{2} \quad \text{and} \quad M_y = \frac{S_y + T_y}{2}$

Row 1: S(–3, 0), M(0, 0), T(3, 0)

Check x: (–3 + 3)/2 = 0/2 = 0 ✓ Check y: (0 + 0)/2 = 0 ✓

M IS the midpoint of ST. ✓

Row 2: S(2, 3), M(3, 4), T(4, 5)

Check x: (2 + 4)/2 = 6/2 = 3 ✓ Check y: (3 + 5)/2 = 8/2 = 4 ✓

M IS the midpoint of ST. ✓

Row 3: S(0, 0), M(0, 5), T(0, –10)

Check x: (0 + 0)/2 = 0 ✓ Check y: (0 + (–10))/2 = –10/2 = –5 ≠ 5 ✗

M is NOT the midpoint of ST. ✗ (The actual midpoint would be (0, –5))

Row 4: S(–8, 7), M(0, –2), T(6, –3)

Check x: (–8 + 6)/2 = –2/2 = –1 ≠ 0 ✗

M is NOT the midpoint of ST. ✗ (The actual midpoint would be (–1, 2))

Summary Table:

Connection found: When M is the midpoint of ST: $M_x = \frac{S_x + T_x}{2} \quad \text{and} \quad M_y = \frac{S_y + T_y}{2}$ The coordinates of M are the averages of the corresponding coordinates of S and T.