Chapter 1: Orienting Yourself: The Use of Coordinates

According to the diagram, point D₁ is located at the +8 mark on the x-axis.

Therefore, the door is 8 units away from the left wall (the y-axis).

Because the door lies exactly on the x-axis, its distance from the x-axis is 0 units

.

From Fig. 1.3, point D₁ lies on the x-axis (the bottom wall of the room).

coordinates of D₁ are (8, 0).

Width of the Door

Given:

• D₁ = (8, 0)
• R₁ = (11.5, 0)

Width of door:

$\text{Width} = x_{R_1} - x_{D_1} = 11.5 - 8 = 3.5 \text{ feet}$

The door is 3.5 feet wide.

Is this a comfortable width?

• Standard room door width = 2.5 to 3 feet (30–36 inches)
• 3.5 feet = 42 inches — slightly wider than standard, but comfortable and usable.

For a wheelchair user:

• Wheelchair-accessible door minimum = 32–36 inches (2.7–3 feet)
• 3.5 feet = 42 inches — Yes, a wheelchair user can enter comfortably. ✓

Width of bathroom door:

Given: $B_1 = (0, 1.5)$ and $B_2 = (0, 4)$ — both points lie on the y-axis.

$\text{Bathroom door width} = 4 - 1.5 = 2.5 \text{ feet}$

Width of room door (from part iii):

$\text{Room door width} = 11.5 - 8 = 3.5 \text{ feet}$

Comparison:

$2.5 \text{ ft} < 3.5 \text{ ft}$

Therefore, the bathroom door is narrower than the room door by 1 foot.

Standard Door Widths

• Standard residential room door: approximately 0.9 m (90 cm) wide.
• Standard bathroom door: approximately 0.7 m to 0.8 m wide.
• School/public building doors: typically 0.9 m to 1.0 m wide.

You can verify this by measuring doors in your home and school with a measuring tape.

Activity: Measure 3–4 doors around you and record the widths in a table.

Wheelchair Accessibility of School Doors

• A wheelchair-accessible door requires a minimum clear width of 0.9 m (90 cm), as per accessibility standards.
• Most standard school doors are about 0.9 m to 1.0 m wide, which should be sufficient.
• However, older school buildings may have narrower doors that are not wheelchair-friendly.

What to check:

• Width of the door (should be ≥ 0.9 m)
• Presence of ramps instead of steps at entrances
• Door handles that can be operated easily

Conclusion: Students should physically check their school doors and reflect on inclusivity and accessibility for differently-abled individuals.

x-coordinate of a Point on the y-axis

The y-axis is the vertical axis. Every point on the y-axis has no horizontal distance from itself — it lies exactly on the y-axis.

• The x-coordinate represents the perpendicular distance from the y-axis, measured along the x-axis.
• Since a point on the y-axis has zero distance from the y-axis:

$\boxed{x = 0}$

Conclusion: The x-coordinate of any point on the y-axis is always 0.

Example: Points like (0, 2), (0, −3), (0, 5) all lie on the y-axis.

y-coordinate of a Point on the x-axis

Yes! There is a similar generalisation.

• The y-coordinate represents the perpendicular distance from the x-axis, measured along the y-axis.
• A point lying on the x-axis has zero vertical distance from the x-axis.

$\boxed{y = 0}$

Conclusion: The y-coordinate of any point on the x-axis is always 0.

Example: Points like (3, 0), (−5, 0), (7, 0) all lie on the x-axis.

When Does Q(y, x) Coincide with P(x, y)?

Two points coincide when they have the same coordinates.

For P(x, y) and Q(y, x) to coincide: $x = y \quad \text{and} \quad y = x$

Both conditions reduce to the same condition: x = y.

Conclusion:

• Yes, Q(y, x) coincides with P(x, y) if and only if x = y.
• If x ≠ y, then P and Q are different points.

Examples:

• If x = 3, y = 3: P = (3, 3) and Q = (3, 3) → same point ✓
• If x = 2, y = 5: P = (2, 5) and Q = (5, 2) → different points ✗

Geometrically, all points where x = y lie on the line y = x, so P and Q coincide only when the point lies on this line.

Verifying the Claim About Ordered Pairs

Claim: (x, y) ≠ (y, x) when x ≠ y, and (x, y) = (y, x) if and only if x = y.

Understanding Ordered Pairs:

In an ordered pair (a, b), the order matters. Two ordered pairs are equal only when their corresponding components are equal: $(a, b) = (c, d) \iff a = c \text{ and } b = d$

Proof:

Case 1: If x = y

• (x, y) becomes (x, x) and (y, x) becomes (x, x).
• Clearly, (x, x) = (x, x). ✓

Case 2: If x ≠ y

• For (x, y) = (y, x), we need x = y (first components equal) and y = x (second components equal).
• Both conditions require x = y, which contradicts our assumption that x ≠ y.
• Therefore, (x, y) ≠ (y, x). ✓

Conclusion:

The claim is TRUE.

• (x, y) = (y, x) if and only if x = y.
• (x, y) ≠ (y, x) whenever x ≠ y.

Example:

• (2, 5) ≠ (5, 2) since 2 ≠ 5
• (4, 4) = (4, 4) since 4 = 4

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, fourth foot is at 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.

From Fig. 1.5, the four corners are $O = (0,\ 0)$, $F = (0,\ 9)$, $R = (-6,\ 9)$, and $P = (-6,\ 0)$.

Reference: Chapter $1$, $\text{Page 7–8}$ (Exercise Set 1.2, Fig. 1.5)

$$\text{Bathroom Corners} = \begin{cases} O = (0,\ 0) & \text{bottom-right corner (origin)} \\ F = (0,\ 9) & \text{top-right corner (on } y\text{-axis)} \\ R = (-6,\ 9) & \text{top-left corner} \\ P = (-6,\ 0) & \text{bottom-left corner (on } x\text{-axis)} \end{cases}$$

Answer: The showering area $\text{SHWR}$ is a $\textbf{Trapezium}$ (one pair of opposite sides is parallel).

Reference: Chapter $1$, $\text{Page 7–8}$ (Exercise Set 1.2, Fig. 1.5)

$$\text{Showering Area Corners} = \begin{cases} S = (-6,\ 5) & \text{bottom-left} \\ H = (-3,\ 5) & \text{bottom-right} \\ W = (-2,\ 9) & \text{top-right} \\ R = (-6,\ 9) & \text{top-left} \end{cases}$$ $$\text{Shape of SHWR} = \textbf{Trapezium} \quad \because \quad \overline{SR} \parallel \overline{HW} \ (\text{both vertical}),\ \text{but } SR \neq HW$$

Real fact: Trapezoidal showering areas are common in modern bathroom design — the angled wall allows more floor space near the entry while keeping the wet zone compact, a technique widely used in space-efficient urban apartments.

Answer: Washbasin is placed at the bottom-left of the bathroom; toilet is placed directly above it.

Reference: Chapter $1$, $\text{Page 7–8}$ (Exercise Set 1.2, Fig. 1.5)

$$\text{Washbasin Space}\ (3\ \text{ft} \times 2\ \text{ft}) = \begin{cases} (-6,\ 0) \\ (-3,\ 0) \\ (-3,\ 2) \\ (-6,\ 2) \end{cases}$$ $$\text{Toilet Space}\ (2\ \text{ft} \times 3\ \text{ft}) = \begin{cases} (-6,\ 2) \\ (-4,\ 2) \\ (-4,\ 5) \\ (-6,\ 5) \end{cases}$$ $$\text{Layout} = \underbrace{(-6,0) \to (-3,0) \to (-3,2) \to (-6,2)}_{\text{Washbasin: } 3\text{ft} \times 2\text{ft}} \quad \text{stacked below} \quad \underbrace{(-6,2) \to (-4,2) \to (-4,5) \to (-6,5)}_{\text{Toilet: } 2\text{ft} \times 3\text{ft}}$$

Real fact: The standard clearance space required in front of a toilet as per the National Building Code of India is at least $2\ \text{ft}$, which is why compact $2\ \text{ft} \times 3\ \text{ft}$ toilet allocations are the minimum acceptable size in small bathrooms.

Answer: The dining room extends from $P = (-6,\ 0)$ to $A = (12,\ 0)$ along the length, and $15\ \text{ft}$ downward (into Quadrant III and IV).

Reference: Chapter $1$, $\text{Page 8}$ (Exercise Set 1.2, Question 4)

$$\text{Length} = |x_A - x_P| = |12 - (-6)| = 18\ \text{ft} \checkmark$$ $$\text{Dining Room Corners} = \begin{cases} P = (-6,\ 0) & \text{top-left corner} \\ A = (12,\ 0) & \text{top-right corner} \\ Q = (12,\ -15) & \text{bottom-right corner} \\ S = (-6,\ -15) & \text{bottom-left corner} \end{cases}$$ $$\text{Dining Room} = \underbrace{18\ \text{ft}}_{\text{length: P to A along x-axis}} \times \underbrace{15\ \text{ft}}_{\text{width: downward along y-axis}}$$

Real fact: In Indian residential architecture, a standard dining room for a family of 4–6 is typically $12\ \text{ft} \times 14\ \text{ft}$ to $15\ \text{ft} \times 18\ \text{ft}$; Reiaan's dining room at $18\ \text{ft} \times 15\ \text{ft}$ is a generously sized family dining space.

Answer: Centre of dining room is $(3,\ -7.5)$; Table feet are at $(0.5,\ -9)$, $(5.5,\ -9)$, $(5.5,\ -6)$, and $(0.5,\ -6)$.

Reference: Chapter $1$, $\text{Page 8}$ (Exercise Set 1.2, Question 4)

Explanation

Step 1: Find the centre of the dining room

$$x_{\text{centre}} = \frac{-6 + 12}{2} = \frac{6}{2} = 3$$ $$y_{\text{centre}} = \frac{0 + (-15)}{2} = \frac{-15}{2} = -7.5$$ $$\therefore \text{Centre} = (3,\ -7.5)$$

Step 2: Find half-dimensions of the table

$$\text{Half-length} = \frac{5}{2} = 2.5\ \text{ft}, \qquad \text{Half-width} = \frac{3}{2} = 1.5\ \text{ft}$$

Step 3: Calculate all four feet coordinates

$$\text{Table Feet} = \begin{cases} (3 - 2.5,\ -7.5 - 1.5) = (0.5,\ -9) & \text{bottom-left foot} \\ (3 + 2.5,\ -7.5 - 1.5) = (5.5,\ -9) & \text{bottom-right foot} \\ (3 + 2.5,\ -7.5 + 1.5) = (5.5,\ -6) & \text{top-right foot} \\ (3 - 2.5,\ -7.5 + 1.5) = (0.5,\ -6) & \text{top-left foot} \end{cases}$$ $$\text{Four Feet} = \{(0.5,\ -9),\ (5.5,\ -9),\ (5.5,\ -6),\ (0.5,\ -6)\}$$ $$\text{Verification: Width} = |5.5 - 0.5| = 5\ \text{ft} \checkmark \quad \text{Breadth} = |-6 - (-9)| = 3\ \text{ft} \checkmark$$

Real fact: A standard $5\ \text{ft} \times 3\ \text{ft}$ ($150\ \text{cm} \times 90\ \text{cm}$) rectangular dining table comfortably seats 4–6 people and is the most commonly sold dining table size in India, fitting well in rooms of $12\ \text{ft} \times 12\ \text{ft}$ or larger.

Answer: The x-axis and y-axis intersect at the $\textbf{Origin}$, so $x\text{-coordinate} = 0$ and $y\text{-coordinate} = 0$.

Reference: Chapter 1, $\text{Page 8 (Solutions PDF)}$

$\text{Point of Intersection} = \underbrace{(0,\ 0)}_{\text{Origin}}$

Explanation

The x-axis and y-axis intersect at exactly one point — the origin. By the definition of the coordinate system, both coordinates of the origin are zero.

$\text{x-coordinate} = 0, \qquad \text{y-coordinate} = 0$

$\therefore \text{ Point of intersection} = (0,\ 0)$

Real fact: The concept of the origin as $(0, 0)$ was made possible by Brahmagupta (c. 628 CE), who formalised zero as an algebraic entity — without which the four-quadrant Cartesian plane cannot exist.

If point W has x-coordinate $-5$, then any point on the line through W parallel to the y-axis will also have x-coordinate $-5$.

Therefore, the coordinates of H will be of the form:

$H = (-5,\ y) \quad \text{where } y \text{ can be any real number}$

Now, depending on the value of $y$:

• If $y > 0$, then H lies in Quadrant II
• If $y < 0$, then H lies in Quadrant III
• If $y = 0$, then H lies on the x-axis

So, H can lie in Quadrant II, Quadrant III, or on the x-axis.

Solution:

Let us observe:

• AM joins $A(0, -2)$ to $M(-5, -2)$ → both have the same y-coordinate → AM is horizontal
• MP joins $M(-5, -2)$ to $P(-5, 2)$ → both have the same x-coordinate → MP is vertical

A horizontal line and a vertical line are always perpendicular.

Therefore: $AM \perp MP$

The two perpendicular sides are AM and MP.

Solution:

• AM is parallel to the x-axis because both points A and M have the same y-coordinate $(−2)$.
• MP is parallel to the y-axis because both points M and P have the same x-coordinate $(−5)$.

One side parallel to an axis is: AM (parallel to x-axis) or MP (parallel to y-axis).

Solution:

Comparing $M(-5, -2)$ and $P(-5, 2)$:

• They have the same x-coordinate
• Their y-coordinates are equal in magnitude but opposite in sign

So, they are mirror images of each other in the x-axis.

The axis is the x-axis.

(Comment: Answers may differ from person to person.)

Solution:

To form a right-angled triangle easily, take:

• $I = (5, 0)$ on the x-axis
• $N = (0, -6)$ on the y-axis

Then triangle IZN is right-angled at Z because:

• IZ is vertical
• ZN is horizontal

Coordinates of the points:

$I = (5,\ 0), \quad Z = (5,\ -6), \quad N = (0,\ -6)$

Lengths of the sides:

IZ = Distance between $(5, 0)$ and $(5, -6)$

$= 0 - (-6) = \mathbf{6 \text{ units}}$

ZN = Distance between $(5, -6)$ and $(0, -6)$

$= 5 - 0 = \mathbf{5 \text{ units}}$

IN = Using distance formula:

$IN = \sqrt{(5-0)^2 + (0-(-6))^2} = \sqrt{25 + 36} = \mathbf{\sqrt{61} \text{ units}}$

Solution:

If we did not have negative numbers, then coordinates could only be zero or positive. So:

• On the x-axis, we could mark only the points to the right of the origin.
• On the y-axis, we could mark only the points above the origin.

This means we could locate points only in:

• Quadrant I
• The positive part of the x-axis
• The positive part of the y-axis
• The origin

We would not be able to locate:

• Points in Quadrant II
• Points in Quadrant III
• Points in Quadrant IV
• Points on the negative parts of the axes

Therefore, such a system would not allow us to locate all the points on a 2-D plane.

Solution:

We can use the distance formula to check.

If three points are collinear, then the sum of the two smaller distances equals the largest distance.

$MA = \sqrt{(0-(-3))^2 + (0-(-4))^2} = \sqrt{9+16} = \sqrt{25} = 5$

$AG = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36+64} = \sqrt{100} = 10$

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

Now checking: $MA + AG = 5 + 10 = 15 = MG$ ✓

Since the sum of the two distances equals the third, the points M, A and G lie on the same straight line.

Solution:

$RB = \sqrt{(-2-(-5))^2 + (-5-(-1))^2} = \sqrt{9+16} = \sqrt{25} = 5$

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

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

Now checking: $RB + BC = 5 + \sqrt{85}$

$5 + \sqrt{85} \neq \sqrt{202}$

Since the sum of the two distances does not equal the third, the points R, B and C do not lie on the same straight line.

Solution:

One possible set of vertices:

$O = (0,\ 0), \quad A = (4,\ 0), \quad B = (0,\ 4)$

Because:

• $OA = 4$ units
• $OB = 4$ units → isosceles
• OA is along the x-axis, OB is along the y-axis → $OA \perp OB$ → right angle at O

So triangle OAB is a right-angled isosceles triangle.

Solution:

One possible set of vertices:

$O = (0,\ 0), \quad P = (-3,\ -4), \quad Q = (3,\ -4)$

Because:

• $OP = \sqrt{9+16} = 5$ units
• $OQ = \sqrt{9+16} = 5$ units → $OP = OQ$ → isosceles
• P lies in Quadrant III, Q lies in Quadrant IV ✓

So triangle OPQ is an isosceles triangle.

Solution:

We use the distance formula to check if $SM = MT$.

Row 1: $S(-3, 0),\ M(0, 0),\ T(3, 0)$

$SM = \sqrt{9+0} = 3, \quad MT = \sqrt{9+0} = 3$

$SM = MT$ → Yes, M is the midpoint.

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

$SM = \sqrt{1+1} = \sqrt{2}, \quad MT = \sqrt{1+1} = \sqrt{2}$

$SM = MT$ → Yes, M is the midpoint.

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Row 3: $S(0, 0),\ M(0, 5),\ T(0, -10)$

$SM = \sqrt{0+25} = 5, \quad MT = \sqrt{0+225} = 15$

$SM \neq MT$ → No, M is not the midpoint.

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Row 4: $S(-8, 7),\ M(0, -2),\ T(6, -3)$

$SM = \sqrt{64+81} = \sqrt{145}, \quad MT = \sqrt{36+1} = \sqrt{37}$

$SM \neq MT$ → No, M is not the midpoint.

Connection found:

When M is the midpoint of ST:

$x_M = \frac{x_S + x_T}{2} \qquad \text{and} \qquad y_M = \frac{y_S + y_T}{2}$

Solution:

Using the midpoint formula:

For x-coordinate:

Distance from A to M $= -7 - 3 = -10$

So distance from M to B is also $-10$.

$x = -7 + (-10) = -17$

For y-coordinate:

Distance from A to M $= 1 - (-4) = 5$

So distance from M to B is also $5$.

$y = 1 + 5 = 6$

Therefore, the coordinates of B are $\mathbf{(-17,\ 6)}$.

Verification:

$\frac{3 + (-17)}{2} = -7 \checkmark \qquad \frac{-4 + 6}{2} = 1 \checkmark$

Solution:

P is the midpoint of A and Q. Q is the midpoint of P and B.

Finding x-coordinates:

$x_P = \frac{4 + x_Q}{2} \quad \cdots (1) \qquad x_Q = \frac{x_P + 16}{2} \quad \cdots (2)$

Substituting (2) into (1):

$x_P = \frac{4 + \frac{x_P + 16}{2}}{2} \implies 4x_P = x_P + 36 \implies x_P = 8$

$x_Q = \frac{8 + 16}{2} = 12$

Finding y-coordinates:

$y_P = \frac{7 + y_Q}{2} \quad \cdots (3) \qquad y_Q = \frac{y_P + (-2)}{2} \quad \cdots (4)$

Substituting (4) into (3):

$4y_Q = y_Q + 3 \implies y_Q = 1$

$y_P = \frac{7 + 1}{2} = 4$

Therefore, $\mathbf{P = (8,\ 4)}$ and $\mathbf{Q = (12,\ 1)}$.

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.

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(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).

Solution:

$OD = \sqrt{(-5)^2 + 6^2} = \sqrt{25+36} = \sqrt{61}$

$OE = \sqrt{0^2 + 9^2} = \sqrt{81} = 9$

Comparing with radius $\sqrt{65} \approx 8.06$:

• $\sqrt{61} \approx 7.81 < \sqrt{65}$ → D lies inside the circle
• $9 > \sqrt{65}$ → E lies outside the circle

Solution:

Let A = $(x_1, y_1)$, B = $(x_2, y_2)$, C = $(x_3, y_3)$

D(5, 1) is midpoint of BC, so: $x_2 + x_3 = 10 \quad \cdots(1)$ $y_2 + y_3 = 2 \quad \cdots(2)$

E(6, 5) is midpoint of CA, so: $x_3 + x_1 = 12 \quad \cdots(3)$ $y_3 + y_1 = 10 \quad \cdots(4)$

F(0, 3) is midpoint of AB, so: $x_1 + x_2 = 0 \quad \cdots(5)$ $y_1 + y_2 = 6 \quad \cdots(6)$

Solving for x-coordinates:

From (5): $x_2 = -x_1$

Substitute into (1): $-x_1 + x_3 = 10$ $\cdots(7)$

Substitute into (3): $(10 + x_1) + x_1 = 12$

$\Rightarrow 2x_1 = 2 \Rightarrow x_1 = 1$

Then: $x_2 = -1$ and $x_3 = 10 + 1 = 11$

Solving for y-coordinates:

From (6): $y_2 = 6 - y_1$

Substitute into (2): $(6 - y_1) + y_3 = 2 \Rightarrow y_3 = y_1 - 4$ $\cdots(8)$

Substitute into (4): $(y_1 - 4) + y_1 = 10$

$\Rightarrow 2y_1 = 14 \Rightarrow y_1 = 7$

Then: $y_2 = 6 - 7 = -1$ and $y_3 = 7 - 4 = 3$

Therefore, the coordinates of the points are:

$\boxed{A = (1,\ 7), \quad B = (-1,\ -1), \quad C = (11,\ 3)}$

Solution:

Since scale is 1 cm = 200 m, and each pair of adjacent streets is 200 m apart:

• There are 10 streets in the N–S direction → draw 10 vertical parallel lines, each 1 cm apart
• There are 10 streets in the E–W direction → draw 10 horizontal parallel lines, each 1 cm apart

This forms a square grid of 10 × 10 lines.

The two main roads (N–S and E–W) cross at the centre of the city.

(Draw the grid in your notebook with lines numbered 1–10 in each direction.)

Solution:

(a) A street intersection is named as: (N–S street number, E–W street number)

• (4, 3) means the 4th N–S street and the 3rd E–W street
• One N–S street and one E–W street can meet at only one point

Therefore: Only 1 intersection can be called (4, 3).

(b) - (3, 4) means the 3rd N–S street and the 4th E–W street

• Again, one N–S street and one E–W street meet at only one point

Therefore: Only 1 intersection can be called (3, 4).

Note: (4, 3) and (3, 4) are two different intersections — just like how the points (4, 3) and (3, 4) are two different points on a coordinate plane.

Solution: (i)

For Circle A with centre (100, 150) and radius 80:

• Left edge: $100 - 80 = 20 > 0$ ✓
• Right edge: $100 + 80 = 180 < 800$ ✓
• Bottom edge: $150 - 80 = 70 > 0$ ✓
• Top edge: $150 + 80 = 230 < 600$ ✓

Circle A lies completely inside the screen.

For Circle B with centre (250, 230) and radius 100:

• Left edge: $250 - 100 = 150 > 0$ ✓
• Right edge: $250 + 100 = 350 < 800$ ✓
• Bottom edge: $230 - 100 = 130 > 0$ ✓
• Top edge: $230 + 100 = 330 < 600$ ✓

Circle B lies completely inside the screen.

No, each of the two circles lies completely inside the screen.

Solution: (ii)

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

Since $AB = 170 < 180$ (sum of radii), the two circles intersect each other.

Yes, the two circles intersect each other.

Solution:

We check by finding the lengths of all four sides and both diagonals.

Finding the lengths of all sides:

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

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

$CD = \sqrt{(1-(-2))^2 + (-2-(-1))^2} = \sqrt{9+1} = \sqrt{10}$

$DA = \sqrt{(2-1)^2 + (1-(-2))^2} = \sqrt{1+9} = \sqrt{10}$

All four sides are equal. ✓

Finding the 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. ✓

Since all four sides are equal and both diagonals are equal, ABCD is a square.

Area of ABCD:

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