Class 9 Maths Chapter 1 NCERT Solutions: Orienting Yourself: The Use of Coordinates

1.1 INTRODUCTION

This section provides historical background on the evolution of coordinate systems in ancient Bharat (such as grid-planned cities in the Sindhu-Sarasvatī Civilisation, Baudhāyana's geometry, and Āryabhaṭa's celestial sines) and their formalisation by René Descartes in Europe. There are no problem exercises in this introductory section.

1.2 SETTLING IN

In-Text Question (from text discussing Fig. 1.1)

Question

Let us examine Fig. 1.1 to understand the layout of the room. Notice that this only shows the map of the floor. Do you see why the position of the windows cannot be marked on this map?

Solution

Reasoning: A floor plan like Fig. 1.1 is a two-dimensional (2-D) top-down view. It only measures horizontal lengths and widths along the flat floor. Windows are built vertically up on the walls, elevated above the floor level.
Conclusion: Because windows exist in the third dimension (height above the floor), their positions cannot be shown directly on a flat 2-D floor map.

1.3 THE 2-D CARTESIAN COORDINATE SYSTEM

Fig. 1.2: Structure of the coordinate plane

EXERCISE SET 1.1

Question 1 (i)

Question

Referring to Fig. 1.3, answer the following questions: If

𝐷 1 𝑅 1

represents the door to Reiaan's room, how far is the door from the left wall (the y-axis) of the room? How far is the door from the x-axis?

Solution

Distance from the left wall (y-axis): In Fig. 1.3, the door starts at point $𝐷 1$ , which is located at $𝑥 = 8$ on the horizontal axis. Since each grid unit represents 1 foot, the door is 8 feet away from the left wall (the y-axis).
Distance from the x-axis: Both endpoints of the door, $𝐷 1$ and $𝑅 1$ , lie directly on the horizontal bottom wall, which is the x-axis ( $𝑦 = 0$ ). Therefore, the door is 0 feet from the x-axis.

Question 1 (ii)

Question

What are the coordinates of

𝐷 1

Solution

Reasoning: Point $𝐷 1$ lies on the horizontal x-axis, 8 units to the right of the origin $𝑂 (0, 0)$ .
Conclusion: The x-coordinate is 8 and the y-coordinate is 0. Its coordinates are $(8, 0)$ .

Question 1 (iii)

Question

𝑅 1

is the point

(11.5, 0)

, how wide is the door? Do you think this is a comfortable width for the room door? If a person in a wheelchair wants to enter the room, will he/she be able to do so easily?

Solution

Width of the door: The door spans along the x-axis from $𝐷 1 (8, 0)$ to $𝑅 1 (11.5, 0)$ . We subtract the x-coordinates to find the width:

W i d t h = 11.5 - 8 = 𝟑 . 𝟓 f e e t

Comfortable width: Yes, 3.5 feet (which is 42 inches) is very comfortable and spacious for a residential bedroom door.
Wheelchair accessibility: Yes, a person in a wheelchair will be able to enter easily. Standard wheelchairs are around 2 to 2.5 feet (24 to 30 inches) wide. A 3.5-foot doorway provides plenty of extra clearance on both sides.

Question 1 (iv)

Question

𝐵 1 (0, 1.5)

and

𝐵 2 (0, 4)

represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?

Solution

Width of bathroom door: Both endpoints lie on the vertical y-axis. We subtract their y-coordinates to find the width:

W i d t h = 4 - 1.5 = 𝟐 . 𝟓 f e e t

Comparison: The room door is 3.5 feet wide, while the bathroom door is only 2.5 feet wide. Therefore, the bathroom door is narrower than the room door.

Think and Reflect (Page 5)

Question 1

Question

What are the standard widths for a room door? Look around your home and in school.

Solution

Home doors: In typical homes, standard bedroom doors are usually 2.5 feet to 3 feet (30 to 36 inches) wide. Bathroom doors are often slightly narrower, around 2 feet to 2.5 feet (24 to 30 inches).
School doors: Classroom and main entrance doors in schools are much wider to allow crowds of students to pass safely, typically ranging between 3 feet and 3.5 feet (36 to 42 inches) wide. (Note: Individual measurements at your home or school may vary slightly).

Question 2

Question

Are the doors in your school suitable for people in wheelchairs?

Solution

Reasoning: Modern building accessibility standards require doorways to have a clear opening of at least 32 inches (about 2.7 feet), though 36 inches (3 feet) is ideal for wheelchair users to maneuver comfortably without scraping their hands or wheels.
Conclusion: If your school doors are around 3 feet wide or more with level thresholds (no high steps), then yes, they are suitable for wheelchair access. Students should use a measuring tape at school to verify their classroom doors!

In-Text Activity / Task (Page 7)

Fig. 1.4 (with sample points P and R marked)

Task

Copy Fig. 1.4 and mark

𝑆

and

𝑄

in your diagram. Mark any point

𝑃

in Quadrant I and any point

𝑅

in Quadrant III, and write down their coordinates.

Solution

Marking S and Q: In Fig. 1.4, point $𝑄$ is located at $(- 5, 3)$ in Quadrant II (5 units left, 3 units up). Point $𝑆$ is located at $(3, - 5)$ in Quadrant IV (3 units right, 5 units down).
Selecting point P in Quadrant I: In Quadrant I, both coordinates must be positive ( $+, +$ ). Let us choose $𝑥 = 4$ and $𝑦 = 2$ . Plot point $𝑃 (4, 2)$ by moving 4 units right from the origin along the x-axis and 2 units straight up.
Selecting point R in Quadrant III: In Quadrant III, both coordinates must be negative ( $-, -$ ). Let us choose $𝑥 = - 3$ and $𝑦 = - 4$ . Plot point $𝑅 (- 3, - 4)$ by moving 3 units left from the origin along the x-axis and 4 units straight down.

Think and Reflect (Page 7)

Question 1

Question

What is the x-coordinate of a point on the y-axis?

Solution

Reasoning: The x-coordinate measures horizontal distance to the left or right of the vertical y-axis. If a point lies directly on the y-axis, it has not moved left or right at all.
Conclusion: The x-coordinate of any point on the y-axis is always $0$ . Its coordinates always look like $(0, 𝑦)$ .

Question 2

Question

Is there a similar generalisation for a point on the x-axis?

Solution

Reasoning: Yes! The y-coordinate measures vertical distance above or below the horizontal x-axis. If a point lies directly on the horizontal x-axis, it has not moved up or down at all.
Conclusion: The y-coordinate of any point on the x-axis is always $0$ . Its coordinates always look like $(𝑥, 0)$ .

Question 3

Question

Does point

𝑄 (𝑦, 𝑥)

ever coincide with point

𝑃 (𝑥, 𝑦)

? Justify your answer.

Solution

When they coincide: They coincide if and only if $𝑥 = 𝑦$ . For example, if $𝑥 = 4$ and $𝑦 = 4$ , then both $𝑃$ and $𝑄$ are at the exact same point $(4, 4)$ .
When they do not coincide: If $𝑥 \neq 𝑦$ , they represent completely different locations. For example, if $𝑥 = 2$ and $𝑦 = 5$ , then point $𝑃 (2, 5)$ is 2 units right and 5 units up (Quadrant I), while point $𝑄 (5, 2)$ is 5 units right and 2 units up. Therefore, they do not coincide unless their coordinates are equal.

Question 4

Question

𝑥 \neq 𝑦

then

(𝑥, 𝑦) \neq (𝑦, 𝑥)

and

(𝑥, 𝑦) = (𝑦, 𝑥)

if and only if

𝑥 = 𝑦

. Is this claim true?

Solution

Conclusion: Yes, this claim is completely true.
Reasoning: In Cartesian coordinate geometry, $(𝑥, 𝑦)$ is an ordered pair, meaning the order of numbers is critical. The first number always specifies horizontal displacement along the x-axis, and the second specifies vertical displacement along the y-axis. Swapping two unequal numbers changes the physical location of the point in the plane.

EXERCISE SET 1.2 (Using Fig. 1.5)

Question 1 (i)

Question

Place Reiaan's rectangular study table with three of its feet at the points

(8, 9)

(11, 9)

and

(11, 7)

. Where will the fourth foot of the table be?

Solution

Reasoning: In a rectangle, opposite sides must be equal in length and parallel to the coordinate axes.
The top side connects $(8, 9)$ and $(11, 9)$ , which means the table has a horizontal length (width) of $11 - 8 = 3$ units along the x-direction.
The right side connects $(11, 9)$ and $(11, 7)$ , which means the table has a vertical depth of $9 - 7 = 2$ units along the y-direction.
To form the bottom-left corner, the fourth foot must align vertically with $𝑥 = 8$ and horizontally with $𝑦 = 7$ .
Conclusion: The fourth foot of the table will be at the point $(8, 7)$ .

Question 1 (ii)

Question

Is this a good spot for the table?

Solution

Reasoning: Let us examine the room layout in Fig. 1.5:
The bed occupies the space from $𝑥 = 0$ to $𝑥 = 7$ and $𝑦 = 5$ to $𝑦 = 8$ .
The room door is located at the bottom wall from $𝑥 = 8$ to $𝑥 = 11.5$ along $𝑦 = 0$ .
The wardrobe is placed against the bottom wall from $𝑥 = 3$ to $𝑥 = 7$ along $𝑦 = 0$ to $𝑦 = 2$ .
Conclusion: Yes, this is a very good spot! Placing the table between $𝑥 = 8$ to $11$ and $𝑦 = 7$ to $9$ puts it in the quiet upper-right corner of the bedroom. It is well away from the swinging room door, does not block access to the wardrobe or bed, and sits near the plant at corner $𝐵 (12, 10)$ .

Question 1 (iii)

Question

What is the width of the table? The length? Can you make out the height of the table?

Solution

Width (horizontal span): Difference in x-coordinates between $(8, 9)$ and $(11, 9)$ :

W i d t h = 11 - 8 = 𝟑 f e e t

Length / Depth (vertical span): Difference in y-coordinates between $(11, 9)$ and $(11, 7)$ :

L e n g t h = 9 - 7 = 𝟐 f e e t

Height: No, you cannot make out the height of the table. A Cartesian floor plan is a 2-D map showing only horizontal x and y floor measurements. Height extends vertically out of the page in the third dimension (z-axis), which cannot be read from Fig. 1.5.

Question 2

Question

If the bathroom door has a hinge at

𝐵 1

and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?

Solution

Will it hit the wardrobe? No, it will not hit the wardrobe.
Proof: In Fig. 1.5, the bathroom door is located between $𝐵 1 (0, 1.5)$ and $𝐵 2 (0, 4)$ , meaning its width is $4 - 1.5 = 2.5$ feet. If hinged at $𝐵 1 (0, 1.5)$ and swung open into the bedroom, its outer edge will reach a maximum horizontal distance of $𝑥 = 2.5$ feet from the left wall. The wardrobe starts at point $𝑊 4 (3, 2)$ , which is at $𝑥 = 3$ feet. Since $2.5 < 3$ , there is a safe clearance of $0.5$ feet (6 inches) between the open door and the wardrobe.
Suggestions if the door is made wider: If the door is widened to 3 feet or more, swinging it open from hinge $𝐵 1$ would cause it to strike the side wall of the wardrobe at $𝑊 4 (3, 2)$ . To prevent this, we suggest three practical solutions:

Move the door hinges to point $𝐵 2 (0, 4)$ so that the door swings upwards against the left wall, away from the wardrobe.
Change the door design so that it opens inwards into the bathroom rather than outwards into the bedroom.
Shift the wardrobe further to the right along the bottom wall (e.g., starting at $𝑥 = 3.5$ or $4$ feet).

Question 3 (i)

Question

What are the coordinates of the four corners O, F, R, and P of the bathroom?

Solution

Let us read the coordinates from Fig. 1.5, keeping in mind that the bathroom lies to the left of the y-axis (negative x-values) and above the x-axis (positive y-values):
Corner O (Origin, bottom-right of bathroom): $(0, 0)$
Corner F (Top-right of bathroom on y-axis): $(0, 10)$
Corner R (Top-left of bathroom): Located at $𝑥 = - 6$ and $𝑦 = 10$ , so its coordinates are $(- 6, 10)$ .
Corner P (Bottom-left of bathroom on x-axis): Located at $𝑥 = - 6$ and $𝑦 = 0$ , so its coordinates are $(- 6, 0)$ .

Question 3 (ii)

Question

What is the shape of the showering area SHWR in Reiaan's bathroom? Write the coordinates of the four corners.

Solution

Coordinates of the four corners:
Corner S: Lies on the left wall along line PR ( $𝑥 = - 6$ ). Looking across to the y-axis, horizontal line SH sits at $𝑦 = 6$ . Therefore, $𝑆 = (- 6, 6)$ .
Corner H: Moving horizontally right from S along $𝑦 = 6$ , point H aligns vertically with grid line $𝑥 = - 2$ . Therefore, $𝐻 = (- 2, 6)$ .
Corner W: Lies on the top bathroom wall ( $𝑦 = 10$ ). Looking down at the grid, point W aligns vertically with grid line $𝑥 = - 1$ . Therefore, $𝑊 = (- 1, 10)$ .
Corner R: The top-left corner of the bathroom, $𝑅 = (- 6, 10)$ .
Shape of SHWR: Notice that side SH (along $𝑦 = 6$ ) and side RW (along $𝑦 = 10$ ) are both horizontal, which means they are strictly parallel. However, side SR is vertical ( $𝑥 = - 6$ ) while side HW is slanted diagonally! A four-sided polygon (quadrilateral) with exactly one pair of parallel sides is called a trapezium (or trapezoid).

Question 3 (iii)

Question

Mark off a

3 f t \times 2 f t

space for the washbasin and a

2 f t \times 3 f t

space for the toilet. Write the coordinates of the corners of these spaces.

Solution

Washbasin ( $3 f t \times 2 f t$ space): In Fig. 1.5, the washbasin is drawn at the bottom-left corner near point $𝑃 (- 6, 0)$ . Let us allocate a space extending 2 feet horizontally along the bottom wall (from $𝑥 = - 6$ to $𝑥 = - 4$ ) and 3 feet vertically along the left wall (from $𝑦 = 0$ to $𝑦 = 3$ ). The four corners of this washbasin space are:

(- 𝟔, 𝟎), (- 𝟒, 𝟎), (- 𝟒, 𝟑), a n d (- 𝟔, 𝟑)

Toilet ( $2 f t \times 3 f t$ space): In Fig. 1.5, the toilet is located along the left wall directly above the washbasin. Let us allocate a space extending 2 feet horizontally into the room (from $𝑥 = - 6$ to $𝑥 = - 4$ ) and 3 feet vertically along the left wall (from $𝑦 = 3$ to $𝑦 = 6$ , which brings it right up to line SH of the shower). The four corners of this toilet space are:

(- 𝟔, 𝟑), (- 𝟒, 𝟑), (- 𝟒, 𝟔), a n d (- 𝟔, 𝟔)

(Note: As long as your chosen corners form a $3 \times 2$ rectangle for the basin and a $2 \times 3$ rectangle for the toilet against the appropriate walls, your answer is completely valid!).

Question 4 (i)

Question

Reiaan's room door leads from the dining room which has the length 18 ft and width 15 ft. The length of the dining room extends from point P to point A. Sketch the dining room and mark the coordinates of its corners.

Sketch: dining room below the x-axis with centred table

Solution

Reasoning: Let us check the horizontal distance from $𝑃 (- 6, 0)$ to $𝐴 (12, 0)$ along the x-axis:

L e n g t h = 12 - (- 6) = 18 f e e t

This matches the stated 18 ft length exactly! Since Reiaan's bedroom and bathroom lie above the x-axis ( $𝑦 \geq 0$ ), and the doorway leads from the dining room into Reiaan's room across the x-axis ( $𝑦 = 0$ ), the dining room must be located directly below the x-axis (in Quadrants III and IV, where y-coordinates are negative).

Coordinates of the dining room corners: Since the width is 15 ft downwards, the y-coordinates will extend from $𝑦 = 0$ down to $𝑦 = - 15$ . The four corners are:
Top-left corner (at point P): $(- 6, 0)$
Top-right corner (at point A): $(12, 0)$
Bottom-right corner: $(12, - 15)$
Bottom-left corner: $(- 6, - 15)$
Sketch instructions for students: On your graph paper, draw a large rectangle below the x-axis connecting the four coordinate points listed above.

Question 4 (ii)

Question

Place a rectangular

5 f t \times 3 f t

dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.

Solution

Step 1: Find the exact centre of the dining room.
The x-coordinates range from $- 6$ to $12$ . The horizontal midpoint is:

𝑥 c e n t r e = - 6 + 12 2 = 62 = 3

The y-coordinates range from $0$ to $- 15$ . The vertical midpoint is:

𝑦 c e n t r e = 0 + (- 15) 2 = - 7.5

Thus, the centre of the dining room is at the point $(3, - 7.5)$ .
Step 2: Place the $5 f t \times 3 f t$ table centered at $(3, - 7.5)$ .
Let us orient the 5 ft length horizontally (parallel to the x-axis) and the 3 ft width vertically (parallel to the y-axis).
Half of the 5 ft length is $2.5$ ft. We move $2.5$ units left and right from $𝑥 = 3$ :

𝑥 l e f t = 3 - 2.5 = 0.5 a n d 𝑥 r i g h t = 3 + 2.5 = 5.5

Half of the 3 ft width is $1.5$ ft. We move $1.5$ units up and down from $𝑦 = - 7.5$ :

𝑦 t o p = - 7.5 + 1.5 = - 6 a n d 𝑦 b o t t o m = - 7.5 - 1.5 = - 9

Conclusion: The coordinates of the four feet of the dining table are:

(𝟎 . 𝟓, - 𝟔), (𝟓 . 𝟓, - 𝟔), (𝟓 . 𝟓, - 𝟗), a n d (𝟎 . 𝟓, - 𝟗)

(Note: If you orient the table vertically so that length is 5 ft along the y-direction and width is 3 ft along the x-direction, the corners will be at $(1.5, - 5), (4.5, - 5), (4.5, - 10),$ and $(1.5, - 10)$ . Both orientations are correct!).

1.4 DISTANCE BETWEEN TWO POINTS IN THE 2-D PLANE

Worked Example (Pages 9–10, Using Fig. 1.6 and Fig. 1.7)

Fig. 1.6: Triangle ADM

Fig. 1.7: Construction for length AD

Fig. 1.8: Distance between two points

Task

Look at triangle ADM in Fig. 1.6 with vertices

𝐴 (3, 4)

𝐷 (7, 1)

, and

𝑀 (9, 6)

. Explain how to find the side lengths

𝐴 𝐷

𝐷 𝑀

, and

𝑀 𝐴

in your own words with full steps.

Full Step-by-Step Solution

To find the distance between any two slanted points in the Cartesian plane, we create a right-angled triangle using horizontal and vertical grid lines, and then apply the Baudhāyana-Pythagoras Theorem ( $H y p o t e n u s e 2 = B a s e 2 + P e r p e n d i c u l a r 2$ ).

Step 1: Calculate the length of side AD.
In Fig. 1.7, draw a horizontal line left from point $𝐷 (7, 1)$ and a vertical line down from point $𝐴 (3, 4)$ . They meet at a right angle at point $𝐶 (3, 1)$ .
Find the horizontal distance along the x-axis ( $𝐶 𝐷$ ):

𝐶 𝐷 = x - c o o r d i n a t e o f 𝐷 - x - c o o r d i n a t e o f 𝐴 = 7 - 3 = 4 u n i t s

Find the vertical distance along the y-axis ( $𝐴 𝐶$ ):

𝐴 𝐶 = y - c o o r d i n a t e o f 𝐴 - y - c o o r d i n a t e o f 𝐷 = 4 - 1 = 3 u n i t s

Now, apply the Baudhāyana-Pythagoras Theorem to right-angled triangle $𝐴 𝐶 𝐷$ :

𝐴 𝐷 = \sqrt 𝐶 𝐷 2 + 𝐴 𝐶 2 = \sqrt 42 + 32 = \sqrt 16 + 9 = \sqrt 25 = 𝟓 u n i t s

Step 2: Calculate the length of side DM.
Find the horizontal shift between $𝐷 (7, 1)$ and $𝑀 (9, 6)$ along the x-axis:

H o r i z o n t a l c h a n g e = | 9 - 7 | = 2 u n i t s

Find the vertical shift along the y-axis:

V e r t i c a l c h a n g e = | 6 - 1 | = 5 u n i t s

Apply the theorem:

𝐷 𝑀 = \sqrt 22 + 52 = \sqrt 4 + 25 = \sqrt 𝟐 𝟗 u n i t s

Step 3: Calculate the length of side MA.
Find the horizontal shift between $𝑀 (9, 6)$ and $𝐴 (3, 4)$ along the x-axis:

H o r i z o n t a l c h a n g e = | 9 - 3 | = 6 u n i t s

Find the vertical shift along the y-axis:

V e r t i c a l c h a n g e = | 6 - 4 | = 2 u n i t s

Apply the theorem:

𝑀 𝐴 = \sqrt 62 + 22 = \sqrt 36 + 4 = \sqrt 𝟒 𝟎 u n i t s (This can also be written in simplified surd form as 2 \sqrt 10 units).

Think and Reflect (Page 9)

Question 1

Question

In moving from

𝐴 (3, 4)

𝐷 (7, 1)

, what distance has been covered along the x-axis? What about the distance along the y-axis?

Solution

Distance along the x-axis (horizontal): We subtract the x-coordinates: $7 - 3 = 𝟒 u n i t s$ .
Distance along the y-axis (vertical): We subtract the y-coordinates: $4 - 1 = 𝟑 u n i t s$ .

Question 2

Question

Can these distances help you find the distance AD?

Solution

Yes! Because horizontal and vertical axes are strictly perpendicular ( $90 \circ$ ), the horizontal distance ( $4$ units) and vertical distance ( $3$ units) form the two perpendicular legs of a right-angled triangle. Using the Baudhāyana-Pythagoras Theorem, we square these two distances, add them together, and take the square root to find the hypotenuse:

𝐴 𝐷 = \sqrt 42 + 32 = \sqrt 25 = 𝟓 u n i t s

Worked Example (Page 11, Using Fig. 1.9)

Fig. 1.9: Reflection of △AMD in the y-axis

Task

In Fig. 1.9, triangle AMD is reflected in the y-axis to form triangle

𝐴' 𝑀' 𝐷'

. Find the coordinates of the reflected vertices and verify the lengths of sides

𝐷' 𝑀'

and

𝑀' 𝐴'

with full steps.

Full Step-by-Step Solution

Rule for reflection in the y-axis: When a point $(𝑥, 𝑦)$ is reflected as a mirror image across the vertical y-axis, its horizontal distance flips to the opposite side. Therefore, its x-coordinate changes sign to become $(- 𝑥)$ , while its vertical height (y-coordinate) remains unchanged:

(𝑥, 𝑦) R e f l e c t i n y - a x i s \leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow\leftarrow \to (- 𝑥, 𝑦)

Finding the reflected coordinates:
$𝐴 (3, 4) \to 𝐀' (- 𝟑, 𝟒)$
$𝐷 (7, 1) \to 𝐃' (- 𝟕, 𝟏)$
$𝑀 (9, 6) \to 𝐌' (- 𝟗, 𝟔)$

Calculating side length $𝐷' 𝑀'$ :
Use the distance formula $\sqrt (𝑥 2 - 𝑥 1) 2 + (𝑦 2 - 𝑦 1) 2$ between $𝐷' (- 7, 1)$ and $𝑀' (- 9, 6)$ :

𝐷' 𝑀' = \sqrt (- 9 - (- 7)) 2 + (6 - 1) 2 = \sqrt (- 2) 2 + 52 = \sqrt 4 + 25 = \sqrt 𝟐 𝟗 u n i t s

Calculating side length $𝑀' 𝐴'$ :
Use the distance formula between $𝑀' (- 9, 6)$ and $𝐴' (- 3, 4)$ :

𝑀' 𝐴' = \sqrt (- 3 - (- 9)) 2 + (4 - 6) 2 = \sqrt 62 + (- 2) 2 = \sqrt 36 + 4 = \sqrt 𝟒 𝟎 u n i t s

Think and Reflect (Page 11)

Question 1

Question

What has remained the same and what has changed with this reflection?

Solution

What remained the same:

The y-coordinates of all vertices remained identical.
The side lengths of the triangle ( $𝐴' 𝐷' = 5$ , $𝐷' 𝑀' = \sqrt 29$ , $𝑀' 𝐴' = \sqrt 40$ ) did not change at all.
The shape, size, and total area of the triangle remained exactly the same.

What changed:

The signs of the x-coordinates were reversed (positive numbers became negative).
The orientation (handedness) of the triangle flipped horizontally, just like looking at a drawing in a mirror.

Question 2

Question

Would these observations be the same if

Δ 𝐴 𝐷 𝑀

is reflected in the x-axis (instead of the y-axis)?

Solution

Geometric properties (Side lengths, shape, area): Yes, these observations would be exactly the same! A reflection in the x-axis is also a rigid geometric transformation, meaning it preserves all distances and side lengths completely.
Coordinate changes: The coordinate behavior would swap! When reflecting across the horizontal x-axis, the x-coordinates remain the same, while the y-coordinates change sign ( $(𝑥, 𝑦) \to (𝑥, - 𝑦)$ ), causing the triangle to flip upside down vertically.

END-OF-CHAPTER EXERCISES

Question 1

Question

What are the x-coordinate and y-coordinate of the point of intersection of the two axes?

Solution

The point where the horizontal x-axis and vertical y-axis intersect is called the origin.
Both its x-coordinate and y-coordinate are $0$ . We write the coordinates of the origin as $(0, 0)$ .

Question 2

Question

Point W has x-coordinate equal to

- 5

. Can you predict the coordinates of point H which is on the line through W parallel to the y-axis? Which quadrants can H lie in?

Solution

Predicting coordinates: Any line parallel to the vertical y-axis is a vertical straight line. Every single point on a vertical line shares the exact same horizontal position (x-coordinate). Since point W lies on this line and has an x-coordinate of $- 5$ , point H must also have an x-coordinate of $- 5$ . Therefore, the coordinates of H will be of the form $(- 5, 𝑦)$ , where $𝑦$ can be any real number.
Quadrants: Because the x-coordinate is strictly negative ( $- 5$ ), point H can lie in:
Quadrant II (if its y-coordinate is positive, e.g., $(- 5, 3)$ ).
Quadrant III (if its y-coordinate is negative, e.g., $(- 5, - 4)$ ).

(Note: If $𝑦 = 0$ , point $𝐻 (- 5, 0)$ lies directly on the negative x-axis, on the boundary between Quadrants II and III).

Question 3

Question

Consider the points

𝑅 (3, 0)

𝐴 (0, - 2)

𝑀 (- 5, - 2)

and

𝑃 (- 5, 2)

. If they are joined in the same order, predict:

Plot of points R, A, M, P

(i) Two sides of RAMP that are perpendicular to each other. (ii) One side of RAMP that is parallel to one of the axes. (iii) Two points that are mirror images of each other in one axis. Which axis will this be? Now plot the points and verify your predictions.

Solution

Let us analyze the coordinates before plotting:

Notice that $𝐴 (0, - 2)$ and $𝑀 (- 5, - 2)$ both share the same y-coordinate of $- 2$ . This means segment $𝐴 𝑀$ is a flat horizontal line.
Notice that $𝑀 (- 5, - 2)$ and $𝑃 (- 5, 2)$ both share the same x-coordinate of $- 5$ . This means segment $𝑀 𝑃$ is a straight vertical line.

Now let us answer the specific questions:

(i) Two perpendicular sides: Sides $𝐴 𝑀$ and $𝑀 𝑃$ are perpendicular to each other. This is because horizontal lines (parallel to the x-axis) and vertical lines (parallel to the y-axis) always meet at a $90 \circ$ right angle at vertex $𝑀 (- 5, - 2)$ .
(ii) Side parallel to an axis: Side $𝐴 𝑀$ is parallel to the x-axis, and side $𝑀 𝑃$ is parallel to the y-axis. (Providing either one is correct!)
(iii) Mirror images: Points $𝑀 (- 5, - 2)$ and $𝑃 (- 5, 2)$ are mirror images of each other across the x-axis. Notice that their horizontal x-coordinates are identical ( $- 5$ ), while their vertical y-coordinates are exact opposites ( $- 2$ and $+ 2$ ), meaning they sit symmetrically below and above the horizontal x-axis.

Verification by plotting: On your graph paper:

Plot $𝑅 (3, 0)$ on the positive x-axis.
Plot $𝐴 (0, - 2)$ on the negative y-axis.
Plot $𝑀 (- 5, - 2)$ in Quadrant III.
Plot $𝑃 (- 5, 2)$ in Quadrant II.
Connect $𝑅 \to 𝐴 \to 𝑀 \to 𝑃 \to 𝑅$ with a ruler. You will visually see the sharp right angle at corner M, the horizontal line AM, the vertical line MP, and the vertical symmetry of M and P across the central horizontal x-axis!

Question 4

Question

Plot point

𝑍 (5, - 6)

on the Cartesian plane. Construct a right-angled triangle IZN and find the lengths of the three sides. (Comment: Answers may differ from person to person.)

Right-angled triangle IZN

Solution

Since you may choose the locations of points I and N to form a right angle, let us construct a simple, clear right-angled triangle that makes calculating side lengths easy!

Step 1: Choose vertices I and N.
Let vertex $𝑍$ be $(5, - 6)$ .
Let us pick point $𝐼$ on the vertical y-axis at the same height as Z: $𝐼 (0, - 6)$ .
Let us pick point $𝑁$ at the origin: $𝑁 (0, 0)$ .

Step 2: Check the right angle.
Side $𝑍 𝐼$ connects $(5, - 6)$ and $(0, - 6)$ . Since y-coordinates are equal, $𝑍 𝐼$ is a horizontal line segment.
Side $𝐼 𝑁$ connects $(0, - 6)$ and $(0, 0)$ . Since x-coordinates are equal ( $0$ ), $𝐼 𝑁$ is a vertical line segment along the y-axis.
Because horizontal and vertical lines meet at $90 \circ$ , $∠ 𝑍 𝐼 𝑁 = 90 \circ$ . Triangle $𝐼 𝑍 𝑁$ is a right-angled triangle!

Step 3: Calculate the lengths of the three sides.
Length of side $𝑍 𝐼$ (horizontal):

𝑍 𝐼 = | 5 - 0 | = 𝟓 u n i t s

Length of side $𝐼 𝑁$ (vertical):

𝐼 𝑁 = | 0 - (- 6) | = 𝟔 u n i t s

Length of hypotenuse $𝑍 𝑁$ (using Baudhāyana-Pythagoras Theorem):

𝑍 𝑁 = \sqrt 𝑍 𝐼 2 + 𝐼 𝑁 2 = \sqrt 52 + 62 = \sqrt 25 + 36 = \sqrt 𝟔 𝟏 u n i t s (which is approximately 7.81 units).

Question 5

Question

What would a system of coordinates be like if we did not have negative numbers? Would this system allow us to locate all the points on a 2-D plane?

Solution

What the system would look like: Without negative numbers, our coordinate axes would only start from zero and extend in the positive directions (rightward along the x-axis and upward along the y-axis). We would only have Quadrant I (the top-right quarter of the grid).
Would it locate all points? No, absolutely not! Without negative numbers, we would be completely unable to locate any points to the left of the y-axis (Quadrants II and III) or below the x-axis (Quadrants III and IV). Three-quarters of the entire 2-D plane would be impossible to describe! This highlights why Brahmagupta's formal introduction of zero and negative numbers in the 7th century CE was essential for creating the full four-quadrant Cartesian plane.

*Question 6

Question

Are the points

𝑀 (- 3, - 4)

𝐴 (0, 0)

and

𝐺 (6, 8)

on the same straight line? Suggest a method to check this without plotting and joining the points.

Collinear points M, A, G

Solution

Method to check without plotting (Distance Method): Three points lie on the same straight line (they are collinear) if the sum of the distances between the two shorter line segments equals the distance of the longest line segment between the two outer points. If $D i s t a n c e 𝑀 𝐴 + D i s t a n c e 𝐴 𝐺 = D i s t a n c e 𝑀 𝐺$ , then the three points must form a straight line!

Step-by-Step Calculation:
Find distance $𝑀 𝐴$ between $𝑀 (- 3, - 4)$ and $𝐴 (0, 0)$ :

𝑀 𝐴 = \sqrt (0 - (- 3)) 2 + (0 - (- 4)) 2 = \sqrt 32 + 42 = \sqrt 9 + 16 = \sqrt 25 = 𝟓 u n i t s

Find distance $𝐴 𝐺$ between $𝐴 (0, 0)$ and $𝐺 (6, 8)$ :

𝐴 𝐺 = \sqrt (6 - 0) 2 + (8 - 0) 2 = \sqrt 62 + 82 = \sqrt 36 + 64 = \sqrt 100 = 𝟏 𝟎 u n i t s

Find distance $𝑀 𝐺$ between $𝑀 (- 3, - 4)$ and $𝐺 (6, 8)$ :

𝑀 𝐺 = \sqrt (6 - (- 3)) 2 + (8 - (- 4)) 2 = \sqrt 92 + 122 = \sqrt 81 + 144 = \sqrt 225 = 𝟏 𝟓 u n i t s

Conclusion: Check the sum:

𝑀 𝐴 + 𝐴 𝐺 = 5 + 10 = 15 u n i t s

Since $𝑀 𝐴 + 𝐴 𝐺 = 𝑀 𝐺$ ( $5 + 10 = 15$ ), yes, the points M, A, and G lie on the exact same straight line! (Alternative slope/ratio method: Notice that for M, $- 4 - 3 = 43$ , and for G, $86 = 43$ . Since both points share the same ratio of y to x from the origin A(0,0), they lie on the same straight line passing through the origin!).

*Question 7

Question

Use your method (from Problem 6) to check if the points

𝑅 (- 5, - 1)

𝐵 (- 2, - 5)

and

𝐶 (4, - 12)

are on the same straight line. Now plot both sets of points and check your answers.

Points R, B, C (not collinear)

Solution

Let us calculate the three pairwise distances between $𝑅 (- 5, - 1)$ , $𝐵 (- 2, - 5)$ , and $𝐶 (4, - 12)$ :

Distance RB:

𝑅 𝐵 = \sqrt (- 2 - (- 5)) 2 + (- 5 - (- 1)) 2 = \sqrt 32 + (- 4) 2 = \sqrt 9 + 16 = \sqrt 25 = 𝟓 u n i t s

Distance BC:

𝐵 𝐶 = \sqrt (4 - (- 2)) 2 + (- 12 - (- 5)) 2 = \sqrt 62 + (- 7) 2 = \sqrt 36 + 49 = \sqrt 𝟖 𝟓 u n i t s \approx 9.22 u n i t s

Distance RC:

𝑅 𝐶 = \sqrt (4 - (- 5)) 2 + (- 12 - (- 1)) 2 = \sqrt 92 + (- 11) 2 = \sqrt 81 + 121 = \sqrt 𝟐 𝟎 𝟐 u n i t s \approx 14.21 u n i t s

Conclusion: Let us check if the two shorter distances add up to the longest distance:

𝑅 𝐵 + 𝐵 𝐶 = 5 + \sqrt 85 \approx 5 + 9.2195 = 14.2195 u n i t s

Since $14.2195 \neq 14.2127$ , we see that $𝑅 𝐵 + 𝐵 𝐶 \neq 𝑅 𝐶$ . Because the sum of the shorter segments is strictly greater than the direct distance between R and C, the points form a triangle. No, the points R, B, and C do NOT lie on the same straight line! (Verification by plotting: When you plot both sets on graph paper, a ruler will align perfectly through M, A, and G. However, when you align a ruler between R and B, you will see that the line misses point C slightly!).

*Question 8

Question

Using the origin as one vertex, plot the vertices of:

Isosceles triangles with a vertex at the origin

(i) A right-angled isosceles triangle. (ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Solution

An isosceles triangle is a triangle that has two sides of equal length. A right-angled triangle has one $90 \circ$ angle.

(i) Right-angled isosceles triangle:
Let Vertex 1 be the origin $𝑂 (0, 0)$ .
To make a right angle at the origin with two equal sides, simply pick two equal distances along the positive coordinate axes!
Let Vertex 2 be $𝑃 (4, 0)$ on the x-axis (length $𝑂 𝑃 = 4$ units).
Let Vertex 3 be $𝑄 (0, 4)$ on the y-axis (length $𝑂 𝑄 = 4$ units).
Proof: Since the x and y axes are perpendicular, $∠ 𝑃 𝑂 𝑄 = 90 \circ$ . Because $𝑂 𝑃 = 𝑂 𝑄 = 4$ units, triangle $Δ 𝑃 𝑂 𝑄$ is a right-angled isosceles triangle! Plot these three points and connect them.

(ii) Isosceles triangle with vertices in Quadrants III and IV:
Let Vertex 1 be the origin $𝑂 (0, 0)$ at the top.
To ensure the other two vertices are equal distances from the origin, choose two points that are mirror images of each other across the vertical y-axis!
In Quadrant III (where both coordinates are negative), choose Vertex 2: $𝑉 1 (- 3, - 4)$ .
In Quadrant IV (where x is positive and y is negative), choose its mirror image for Vertex 3: $𝑉 2 (3, - 4)$ .
Proof: Let us calculate the distances from the origin $𝑂 (0, 0)$ :

𝑂 𝑉 1 = \sqrt (- 3 - 0) 2 + (- 4 - 0) 2 = \sqrt 9 + 16 = \sqrt 25 = 5 u n i t s

𝑂 𝑉 2 = \sqrt (3 - 0) 2 + (- 4 - 0) 2 = \sqrt 9 + 16 = \sqrt 25 = 5 u n i t s

Since side $𝑂 𝑉 1 = 𝑂 𝑉 2 = 5$ units, triangle $Δ 𝑂 𝑉 1 𝑉 2$ has two equal sides, making it an isosceles triangle with vertices in Quadrants III and IV! Plot these three points and connect them.

*Question 9

Question

The following table shows the coordinates of points S, M and T. In each case, state whether M is the midpoint of segment ST. Justify your answer.

Solution

A point $𝑀$ is the exact midpoint of a line segment $𝑆 𝑇$ if and only if it sits precisely halfway between S and T on the straight line. This means its coordinates must be the arithmetic mean (averages) of the coordinates of S and T:

𝑥 𝑀 = 𝑥 𝑆 + 𝑥 𝑇 2 a n d 𝑦 𝑀 = 𝑦 𝑆 + 𝑦 𝑇 2

Let us test each row in the table:

Coordinates of S	Coordinates of M	Coordinates of T	Is M the midpoint of ST? (Yes / No)	Reason for your answer
$(- 3, 0)$	$(0, 0)$	$(3, 0)$	Yes	The averages of the coordinates are $- 3 + 3 2 = 0$ and $0 + 0 2 = 0$ , which match $𝑀 (0, 0)$ exactly. Both S and T lie on the x-axis, 3 units away from M on opposite sides.
$(2, 3)$	$(3, 4)$	$(4, 5)$	Yes	The averages of the coordinates are $2 + 4 2 = 3$ and $3 + 5 2 = 4$ , which match $𝑀 (3, 4)$ exactly. Point M lies on line segment ST and divides it into two equal lengths of $\sqrt 2$ units.
$(0, 0)$	$(0, 5)$	$(0, - 10)$	No	The average of the y-coordinates is $0 + (- 10) 2 = - 5$ , but M has a y-coordinate of $+ 5$ . The distance $𝑆 𝑀 = 5$ while $𝑀 𝑇 = 15$ ; they are not equal, and M is not located between S and T.
$(- 8, 7)$	$(0, - 2)$	$(6, - 3)$	No	The average of the x-coordinates is $- 8 + 6 2 = - 1 \neq 0$ , and the average of the y-coordinates is $7 + (- 3) 2 = 2 \neq - 2$ . The coordinates of M do not equal the averages of S and T.

*Question 10

Question

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T? Use the connection you found to find the coordinates of B given that

𝑀 (- 7, 1)

is the midpoint of

𝐴 (3, - 4)

and

𝐵 (𝑥, 𝑦)

Solution

Connection (The Midpoint Formula): As discovered in Question 9, when $𝑀 (𝑥 𝑀, 𝑦 𝑀)$ is the midpoint of segment $𝑆 𝑇$ with endpoints $𝑆 (𝑥 𝑆, 𝑦 𝑆)$ and $𝑇 (𝑥 𝑇, 𝑦 𝑇)$ , each coordinate of M is the exact average of the corresponding coordinates of endpoints S and T:

𝑥 𝑀 = 𝑥 𝑆 + 𝑥 𝑇 2 a n d 𝑦 𝑀 = 𝑦 𝑆 + 𝑦 𝑇 2

Finding coordinates of B(x, y): We are given endpoint $𝐴 (3, - 4)$ and midpoint $𝑀 (- 7, 1)$ . Let us set up the midpoint equations and solve for x and y:
For the x-coordinate:

3 + 𝑥 2 = - 7

Multiply both sides by 2:

3 + 𝑥 = - 14

Subtract 3 from both sides:

𝑥 = - 14 - 3 = - 𝟏 𝟕

For the y-coordinate:

- 4 + 𝑦 2 = 1

Multiply both sides by 2:

- 4 + 𝑦 = 2

Add 4 to both sides:

𝑦 = 2 + 4 = 𝟔

Conclusion: The coordinates of point B are $(- 17, 6)$ .

*Question 11

Question

Let P, Q be points of trisection of AB, with P closer to A, and Q closer to B. Using your knowledge of how to find the coordinates of the midpoint of a segment, how would you find the coordinates of P and Q? Do this for the case when the points are

𝐴 (4, 7)

and

𝐵 (16, - 2)

Solution

How to find P and Q using midpoint concepts:

"Points of trisection" means that points P and Q divide the line segment AB into three segments of equal length ( $𝐴 𝑃 = 𝑃 𝑄 = 𝑄 𝐵$ ). Let us look at the symmetry:

Since $𝐴 𝑃 = 𝑃 𝑄$ , point P is the exact midpoint of segment AQ!
Since $𝑃 𝑄 = 𝑄 𝐵$ , point Q is the exact midpoint of segment PB!

This means that each point of trisection represents taking exactly one-third ( $13$ ) of the total horizontal and vertical distance from endpoint A to endpoint B! Therefore, to find P and Q:

Find the total change in x ( $𝑥 𝐵 - 𝑥 𝐴$ ) and total change in y ( $𝑦 𝐵 - 𝑦 𝐴$ ).
Divide these total changes by 3 to find the size of one trisection step.
Add one step to A's coordinates to find P (which is $13$ of the way).
Add two steps to A's coordinates to find Q (which is $23$ of the way).

Step-by-Step Calculation for $𝐴 (4, 7)$ and $𝐵 (16, - 2)$ :
Step 1: Find total changes from A to B:

H o r i z o n t a l c h a n g e = 16 - 4 = 12 u n i t s

V e r t i c a l c h a n g e = - 2 - 7 = - 9 u n i t s

Step 2: Divide by 3 to find the size of one step:

S t e p i n x - d i r e c t i o n = 123 = 4 u n i t s

S t e p i n y - d i r e c t i o n = - 9 3 = - 3 u n i t s

Step 3: Find coordinates of P (closer to A, 1 step from A):

𝑥 𝑃 = 4 + 4 = 𝟖 a n d 𝑦 𝑃 = 7 + (- 3) = 𝟒

Therefore, $𝑃 = (8, 4)$ .

Step 4: Find coordinates of Q (closer to B, 2 steps from A):

𝑥 𝑄 = 8 + 4 = 𝟏 𝟐 a n d 𝑦 𝑄 = 4 + (- 3) = 𝟏

Therefore, $𝑄 = (12, 1)$ . (Verification by midpoint rule: Let us check if P(8,4) is the midpoint of A(4,7) and Q(12,1). The average of x is $4 + 12 2 = 8$ , and the average of y is $7 + 1 2 = 4$ . Yes! Let us check if Q(12,1) is the midpoint of P(8,4) and B(16,-2). The average of x is $8 + 16 2 = 12$ , and the average of y is $4 + (- 2) 2 = 1$ . Both midpoint checks confirm our answers!).

*Question 12 (i)

Question

Given the points

𝐴 (1, - 8)

𝐵 (- 4, 7)

and

𝐶 (- 7, - 4)

, show that they lie on a circle K whose center is the origin

𝑂 (0, 0)

. What is the radius of circle K?

Circle K with centre O and points A, B, C, D, E

Solution

Reasoning: By geometric definition, a circle is the set of all points in a plane that are at a constant fixed distance (called the radius) from a center point. To prove that points A, B, and C lie on circle K centered at the origin $𝑂 (0, 0)$ , we must calculate their distances from the origin using the formula $\sqrt 𝑥 2 + 𝑦 2$ and show that all three distances are identical!
Step-by-Step Calculation:
Distance from origin to A(1, -8):

𝑂 𝐴 = \sqrt 12 + (- 8) 2 = \sqrt 1 + 64 = \sqrt 𝟔 𝟓 u n i t s

Distance from origin to B(-4, 7):

𝑂 𝐵 = \sqrt (- 4) 2 + 72 = \sqrt 16 + 49 = \sqrt 𝟔 𝟓 u n i t s

Distance from origin to C(-7, -4):

𝑂 𝐶 = \sqrt (- 7) 2 + (- 4) 2 = \sqrt 49 + 16 = \sqrt 𝟔 𝟓 u n i t s

Conclusion: Since $𝑂 𝐴 = 𝑂 𝐵 = 𝑂 𝐶 = \sqrt 65$ units, all three points are equidistant from the origin $𝑂 (0, 0)$ . Therefore, they lie on circle K. The radius of circle K is $\sqrt 65 u n i t s$ (approximately $8.06$ units).

*Question 12 (ii)

Question

Given the points

𝐷 (- 5, 6)

and

𝐸 (0, 9)

, check whether D and E lie within the circle, on the circle, or outside the circle K.

Solution

Rule: To determine where a point lies relative to circle K (which has radius $𝑅 = \sqrt 65$ ), we calculate its distance $𝑑$ from the center $𝑂 (0, 0)$ :
If $𝑑 < \sqrt 65$ , the point lies within (inside) the circle.
If $𝑑 = \sqrt 65$ , the point lies on the circle.
If $𝑑 > \sqrt 65$ , the point lies outside the circle.

Check Point D(-5, 6):

𝑂 𝐷 = \sqrt (- 5) 2 + 62 = \sqrt 25 + 36 = \sqrt 61 u n i t s

Since $61 < 65$ , we know $\sqrt 61 < \sqrt 65$ ( $7.81 < 8.06$ ). Therefore, point D lies within (inside) circle K.

Check Point E(0, 9):

𝑂 𝐸 = \sqrt 02 + 92 = \sqrt 0 + 81 = \sqrt 81 = 9 u n i t s

Since $9 = \sqrt 81$ and $\sqrt 81 > \sqrt 65$ ( $9 > 8.06$ ), point E lies outside circle K.

*Question 13

Question

The midpoints of the sides of triangle ABC are the points D, E, and F. Given that the coordinates of D, E, and F are

(5, 1)

(6, 5)

, and

(0, 3)

, respectively, find the coordinates of A, B and C.

Triangle ABC with midpoints D, E, F

Solution

Let the coordinates of the three vertices of triangle ABC be $𝐴 (𝑥 1, 𝑦 1)$ , $𝐵 (𝑥 2, 𝑦 2)$ , and $𝐶 (𝑥 3, 𝑦 3)$ . In standard triangle notation:

Let $𝐹 (0, 3)$ be the midpoint of side $𝐴 𝐵$ .
Let $𝐷 (5, 1)$ be the midpoint of side $𝐵 𝐶$ .
Let $𝐸 (6, 5)$ be the midpoint of side $𝐶 𝐴$ .

(Note: Whichever midpoint is assigned to which side, the resulting set of three vertex coordinates will be identical).

Using the midpoint formula ( $𝑥 m i d = 𝑥 𝑎 + 𝑥 𝑏 2$ ), we set up equations for the x-coordinates:

From F: $𝑥 1 + 𝑥 2 2 = 0 ⟹ 𝑥 1 + 𝑥 2 = 0$
From D: $𝑥 2 + 𝑥 3 2 = 5 ⟹ 𝑥 2 + 𝑥 3 = 10$
From E: $𝑥 3 + 𝑥 1 2 = 6 ⟹ 𝑥 3 + 𝑥 1 = 12$

Step 1: Solve for the x-coordinates ( $𝑥 1, 𝑥 2, 𝑥 3$ ).
Add all three equations together:

(𝑥 1 + 𝑥 2) + (𝑥 2 + 𝑥 3) + (𝑥 3 + 𝑥 1) = 0 + 10 + 12

2 𝑥 1 + 2 𝑥 2 + 2 𝑥 3 = 22

Divide by 2 to find the sum of all three x-coordinates:

𝑥 1 + 𝑥 2 + 𝑥 3 = 11 - - - (E q u a t i o n S)

Now substitute each pair into Equation S:
Since $𝑥 2 + 𝑥 3 = 10$ : $𝑥 1 + 10 = 11 ⟹ 𝐱 𝟏 = 𝟏$
Since $𝑥 3 + 𝑥 1 = 12$ : $𝑥 2 + 12 = 11 ⟹ 𝐱 𝟐 = - 𝟏$
Since $𝑥 1 + 𝑥 2 = 0$ : $0 + 𝑥 3 = 11 ⟹ 𝐱 𝟑 = 𝟏 𝟏$

Step 2: Solve for the y-coordinates ( $𝑦 1, 𝑦 2, 𝑦 3$ ).

Set up equations for the y-coordinates from the midpoints:

From F: $𝑦 1 + 𝑦 2 2 = 3 ⟹ 𝑦 1 + 𝑦 2 = 6$
From D: $𝑦 2 + 𝑦 3 2 = 1 ⟹ 𝑦 2 + 𝑦 3 = 2$
From E: $𝑦 3 + 𝑦 1 2 = 5 ⟹ 𝑦 3 + 𝑦 1 = 10$

Add all three equations together:

2 𝑦 1 + 2 𝑦 2 + 2 𝑦 3 = 6 + 2 + 10 = 18

Divide by 2:

𝑦 1 + 𝑦 2 + 𝑦 3 = 9 - - - (E q u a t i o n T)

Substitute each pair into Equation T:
Since $𝑦 2 + 𝑦 3 = 2$ : $𝑦 1 + 2 = 9 ⟹ 𝐲 𝟏 = 𝟕$
Since $𝑦 3 + 𝑦 1 = 10$ : $𝑦 2 + 10 = 9 ⟹ 𝐲 𝟐 = - 𝟏$
Since $𝑦 1 + 𝑦 2 = 6$ : $6 + 𝑦 3 = 9 ⟹ 𝐲 𝟑 = 𝟑$

Conclusion: The coordinates of the three vertices of triangle ABC are:

𝐀 = (𝟏, 𝟕), 𝐁 = (- 𝟏, - 𝟏), a n d 𝐂 = (𝟏 𝟏, 𝟑)

(Geometric Shortcut for students: To find any vertex of a triangle from its midpoints, simply add the coordinates of the two midpoints adjacent to that vertex and subtract the opposite midpoint! For example: $𝐴 = 𝐹 + 𝐸 - 𝐷 = (0, 3) + (6, 5) - (5, 1) = (1, 7)$ !).

Question 14 (i)

Question

A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South (N-S) direction and East-West (E-W) direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 10 streets in each direction. Using

1 c m = 200 m

, draw a model of the city in your notebook. Represent the roads/streets by single lines.

Solution

Instructions for drawing the model in your notebook:

Draw the Main Roads: In the middle of your graph sheet, draw two thick perpendicular lines intersecting at the center. Label the vertical line as the North-South (N-S) Main Road (this acts as the y-axis) and the horizontal line as the East-West (E-W) Main Road (this acts as the x-axis). Their intersection is the city centre $(0, 0)$ .
Scale: Because $1 c m = 200 m$ , each 200 m street block is represented by exactly 1 cm on your ruler.
Draw the Streets: Draw 10 thin vertical lines parallel to the N-S Main Road, spaced exactly 1 cm apart from each other. Draw 10 thin horizontal lines parallel to the E-W Main Road, also spaced exactly 1 cm apart. This creates a neat grid representing the city blocks!

Question 14 (ii)

Question

There are street intersections in the model. Each street intersection is formed by two streets—one running in the N-S direction and another in the E-W direction. Each street intersection is referred to in the following manner: If the second street running in the N-S direction and 5th street in the E-W direction meet at some crossing, then we call this street intersection

(2, 5)

. Using this convention, find:

(a) how many street intersections can be referred to as $(4, 3)$ . (b) how many street intersections can be referred to as $(3, 4)$ .

Solution

Understanding the convention: The ordered pair $(𝑥, 𝑦)$ means:
First number ( $𝑥$ ) = The street running in the North-South direction (vertical line).
Second number ( $𝑦$ ) = The street running in the East-West direction (horizontal line).
(a) How many street intersections can be referred to as $(4, 3)$ ?
The coordinate $(4, 3)$ refers specifically to the crossing where the 4th N-S street meets the 3rd E-W street. Because two straight lines intersect at only one single point in a plane, there is exactly $1$ street intersection that can be referred to as $(4, 3)$ .
(b) How many street intersections can be referred to as $(3, 4)$ ?
Similarly, $(3, 4)$ specifies the unique crossing where the 3rd N-S street meets the 4th E-W street. There is exactly $1$ street intersection that can be referred to as $(3, 4)$ .

(Note: This reinforces that $(4, 3)$ and $(3, 4)$ represent two completely different intersections in the city!).

Question 15

Question

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point

𝐴 (100, 150)

. Another circular icon of radius 100 pixels is drawn with its centre at the point

𝐵 (250, 230)

. Determine:

(i) whether any part of either circle lies outside the screen. (ii) whether the two circles intersect each other.

Solution

Because the origin $(0, 0)$ is at the bottom-left corner, the visible screen area consists of all points $(𝑥, 𝑦)$ where horizontal coordinates range from $0$ to $800$ ( $0 \leq 𝑥 \leq 800$ ) and vertical coordinates range from $0$ to $600$ ( $0 \leq 𝑦 \leq 600$ ).

(i) Does any part of either circle lie outside the screen?
Check Circle A: Center $𝐴 (100, 150)$ , radius $𝑟 𝐴 = 80$ pixels.
Leftmost edge reaches: $𝑥 = 100 - 80 = 𝟐 𝟎$ (Since $20 \geq 0$ , it is inside the left edge).
Rightmost edge reaches: $𝑥 = 100 + 80 = 𝟏 𝟖 𝟎$ (Since $180 \leq 800$ , it is inside the right edge).
Bottommost edge reaches: $𝑦 = 150 - 80 = 𝟕 𝟎$ (Since $70 \geq 0$ , it is inside the bottom edge).
Topmost edge reaches: $𝑦 = 150 + 80 = 𝟐 𝟑 𝟎$ (Since $230 \leq 600$ , it is inside the top edge).
Circle A is entirely inside the screen.
Check Circle B: Center $𝐵 (250, 230)$ , radius $𝑟 𝐵 = 100$ pixels.
Leftmost edge reaches: $𝑥 = 250 - 100 = 𝟏 𝟓 𝟎$ ( $\geq 0$ ).
Rightmost edge reaches: $𝑥 = 250 + 100 = 𝟑 𝟓 𝟎$ ( $\leq 800$ ).
Bottommost edge reaches: $𝑦 = 230 - 100 = 𝟏 𝟑 𝟎$ ( $\geq 0$ ).
Topmost edge reaches: $𝑦 = 230 + 100 = 𝟑 𝟑 𝟎$ ( $\leq 600$ ).
Circle B is entirely inside the screen.
Conclusion for (i): No, neither circle has any part lying outside the computer screen.

(ii) Do the two circles intersect each other?
Rule: Two circles intersect if the distance between their centers ( $𝐴 𝐵$ ) is less than or equal to the sum of their radii ( $𝑟 𝐴 + 𝑟 𝐵$ ).
Step 1: Calculate the sum of the radii:

𝑟 𝐴 + 𝑟 𝐵 = 80 + 100 = 𝟏 𝟖 𝟎 p i x e l s

Step 2: Calculate the exact distance between center A(100, 150) and center B(250, 230):

𝐴 𝐵 = \sqrt (250 - 100) 2 + (230 - 150) 2

𝐴 𝐵 = \sqrt 1502 + 802

𝐴 𝐵 = \sqrt 22500 + 6400 = \sqrt 28900

Since $172 = 289$ , we take the square root:

𝐴 𝐵 = 𝟏 𝟕 𝟎 p i x e l s

Step 3: Compare distance to radius sum:

The distance between the centers is $170$ pixels, while the combined reach of their radii is $180$ pixels. Since $170 < 180$ , the circles overlap!

Conclusion for (ii): Yes, the two circles intersect each other (overlapping by 10 pixels).

Question 16

Question

Plot the points

𝐴 (2, 1)

𝐵 (- 1, 2)

𝐶 (- 2, - 1)

, and

𝐷 (1, - 2)

in the coordinate plane. Is ABCD a square? Can you explain why? What is the area of this square?

Square ABCD

Solution

To prove geometrically that a four-sided polygon (quadrilateral) is a square, we must prove two things:

All four outer sides are equal in length.
The two diagonals are equal in length (which confirms that all corner angles are $90 \circ$ right angles).

Step 1: Calculate the lengths of all four outer sides using the distance formula $\sqrt (𝑥 2 - 𝑥 1) 2 + (𝑦 2 - 𝑦 1) 2$ :
Side AB: between $𝐴 (2, 1)$ and $𝐵 (- 1, 2)$

𝐴 𝐵 = \sqrt (- 1 - 2) 2 + (2 - 1) 2 = \sqrt (- 3) 2 + 12 = \sqrt 9 + 1 = \sqrt 𝟏 𝟎 u n i t s

Side BC: between $𝐵 (- 1, 2)$ and $𝐶 (- 2, - 1)$

𝐵 𝐶 = \sqrt (- 2 - (- 1)) 2 + (- 1 - 2) 2 = \sqrt (- 1) 2 + (- 3) 2 = \sqrt 1 + 9 = \sqrt 𝟏 𝟎 u n i t s

Side CD: between $𝐶 (- 2, - 1)$ and $𝐷 (1, - 2)$

𝐶 𝐷 = \sqrt (1 - (- 2)) 2 + (- 2 - (- 1)) 2 = \sqrt 32 + (- 1) 2 = \sqrt 9 + 1 = \sqrt 𝟏 𝟎 u n i t s

Side DA: between $𝐷 (1, - 2)$ and $𝐴 (2, 1)$

𝐷 𝐴 = \sqrt (2 - 1) 2 + (1 - (- 2)) 2 = \sqrt 12 + 32 = \sqrt 1 + 9 = \sqrt 𝟏 𝟎 u n i t s

Result: Since $𝐴 𝐵 = 𝐵 𝐶 = 𝐶 𝐷 = 𝐷 𝐴 = \sqrt 10$ units, all four sides are equal in length!

Step 2: Calculate the lengths of the two interior diagonals AC and BD:
Diagonal AC: between opposite corners $𝐴 (2, 1)$ and $𝐶 (- 2, - 1)$

𝐴 𝐶 = \sqrt (- 2 - 2) 2 + (- 1 - 1) 2 = \sqrt (- 4) 2 + (- 2) 2 = \sqrt 16 + 4 = \sqrt 𝟐 𝟎 u n i t s

Diagonal BD: between opposite corners $𝐵 (- 1, 2)$ and $𝐷 (1, - 2)$

𝐵 𝐷 = \sqrt (1 - (- 1)) 2 + (- 2 - 2) 2 = \sqrt 22 + (- 4) 2 = \sqrt 4 + 16 = \sqrt 𝟐 𝟎 u n i t s

Result: Since diagonal $𝐴 𝐶 = d i a g o n a l 𝐵 𝐷 = \sqrt 20$ units, the diagonals are equal!

Is ABCD a square? Explain why.

Yes, ABCD is a square. Because all four sides are equal in length ( $\sqrt 10$ units) and the diagonals are equal in length ( $\sqrt 20$ units), the shape is mathematically proven to be a square! (Verification of $90 \circ$ angles by Pythagoras: In triangle ABC, $(\sqrt 10) 2 + (\sqrt 10) 2 = 10 + 10 = 20$ . Since the hypotenuse squared is $𝐴 𝐶 2 = (\sqrt 20) 2 = 20$ , the corner angle $∠ 𝐴 𝐵 𝐶$ is exactly $90 \circ$ !).

What is the area of this square?

The area of any square is found by squaring its side length:

A r e a = (S i d e l e n g t h) 2 = (\sqrt 10) 2 = 𝟏 𝟎 s q u a r e u n i t s!

CHAPTER SUMMARY

Cartesian Plane: A 2-D system using two perpendicular number lines—the horizontal x-axis and vertical y-axis—that intersect at the origin $(0, 0)$ .
Quadrants: The axes divide the plane into four quadrants:
Quadrant I: $(+, +)$
Quadrant II: $(-, +)$
Quadrant III: $(-, -)$
Quadrant IV: $(+, -)$
Coordinates: Written as an ordered pair $(𝑥, 𝑦)$ , where $𝑥$ is horizontal distance from the y-axis and $𝑦$ is vertical distance from the x-axis. Points on the x-axis are $(𝑥, 0)$ and points on the y-axis are $(0, 𝑦)$ .
Distance Formula: By the Baudhāyana-Pythagoras Theorem, the distance between any two points $(𝑥 1, 𝑦 1)$ and $(𝑥 2, 𝑦 2)$ is:

D i s t a n c e = \sqrt (𝑥 2 - 𝑥 1) 2 + (𝑦 2 - 𝑦 1) 2