Orienting Yourself: The Use of Coordinates
Grade 9 Mathematics | Ganita Manjari | Part I
Chapter 1 – Colorful Concepts & Complete Book Solutions
1.1 Introduction
📍 What is a Coordinate System?
A system of coordinates is a structured framework that helps us use numbers to describe the exact location of points or objects.
We can think of it like the grid lines on a map or graph paper.
🇮🇳 A Little History
The idea of grid-based thinking has deep roots in Bhārat. The Sindhu-Sarasvatī Civilisation used carefully planned North–South and East–West streets.
Baudhāyana used East–West and North–South lines in geometric constructions. Later, mathematicians such as Āryabhaṭa and Brahmagupta contributed important ideas involving coordinates, zero and negative numbers.
⭐ Important Names
- Baudhāyana: Important geometric constructions and the Baudhāyana–Pythagoras Theorem.
- Āryabhaṭa: Used sines and worked with celestial coordinates.
- Brahmagupta: Formalised zero and negative numbers as algebraic entities.
- Al-Bīrūnī: Used Indian mathematical and trigonometric methods.
- Ömar Khayyām: Used geometry to solve algebraic problems.
- René Descartes: Formalised the use of two numbers to locate a point in a plane.
💡 Key Idea
A point in a two-dimensional plane can be located using two numbers. These numbers are called its coordinates.
1.2 Settling In
Reiaan and his sister Shalini have moved to a new city. Shalini uses a rectangular grid with pins and threads to help Reiaan understand the positions of objects in his room.
The scale used is 1 cm : 1 foot. Pins mark important points and thick wool connects corners of objects.
Why is this useful?
- It helps represent the room accurately.
- It gives exact positions to objects.
- It can help a person understand directions through touch.
🧠 Think About the Map
The sketch represents the floor of the room. Windows cannot be properly marked on this floor map because windows are positioned vertically on walls, not on the floor.
1.3 The 2-d Cartesian Coordinate System
📐 Two Important Lines
- x-axis: The horizontal line.
- y-axis: The vertical line.
- Origin O: The point where the two axes intersect.
Origin = O = (0, 0)
➕ Positive and Negative Directions
- Right of O → positive x-direction
- Left of O → negative x-direction
- Above O → positive y-direction
- Below O → negative y-direction
📍 Coordinates
A point is generally written as (x, y). The first number is the x-coordinate and the second number is the y-coordinate.
Examples from the chapter:
B = (4.5, 0)
G = (0, −4.5)
H = (0, 4)
B = (4.5, 0)
G = (0, −4.5)
H = (0, 4)
Coordinate Plane – Important Examples
A point P = (x, 0) lies on the x-axis.
A point P = (0, y) lies on the y-axis.
x-axis → (x, 0)
y-axis → (0, y)
Origin → (0, 0)
y-axis → (0, y)
Origin → (0, 0)
Easy Trick:
If the point is on the x-axis, its y-coordinate is always 0.
If the point is on the y-axis, its x-coordinate is always 0.
Exercise Set 1.1
(i) If D1R1 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:
From Fig. 1.3, D1 lies at x = 8 and on the x-axis.
Distance from the y-axis = 8 ft.
Distance from the x-axis = 0 ft.
From Fig. 1.3, D1 lies at x = 8 and on the x-axis.
Distance from the y-axis = 8 ft.
Distance from the x-axis = 0 ft.
(ii) What are the coordinates of D1?
Solution:
D1 is 8 units to the right of the origin and lies on the x-axis. Therefore,
D1 is 8 units to the right of the origin and lies on the x-axis. Therefore,
D1 = (8, 0)
(iii) If R1 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:
D1 = (8, 0) and R1 = (11.5, 0).
Door width = 11.5 − 8 = 3.5 ft.
A 3.5 ft door is a fairly wide room door and should provide comfortable entry for a wheelchair compared with a narrow door.
D1 = (8, 0) and R1 = (11.5, 0).
Door width = 11.5 − 8 = 3.5 ft.
A 3.5 ft door is a fairly wide room door and should provide comfortable entry for a wheelchair compared with a narrow door.
(iv) If B1 (0, 1.5) and B2 (0, 4) represent the ends of the bathroom door, is the bathroom door narrower or wider than the room door?
Solution:
Bathroom door width = 4 − 1.5 = 2.5 ft.
Room door width = 3.5 ft.
Therefore, the bathroom door is narrower than the room door.
Bathroom door width = 4 − 1.5 = 2.5 ft.
Room door width = 3.5 ft.
Therefore, the bathroom door is narrower than the room door.
Think and Reflect
1. What are the standard widths for a room door? Look around your home and in school.
Solution:
Door widths can vary depending on the purpose of the room and accessibility requirements. A common room-door width is around 3 ft, while wider doors may be used where easier accessibility is needed.
Door widths can vary depending on the purpose of the room and accessibility requirements. A common room-door width is around 3 ft, while wider doors may be used where easier accessibility is needed.
2. Are the doors in your school suitable for people in wheelchairs?
Sample Answer:
We should check the actual doors in our school. A door should be sufficiently wide and have a suitable approach so that a wheelchair user can enter easily.
We should check the actual doors in our school. A door should be sufficiently wide and have a suitable approach so that a wheelchair user can enter easily.
Cartesian Plane and Four Quadrants
The two coordinate axes divide the plane into four parts called quadrants.
Quadrant I
(+, +)
(+, +)
Quadrant II
(−, +)
(−, +)
Quadrant III
(−, −)
(−, −)
Quadrant IV
(+, −)
(+, −)
Point = (x, y)
x = x-coordinate
y = y-coordinate
x = x-coordinate
y = y-coordinate
Easy Way to Remember
First coordinate → x → horizontal movement.
Second coordinate → y → vertical movement.
Quadrants – Examples
S (3, −5) lies in Quadrant IV.
x = 3 and y = −5.
Q (−5, 3) lies in Quadrant II.
x = −5 and y = 3.
x = 3 and y = −5.
Q (−5, 3) lies in Quadrant II.
x = −5 and y = 3.
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:
S = (3, −5)
Q = (−5, 3)
One possible choice is:
P = (2, 3) in Quadrant I
R = (−2, −3) in Quadrant III
Many other correct coordinates are possible.
S = (3, −5)
Q = (−5, 3)
One possible choice is:
P = (2, 3) in Quadrant I
R = (−2, −3) in Quadrant III
Many other correct coordinates are possible.
Think and Reflect
1. What is the x-coordinate of a point on the y-axis?
Answer: The x-coordinate is 0.
2. Is there a similar generalisation for a point on the x-axis?
Answer: Yes. The y-coordinate of every point on the x-axis is 0.
3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.
Answer:
Yes. They coincide when the two coordinates are equal:
x = y.
For example, (3, 3) and (3, 3) are the same point.
Yes. They coincide when the two coordinates are equal:
x = y.
For example, (3, 3) and (3, 3) are the same point.
4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?
Answer: Yes, the claim is true.
Two ordered pairs are equal only when their corresponding coordinates are equal. Therefore,
(x, y) = (y, x) only when x = y.
Two ordered pairs are equal only when their corresponding coordinates are equal. Therefore,
(x, y) = (y, x) only when x = y.
Exercise Set 1.2
On a graph sheet, mark the x-axis and y-axis and the origin O. Mark points from (−7, 0) to (13, 0) on the x-axis and from (0, −15) to (0, 12) on the y-axis. (Use the scale 1 cm = 1 unit.) Using Fig. 1.5, answer the given questions.
Graph-work instruction:
Draw the x-axis horizontally and the y-axis vertically. Mark O = (0, 0). Use 1 cm = 1 unit and mark the requested ranges exactly.
Draw the x-axis horizontally and the y-axis vertically. Mark O = (0, 0). Use 1 cm = 1 unit and mark the requested ranges exactly.
Exercise Set 1.2 – Solutions
1. Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7).
(i) Where will the fourth foot of the table be?
(ii) Is this a good spot for the table?
(iii) What is the width of the table? The length? Can you make out the height of the table?
(i) Where will the fourth foot of the table be?
(ii) Is this a good spot for the table?
(iii) What is the width of the table? The length? Can you make out the height of the table?
(i)
The three points form three corners of a rectangle. The missing corner is:
(8, 7).
(ii)
Yes, the table can be placed there because the four points form a proper rectangle within the room. Whether it is practically a good spot also depends on the furniture and available space.
(iii)
Horizontal distance = 11 − 8 = 3 ft.
Vertical distance = 9 − 7 = 2 ft.
So the table is 3 ft × 2 ft. Its height cannot be determined from a floor-plan coordinate diagram.
The three points form three corners of a rectangle. The missing corner is:
(8, 7).
(ii)
Yes, the table can be placed there because the four points form a proper rectangle within the room. Whether it is practically a good spot also depends on the furniture and available space.
(iii)
Horizontal distance = 11 − 8 = 3 ft.
Vertical distance = 9 − 7 = 2 ft.
So the table is 3 ft × 2 ft. Its height cannot be determined from a floor-plan coordinate diagram.
2. If the bathroom door has a hinge at B1 and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?
Solution:
The wardrobe starts some distance away from the bathroom door. With the door shown in Fig. 1.5, its swing does not reach the wardrobe.
If the door is made wider, its swing will cover a larger area. A possible improvement would be to change the hinge/opening direction or use a sliding door so that the wardrobe is not obstructed.
The wardrobe starts some distance away from the bathroom door. With the door shown in Fig. 1.5, its swing does not reach the wardrobe.
If the door is made wider, its swing will cover a larger area. A possible improvement would be to change the hinge/opening direction or use a sliding door so that the wardrobe is not obstructed.
3. Look at Reiaan’s bathroom.
(i) What are the coordinates of the four corners O, F,R, and P of the bathroom?
(ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.
(iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
(i) What are the coordinates of the four corners O, F,R, and P of the bathroom?
(ii) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.
(iii) Mark off a 3 ft × 2 ft space for the washbasin and a 2 ft × 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
(i) Coordinates of the bathroom corners:
O = (0, 0)
F = (0, 9)
R = (−6, 9)
P = (−6, 0)
(ii)
SHWR is a trapezium.
One set of coordinates from Fig. 1.5 is:
S = (−6, 6)
H = (−2, 6)
W = (−2, 9)
R = (−6, 9)
(iii)
The exact placement of the washbasin and toilet can vary, provided the required dimensions are maintained and they fit inside the bathroom.
One possible placement:
Washbasin: (−5, 1), (−2, 1), (−2, 3), (−5, 3)
Toilet: (−5, 4), (−2, 4), (−2, 7), (−5, 7)
Other suitable placements are also possible.
O = (0, 0)
F = (0, 9)
R = (−6, 9)
P = (−6, 0)
(ii)
SHWR is a trapezium.
One set of coordinates from Fig. 1.5 is:
S = (−6, 6)
H = (−2, 6)
W = (−2, 9)
R = (−6, 9)
(iii)
The exact placement of the washbasin and toilet can vary, provided the required dimensions are maintained and they fit inside the bathroom.
One possible placement:
Washbasin: (−5, 1), (−2, 1), (−2, 3), (−5, 3)
Toilet: (−5, 4), (−2, 4), (−2, 7), (−5, 7)
Other suitable placements are also possible.
4. Other rooms in the house:
(i) 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.
(ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
(i) 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.
(ii) Place a rectangular 5 ft × 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
(i)
From Fig. 1.5, P = (−6, 0) and A = (12, 0). The dining room has width 15 ft, so the upper corners are:
(−6, 15) and (12, 15).
Thus the four corners are:
P = (−6, 0), A = (12, 0), (12, 15), (−6, 15).
(ii)
The centre of the dining room is:
x = (−6 + 12)/2 = 3
y = 15/2 = 7.5
For a 5 ft × 3 ft table, half-lengths are 2.5 ft and 1.5 ft.
One valid set of corner coordinates is:
(0.5, 6), (5.5, 6), (5.5, 9), (0.5, 9).
From Fig. 1.5, P = (−6, 0) and A = (12, 0). The dining room has width 15 ft, so the upper corners are:
(−6, 15) and (12, 15).
Thus the four corners are:
P = (−6, 0), A = (12, 0), (12, 15), (−6, 15).
(ii)
The centre of the dining room is:
x = (−6 + 12)/2 = 3
y = 15/2 = 7.5
For a 5 ft × 3 ft table, half-lengths are 2.5 ft and 1.5 ft.
One valid set of corner coordinates is:
(0.5, 6), (5.5, 6), (5.5, 9), (0.5, 9).
1.4 Distance Between Two Points in the 2-D Plane
📏 Distance Along the Axes
If two points have the same y-coordinate, their distance is the difference between their x-coordinates.
If two points have the same x-coordinate, their distance is the difference between their y-coordinates.
📐 When the Line Is Not Parallel to an Axis
We can use the Baudhāyana–Pythagoras Theorem. The horizontal and vertical changes form the two perpendicular sides of a right triangle.
Distance = √[(x₂ − x₁)² + (y₂ − y₁)²]
Example: A(3, 4), D(7, 1)
Horizontal distance:
7 − 3 = 4
Vertical distance:
4 − 1 = 3
Therefore:
AD = √(4² + 3²)
AD = √25
AD = 5 units
7 − 3 = 4
Vertical distance:
4 − 1 = 3
Therefore:
AD = √(4² + 3²)
AD = √25
AD = 5 units
More Distance Examples
For D(7,1) and M(9,6):
DM = √[(9−7)² + (6−1)²]
= √(2² + 5²)
= √29 units.
DM = √[(9−7)² + (6−1)²]
= √(2² + 5²)
= √29 units.
For M(9,6) and A(3,4):
MA = √[(9−3)² + (6−4)²]
= √(6² + 2²)
= √40 units.
MA = √[(9−3)² + (6−4)²]
= √(6² + 2²)
= √40 units.
⭐ General Distance Formula
For P(x₁,y₁) and Q(x₂,y₂):
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]
It does not matter whether the changes along the axes are positive or negative because distance is always non-negative.
Distance with Negative Coordinates
The same distance formula works when some or all coordinates are negative.
Reflection in the y-axis
A(3,4) becomes A′(−3,4)
M(9,6) becomes M′(−9,6)
D(7,1) becomes D′(−7,1)
M(9,6) becomes M′(−9,6)
D(7,1) becomes D′(−7,1)
Reflection Rule
When a point is reflected in the y-axis, the x-coordinate changes its sign while the y-coordinate remains unchanged.
(x,y) → (−x,y)
For A′(−3,4) and D′(−7,1):
Horizontal distance = −3 − (−7) = 4
Vertical distance = 4 − 1 = 3
Therefore:
A′D′ = √(4² + 3²) = 5 units.
Horizontal distance = −3 − (−7) = 4
Vertical distance = 4 − 1 = 3
Therefore:
A′D′ = √(4² + 3²) = 5 units.
🔍 What does reflection preserve?
Reflection changes the positions of the points, but it preserves the lengths of corresponding sides.
Think and Reflect
1. What has remained the same and what has changed with this reflection?
Answer:
The x-coordinates change sign. The y-coordinates remain the same. The side lengths and shape of the triangle remain the same.
The x-coordinates change sign. The y-coordinates remain the same. The side lengths and shape of the triangle remain the same.
2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?
Answer:
Yes. Reflection in the x-axis changes the sign of the y-coordinate while the x-coordinate remains unchanged. The lengths and shape remain unchanged.
Yes. Reflection in the x-axis changes the sign of the y-coordinate while the x-coordinate remains unchanged. The lengths and shape remain unchanged.
End-of-Chapter Exercises
1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Answer:
The intersection of the axes is the origin.
O = (0, 0)
Therefore, x-coordinate = 0 and y-coordinate = 0.
The intersection of the axes is the origin.
O = (0, 0)
Therefore, x-coordinate = 0 and y-coordinate = 0.
2. 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?
Answer:
Since H lies on a line parallel to the y-axis through W, its x-coordinate remains −5.
So H has the form:
H = (−5, y).
If y > 0, H lies in Quadrant II.
If y < 0, H lies in Quadrant III.
If y = 0, H lies on the negative x-axis.
Since H lies on a line parallel to the y-axis through W, its x-coordinate remains −5.
So H has the form:
H = (−5, y).
If y > 0, H lies in Quadrant II.
If y < 0, H lies in Quadrant III.
If y = 0, H lies on the negative x-axis.
3. 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:
(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.
(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.
(i)
AM is horizontal and MP is vertical. Therefore, AM ⟂ MP.
(ii)
AM is parallel to the x-axis. Also, MP is parallel to the y-axis.
(iii)
M = (−5, −2) and P = (−5, 2). Their x-coordinates are equal and their y-coordinates are opposites. Therefore M and P are mirror images in the x-axis.
AM is horizontal and MP is vertical. Therefore, AM ⟂ MP.
(ii)
AM is parallel to the x-axis. Also, MP is parallel to the y-axis.
(iii)
M = (−5, −2) and P = (−5, 2). Their x-coordinates are equal and their y-coordinates are opposites. Therefore M and P are mirror images in the x-axis.
4. 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.)
One possible construction:
Take I = (5, 0) and N = (0, −6).
IZ = 6 units
ZN = 5 units
IN = √(6² + 5²) = √61 units
So the three sides are: 5 units, 6 units and √61 units.
Take I = (5, 0) and N = (0, −6).
IZ = 6 units
ZN = 5 units
IN = √(6² + 5²) = √61 units
So the three sides are: 5 units, 6 units and √61 units.
5. 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?
Answer:
Without negative numbers, coordinates could not represent points to the left of the y-axis or below the x-axis.
Therefore, such a system would not allow us to locate all points on a 2-D plane. It would mainly cover the first quadrant and the axes.
Without negative numbers, coordinates could not represent points to the left of the y-axis or below the x-axis.
Therefore, such a system would not allow us to locate all points on a 2-D plane. It would mainly cover the first quadrant and the axes.
*6. 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.
Solution:
Use distances or compare slopes.
MA = √[3² + 4²] = 5
AG = √[6² + 8²] = 10
MG = √[9² + 12²] = 15
Since:
MA + AG = MG
5 + 10 = 15
the three points lie on the same straight line.
Use distances or compare slopes.
MA = √[3² + 4²] = 5
AG = √[6² + 8²] = 10
MG = √[9² + 12²] = 15
Since:
MA + AG = MG
5 + 10 = 15
the three points lie on the same straight line.
*7. 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.
Solution:
RB = √[(−2+5)² + (−5+1)²]
= √(3² + (−4)²)
= 5
BC = √[(4+2)² + (−12+5)²]
= √(6² + (−7)²)
= √85
RC = √[(4+5)² + (−12+1)²]
= √(9² + (−11)²)
= √202
Since RB + BC ≠ RC, the three points are not on the same straight line.
RB = √[(−2+5)² + (−5+1)²]
= √(3² + (−4)²)
= 5
BC = √[(4+2)² + (−12+5)²]
= √(6² + (−7)²)
= √85
RC = √[(4+5)² + (−12+1)²]
= √(9² + (−11)²)
= √202
Since RB + BC ≠ RC, the three points are not on the same straight line.
*8. Using the origin as one vertex, plot the vertices of:
(i) A right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
(i) A right-angled isosceles triangle.
(ii) An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
(i) One possible answer:
Take O = (0,0), A = (4,0), B = (0,4).
OA = OB = 4 and angle AOB = 90°.
So it is a right-angled isosceles triangle.
(ii) One possible answer:
Take O = (0,0), A = (−2,−2), B = (2,−2).
A lies in Quadrant III and B lies in Quadrant IV.
OA = OB = 2√2.
Hence it is isosceles.
Take O = (0,0), A = (4,0), B = (0,4).
OA = OB = 4 and angle AOB = 90°.
So it is a right-angled isosceles triangle.
(ii) One possible answer:
Take O = (0,0), A = (−2,−2), B = (2,−2).
A lies in Quadrant III and B lies in Quadrant IV.
OA = OB = 2√2.
Hence it is isosceles.
*9. 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.
| S | M | T | Is M the midpoint of ST? Yes or No | Reason for your answer |
|---|---|---|---|---|
| (−3, 0) | (0, 0) | (3, 0) | Yes | Midpoint = ((−3+3)/2, (0+0)/2) = (0,0). |
| (2, 3) | (3, 4) | (4, 5) | Yes | Midpoint = ((2+4)/2, (3+5)/2) = (3,4). |
| (0, 0) | (0, 5) | (0, −10) | No | Midpoint = (0,−5), not (0,5). |
| (−8, 7) | (0, −2) | (6, −3) | No | Midpoint = (−1,2), not (0,−2). |
Connection:
When M is the midpoint of ST:
M’s x-coordinate = average of S’s and T’s x-coordinates.
M’s y-coordinate = average of S’s and T’s y-coordinates.
So,
M = ((x₁+x₂)/2, (y₁+y₂)/2).
When M is the midpoint of ST:
M’s x-coordinate = average of S’s and T’s x-coordinates.
M’s y-coordinate = average of S’s and T’s y-coordinates.
So,
M = ((x₁+x₂)/2, (y₁+y₂)/2).
*10. 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:
M = ((xA + xB)/2, (yA + yB)/2)
−7 = (3 + x)/2
−14 = 3 + x
x = −17
1 = (−4 + y)/2
2 = −4 + y
y = 6
Therefore:
M = ((xA + xB)/2, (yA + yB)/2)
−7 = (3 + x)/2
−14 = 3 + x
x = −17
1 = (−4 + y)/2
2 = −4 + y
y = 6
Therefore:
B = (−17, 6)
*11. 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:
A = (4,7), B = (16,−2).
Change in x = 16−4 = 12.
One-third of this change = 4.
Change in y = −2−7 = −9.
One-third = −3.
P is one-third of the way from A to B:
P = (4+4, 7−3)
P = (8,4)
Q is two-thirds of the way from A to B:
Q = (4+8, 7−6)
Q = (12,1)
A = (4,7), B = (16,−2).
Change in x = 16−4 = 12.
One-third of this change = 4.
Change in y = −2−7 = −9.
One-third = −3.
P is one-third of the way from A to B:
P = (4+4, 7−3)
P = (8,4)
Q is two-thirds of the way from A to B:
Q = (4+8, 7−6)
Q = (12,1)
*12. (i) 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?
(ii) 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.
(ii) 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.
(i)
Distance of A from O:
OA = √(1² + (−8)²) = √65.
OB = √((−4)² + 7²) = √65.
OC = √((−7)² + (−4)²) = √65.
Therefore OA = OB = OC = √65.
Hence A, B and C lie on a circle with centre O and radius:
√65 units.
(ii)
OD = √((−5)² + 6²) = √61.
Since √61 < √65, D lies inside the circle.
OE = √(0² + 9²) = 9.
Since 9 > √65, E lies outside the circle.
Distance of A from O:
OA = √(1² + (−8)²) = √65.
OB = √((−4)² + 7²) = √65.
OC = √((−7)² + (−4)²) = √65.
Therefore OA = OB = OC = √65.
Hence A, B and C lie on a circle with centre O and radius:
√65 units.
(ii)
OD = √((−5)² + 6²) = √61.
Since √61 < √65, D lies inside the circle.
OE = √(0² + 9²) = 9.
Since 9 > √65, E lies outside the circle.
*13. 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.
Solution:
Let:
D be midpoint of BC.
E be midpoint of CA.
F be midpoint of AB.
Using the midpoint relationships:
A = E + F − D
B = D + F − E
C = D + E − F
A = (6,5) + (0,3) − (5,1)
A = (1,7)
B = (5,1) + (0,3) − (6,5)
B = (−1,−1)
C = (5,1) + (6,5) − (0,3)
C = (11,3)
Let:
D be midpoint of BC.
E be midpoint of CA.
F be midpoint of AB.
Using the midpoint relationships:
A = E + F − D
B = D + F − E
C = D + E − F
A = (6,5) + (0,3) − (5,1)
A = (1,7)
B = (5,1) + (0,3) − (6,5)
B = (−1,−1)
C = (5,1) + (6,5) − (0,3)
C = (11,3)
14. 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.
(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
(ii) 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).
(i) Using 1 cm = 200 m, draw a model of the city in your notebook. Represent the roads/streets by single lines.
(ii) 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).
(i)
Use a scale of 1 cm = 200 m. Draw the N–S streets vertically and E–W streets horizontally. Keep adjacent streets 1 cm apart.
(ii)
An intersection is identified by the street numbers in the two directions. Therefore:
(a) (4,3) identifies one specific intersection.
Answer: 1
(b) (3,4) also identifies one specific intersection.
Answer: 1
Use a scale of 1 cm = 200 m. Draw the N–S streets vertically and E–W streets horizontally. Keep adjacent streets 1 cm apart.
(ii)
An intersection is identified by the street numbers in the two directions. Therefore:
(a) (4,3) identifies one specific intersection.
Answer: 1
(b) (3,4) also identifies one specific intersection.
Answer: 1
15. 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.
(i) whether any part of either circle lies outside the screen.
(ii) whether the two circles intersect each other.
(i) Circle A
Centre = (100,150), radius = 80.
Left = 100−80 = 20
Right = 100+80 = 180
Bottom = 150−80 = 70
Top = 150+80 = 230
All are within 0 to 800 horizontally and 0 to 600 vertically. So Circle A is completely inside the screen.
Circle B
Centre = (250,230), radius = 100.
Left = 150
Right = 350
Bottom = 130
Top = 330
Circle B is also completely inside the screen.
Answer: No part of either circle lies outside the screen.
(ii)
Distance between centres:
AB = √[(250−100)² + (230−150)²]
= √(150² + 80²)
= √28900
= 170 pixels.
Sum of radii = 80 + 100 = 180 pixels.
Difference of radii = 100 − 80 = 20 pixels.
Since:
20 < 170 < 180
the two circles intersect each other.
Centre = (100,150), radius = 80.
Left = 100−80 = 20
Right = 100+80 = 180
Bottom = 150−80 = 70
Top = 150+80 = 230
All are within 0 to 800 horizontally and 0 to 600 vertically. So Circle A is completely inside the screen.
Circle B
Centre = (250,230), radius = 100.
Left = 150
Right = 350
Bottom = 130
Top = 330
Circle B is also completely inside the screen.
Answer: No part of either circle lies outside the screen.
(ii)
Distance between centres:
AB = √[(250−100)² + (230−150)²]
= √(150² + 80²)
= √28900
= 170 pixels.
Sum of radii = 80 + 100 = 180 pixels.
Difference of radii = 100 − 80 = 20 pixels.
Since:
20 < 170 < 180
the two circles intersect each other.
16. 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?
Solution:
Calculate the four side lengths.
AB = √[(−1−2)² + (2−1)²]
= √(9+1)
= √10
BC = √[(−2+1)² + (−1−2)²]
= √(1+9)
= √10
CD = √[(1+2)² + (−2+1)²]
= √(9+1)
= √10
DA = √[(2−1)² + (1+2)²]
= √(1+9)
= √10
All four sides are equal. The diagonals are also equal:
AC = √20
BD = √20
Therefore, ABCD is a square.
Side² = (√10)² = 10.
Therefore:
Area = 10 square units.
Calculate the four side lengths.
AB = √[(−1−2)² + (2−1)²]
= √(9+1)
= √10
BC = √[(−2+1)² + (−1−2)²]
= √(1+9)
= √10
CD = √[(1+2)² + (−2+1)²]
= √(9+1)
= √10
DA = √[(2−1)² + (1+2)²]
= √(1+9)
= √10
All four sides are equal. The diagonals are also equal:
AC = √20
BD = √20
Therefore, ABCD is a square.
Side² = (√10)² = 10.
Therefore:
Area = 10 square units.
Chapter Summary
- To locate the position of an object or a point in a plane, we require two perpendicular lines.
- The plane is called the Cartesian plane, coordinate plane or xy-plane.
- The horizontal line is called the x-axis.
- The vertical line is called the y-axis.
- The coordinate axes divide the plane into four quadrants.
- The point of intersection of the axes is called the origin.
- The distance of a point from the y-axis is its x-coordinate.
- The distance of a point from the x-axis is its y-coordinate.
- The coordinates of a point are written as (x, y).
- A point on the x-axis has coordinates (x, 0).
- A point on the y-axis has coordinates (0, y).
- The origin is (0, 0).
- Quadrant I → (+, +).
- Quadrant II → (−, +).
- Quadrant III → (−, −).
- Quadrant IV → (+, −).
- If x = y, then (x, y) = (y, x).
- If x ≠ y, then (x, y) ≠ (y, x).
- Distance between (x₁,y) and (x₂,y) is |x₂−x₁|.
- Distance between (x,y₁) and (x,y₂) is |y₂−y₁|.
Distance between (x₁,y₁) and (x₂,y₂)
= √[(x₂ − x₁)² + (y₂ − y₁)²]
= √[(x₂ − x₁)² + (y₂ − y₁)²]
⭐ Quick Revision
| Concept | Remember |
|---|---|
| Origin | (0,0) |
| x-axis | (x,0) |
| y-axis | (0,y) |
| Quadrant I | (+, +) |
| Quadrant II | (−, +) |
| Quadrant III | (−, −) |
| Quadrant IV | (+, −) |
| Distance | √[(x₂−x₁)² + (y₂−y₁)²] |
🧠 Easy Memory Trick
First x, then y.
Move left/right first according to x, then move up/down according to y.
🎯 Chapter 1 Complete
Orienting Yourself: The Use of Coordinates
Grade 9 Mathematics | Ganita Manjari | Part I