M 9101. Orienting Yourself: The Use of Coordinates

Orienting Yourself: The Use of Coordinates

The 2-D Cartesian Coordinate System

Cartesian plane –> The Cartesian plane is a two-dimensional coordinate system used in mathematics to locate points and graph equations.

It was named after the mathematician , who helped develop this system.

We studied about the number line which is one-dimensional. For Cartesian system we took two number lines on a plane. This number line maybe horizontal and vertical.

We know about the number line and a plane in mathematics, both of them has no end and they are spreaded in all directions infinitively. In the ‘Number System’ the distances from a fixed point are marked in equal units positively in one direction and negatively in the other on the number line. The fixed point from which the distances are marked is called the origin.

The two-dimensional coordinate system uses two lines one horizontal and one vertical at right angles to each other joined 0 to 0 marked 'O'.

Exercise Set 1.1VM

1. Draw the Number Lines:

(A). Draw Horizontal Number lines starting with –5 to + 5.

(B). Draw Horizontal Number lines starting with –7 to + 7.

(C). Draw Horizontal Number lines starting with –9 to + 9.

(D) Draw Vertical Number lines starting with –6 to + 6.

(E) Draw Vertical Number lines starting with –7 to + 7.

(F) Draw Vertical Number lines starting with –8 to + 8.

2. Show the following points on the Horizontal Number Line separately.

A. (1) B. (4.5) C. (–3) D. (–1)

E. (–2) F. (7) G. (– 5) H. (5)

I. (7) J. (–9) K. (3.5) L. (–2.5)

M. (–2.3) N. (7.4) O. (– 5.5) P. (5)

3. Show the following points on the Vertical Number Line separately.

A. (1) B. (4.5) C. (–3) D. (–1)

E. (–2) F. (7) G. (– 5) H. (5)

I. (7) J. (–9) K. (3.5) L. (–2.5)

M. (–2.3) N. (7.4) O. (– 5.5) P. (5)

The Horizontal Number line is called x-axis or Abscissa and Vertical Number Line is called y-axis or Ordinates.

The plane in which the axes are situated is called the Cartesian Plane, the Coordinate Plane or the xy-plane.

Distances are marked off in equal units, on both the axes. Mind that the fistances to the right of O or upwards from O are considered positive, and distances to the left of O or downwards from O are considered negative.

For Cartesian system we took two number lines on a plane. The Horizontal Number line is called x-axis or Abscissa and Vertical number line is called x-axis or Ordinates. Now combine both the lines in such a way that the two lines cross each other at their zeroes, or origins at right angles. It gives a cross point like the above figure. The point of intersection of the x-axis and y-axis is called the origin O. its coordinates are (0, 0).

Coordinate axes (the plural of ‘axis’) help us to locate any point in 2-D space using the point’s ‘coordinates’. [2-D space –> two-dimensional space].

Since the positive numbers lie on the directions OX and OY, so OX and OY are called the positive directions of the x - axis and the y - axis, respectively. Similarly, OX' and OY' are called the negative directions of the x - axis and the y - axis, respectively.

x-axis, y-axis and origin

★ x-axis or Abscissa –> In a Cartesian plane the the horizontal lines XOX', is called x-axis or Abscissa.

★ y-axis or ordinate. –> In a Cartesian plane the vertical line YOY' is called y-axis or ordinate.

★ Origin –> The point of intersection of the x-axis and y-axis is called the origin and denoted by 'O'. Its coordinates are (0, 0).

Exercise Set 1.2 VM

1. See Fig. and complete the following statements:

(i) The abscissa and the ordinate of the point B are ___ and ___ respectively. Hence, the coordinates of B are (__, ___)

(ii) The x-coordinate and the y-coordinate of the point M are ___ and ___ respectively. Hence, the coordinates of M are ( ___ , ___).

(iii) The x-coordinate and the y-coordinate of the point L are ___ and ___, respectively. Hence, the coordinates of L are (___ , ___ ).

(iv) The x-coordinate and the y-coordinate of the point S are and __, respectively. Hence, the coordinates of S are (___ , ___ )

नीचे दी गई आकृति को देखकर निम्नलिखित कथनों को पूरा कीजिएः

(i) बिन्दु B का भुज और कोटि क्रमशः ___ और ___ हैं। अतः B के निर्देशांक (___ , ___ ) हैं।

(ii) बिन्दु M के x-निर्देशांक और y-निर्देशांक क्रमशः ___ और ___ हैं। अतः M के निर्देशांक (___ , ___ ) हैं।

(iii) बिन्दु L के x-निर्देशांक और y-निर्देशांक क्रमशः ___ और ___ हैं। अतः L के निर्देशांक (___ , ___ ) हैं।

(iv) बिन्दु S के x-निर्देशांक और y-निर्देशांक क्रमशः ___ और _____ हैं। अतः S के निर्देशांक (___ , ___ ) हैं।

(2) Write the equation of x axis, y axis and coordinates of the origin.

x-अक्ष तथा y-अक्ष की समीकरण लिखो तथा मूल बिंदू के निर्देशांक भी लिखो ।

(3): Write the coordinates of the points marked on the axes in Fig.

(आकृति में अक्षों पर अंकित बिन्दुओं के निर्देशांक लिखिएः)

Sol. : You can see that :	हल: आप यहाँ देख सकते हैं कि :
(i) The point A is at a distance of + 4 units from the y - axis and at a distance zero from the x - axis. Therefore, the x - coordinate of A is 4 and the y - coordinate is 0. Hence, the coordinates of A are (4, 0).	(i) बिन्दु A, y-अक्ष +4 एकक की दूरी पर है और x-अक्ष से दूरी 0 पर है। अतः बिन्दु A का x-निर्देशांक 4 है और -निर्देशांक 0 है। इसलिए A के निर्देशांक (4,0) हैं।
(ii) The coordinates of B are (0, 3). Why?	(ii) B के निर्देशांक (0, 3) हैं। क्यों?
(iii) The coordinates of C are (– 5, 0). Why?	(iii) C के निर्देशांक (5,0) हैं। क्यों?
(iv) The coordinates of D are (0, – 4). Why?	(iv) D के निर्देशांक (0,4) हैं। क्यों?
(v) The coordinates of E are ( ⅔, 0) . Why?	(v) E के निर्देशांक (⅔, 0) है। क्यों ?

(4). See Fig.3.14, and write the following:

1. The coordinates of B.

2. The coordinates of C.

3. The point identified by the coordinates (-3, -5).

4. The point identified by the coordinates (2, -4).

5. The abscissa of the point D.

6. The ordinate of the point H.

7. The coordinates of the point L.

8. The coordinates of the point M.

(5). Write the answer of each of the following questions:

(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?

(ii) What is the name of each part of the plane formed by these two lines?

(iii) Write the name of the point where these two lines intersect.

(6). How will you described the position of a table lamp on your table to another person.

(7) What is abscissa and ordinate of the point ?

(8). Write the name of the point in the Cartesian plane where the horizontal and the vertical lines intersect and what is the name of each part of the plane formed by these two lines.

(9). In which quadrant do each of the points lie and very your answer by location them on the Cartesian plane.

(10). In which quadrant do each of the points lie and very your answer by location them on the Cartesian plane.

(11). From the given graph, Write:

(a) The coordinates of the Points B and F.

(b) The abscissa of points D and H.

(d) The perpendicular distance of the points G from the x-axis.

दिए हुए ग्राफ से लिखिएः

(क) बिन्दु B और F के निर्देशांक ।

(ख) बिन्दु D और H के भुज।

(ग) बिन्दु A और C के कोटि ।

(घ) बिन्दु G को x- अक्ष से लाम्बिक दूरी।

Quadrants: - We observe that either x-axis or Abscissa and y-axis or Ordinates axes (plural of the word ‘axis’) divide the plane into four parts.

These axes divide the cartisan plane into four parts, called quadrants.

These all four parts are called the quadrants (one fourth part of a whole), numbered I, II, III and IV anticlockwise from OX.

* In first quadrant x and y both are positive as (x, y)

* In Second quadrant x is Negative and y is positive as (– x, y)

* In Third Quadrant x and y both are Negative as (– x, – y)

* In forth Quadrant is negative and y is positive as (x, – y)

* प्रथम तल में x और y दोनो धनात्मक होते हैं।

* दितीय तल में x ऋणत्मक तथा y थनात्मक होता है।* तृतीय तल में x और y दोनो ऋणात्मक होते हैं।

* चतुर्थ तल में x धनात्मक तथा y ऋणत्मक होता है।

The plane is divided into 4 sections:

Quadrant I: (+, +)

Quadrant II: (−, +)

Quadrant III: (−, −)

Quadrant IV: (+, −)

What are coordinates? How they are written?

★ Coordinates –> Co-ordinate is a number denoted by Abscissa and Ordinate in a bracket separated by comma.

This Coordinates are numbers that show the exact position of a point on the Cartesian plane.

They are written in the form: (x, y)

Here:

The first number is called the x-coordinate

The second number is called the y-coordinate

Coordinates are always written:

* inside brackets

* separated by a comma

* Write the x-coordinate first and the y-coordinate second

Q. What did the value of co-ordinate (x, y) tell or show to us?

x-value tells us how far left or right you move

y-value tells us how far up or down you move

👉 Example: The point (3, 2) means:

Here:

3 is the x-coordinate which is positive.

2 is the y-coordinate which is also positive.

Move 3 units right (x = 3)

Move 2 units up (y = 2)

Plotting the coordinates: - In the coordinate geometry the both axis are written in bracket [like(x,y)] and the first number shows x and the second number shows the y. First of all count x with sign on the Horizontal number line from origin and then count y upward or downward According to their sign on the vertical number line.

Exercise Set 1.3 VM

(1) Name the quadrant in which the point lies :

A. (1,1) B. (2,4) C. (–3, – 10)

D. (–1, 2) E. (1, –2) F. (–3,10)

G. (–2, – 4) H. (5,4) I. (7, – 4)

J. (–1, – 3). K. (–1, 5) L. (2, –2)

M. (–3, –3)

(2). In which quadrant or axis do the following points lie?

(a) (1996, 1965) (b) (1997,0)

(3) Plots or locate the points on the graph paper:

A. (1,1) B. (2,4) C. (–3, – 10)

D. (–1, 2) E. (1, –2) F. (–3,10)

G. (–2, – 4) H. (5,4) I. (7, – 4)

J. (–1, – 3) K. (–1, 5) L. (2, –2)

M. (–3, –3)

(4). Locate the points (5, 0), (0, 5), (2, 5), (5, 2), (–3, 5), (–3, –5), (5, –3) and (6, 1) in the Cartesian plane.

Sol.

Draw the figures with the help of the coordinates

Exercise Set 1.4 VM

1. Draw the figures with the help of the coordinates.

(1) Plot the points A. (2,0), and В. (2,2) on cartesian plan and join the points A and B what figure do you obtain?

इन A. (2,0), और B. (2,2) बिन्दुओं को ग्राफ पर दर्शाओं तथा बिन्दु A और B को जोड़ो । इससे कौन सी आकृति बनती है।

(2) Plot the points M. (2,2), and N. (5,5) on cartesian plan and join the points M and N what figure do you obtain?

इन M. (2,2), और N. (5,5) बिन्दुओं को ग्राफ पर दर्शाओं तथा बिन्दु M और N को जोड़ो । इससे कौन सी आकृति बनती है।

(3) Plot the points X. (2,3), and Y. (5,6) on cartesian plan and join the points X and Y what figure do you obtain?

इन X. (2, 3), और Y. (5, 6) बिन्दुओं को ग्राफ पर दर्शाओं तथा बिन्दु X और Y को जोड़ो । इससे कौन सी आकृति बनती है।

(4) Plot the points M. (2,2), and N. (–2, –2) on cartesian plan and join the points M and N what figure do you obtain?

इन M. (2,2), और N. (–2, –2) बिन्दुओं को ग्राफ पर दर्शाओं तथा बिन्दु M और N को जोड़ो । इससे कौन सी आकृति बनती है।

(5) Plot the points X. (2,–3), and Y. (–5,6) on cartesian plan and join the points X and Y what figure do you obtain?

इन X. (2, –3), और Y. (–5, 6) बिन्दुओं को ग्राफ पर दर्शाओं तथा बिन्दु X और Y को जोड़ो । इससे कौन सी आकृति बनती है।

(6) Plots the points A. (4,4) B. (-4,4) and join the line OA, OB and BA what figure do you obtain?

बिन्दु A. (4,4), B. (-4,4) को ग्राफ पर दर्शाओं तथा रेखाखंड OA, AB व BO खींचो । इससे कौन सी आकृति बनती है।

2. Plot the following points at the various quadratic of graph with help of coordinate geometry.

निम्न बिंदुओं को समकोणिक निर्देशांक पद्धति के अनुसार विभिन्न चर्तुथांशों में व्यक्त कीजिए।

1) (3, 2), (–2,–3) and (2, 3)

2) (1, 5), (2, 3) and (–2, 11)

3) (1, 7), (4, 2), (–1,–1) and (– 4, 4)

4) (–1, 2), (1, 0), (1, 2) and (–3, 0)

5) (2, 3), (4, 1) (–5, 7) and (1, –3)

6) (–3, 5), (3, 1), (0, 3) and (–1,–4)

Exercise Set 1.1

Figure shows Reiaan’s room with points OABC marking its corners. The x- and y-axes are marked in the figure. Point O is the origin.

Referring to Figure, answer the following questions:

(i) If D₁R₂ 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?

Sol.

(i) The door is represented by points and .

Distance from the left wall (y-axis) = the x-coordinate of the point.

Distance from the x-axis = the y-coordinate of the point.

If D₁ = (10, 0), then:

Distance from the y-axis = 10 units

Distance from the x-axis = 0 units

(ii) What are the coordinates of D₁ ?

Sol. (ii) The coordinates of are: (10, 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?

Sol.

(iii) Given:

R₁ = (11.5, 0)

and

D₁ = (10, 0)

Width of the door:

= 11.5 - 10 = 1.5

So, the door is 1.5 units wide.

This is a comfortable width for a room door. A wheelchair user can also enter easily because wheelchairs usually need a wider doorway.

(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?

Sol.

(iv) Bathroom door endpoints:

B₁ = (0, 1.5)

B₂ = (0, 4)

Width of bathroom door:

= 4 – 1.5 = 2.5

So, the bathroom door is wider than the room door because:

2.5 > 1.5

Think and Reflect

1. What are the standard widths for a room door?

Sol. The standard width of a room door is usually about 0.8 m to 1 m. Main doors are often wider.

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

Sol. Many school doors are wide enough for wheelchairs, but some narrow doors may make movement difficult. Schools should have wider doors and ramps to help wheelchair users move comfortably.!

Think and Reflect

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

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

3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

4. If x ≠ y, then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?

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.

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?

Sol.

1. Study Table

Three feet of the table are at:

(8,9), (11,9), (11,7)

(i) Where will the fourth foot of the table be?

Since it is a rectangle:

x-coordinate matches the left side →

y-coordinate matches the bottom side →

So the fourth foot is at:

(8,7)

(ii) Is this a good spot for the table?

Yes, it appears to fit properly in the room without crossing walls or blocking space.

(iii) What is the width and length of the table?Horizontal distance:

11 – 8 = 3 ft

Vertical distance:

9 – 7 = 2 ft

Sol.

Length = 3 ft

Width = 2 ft

The height cannot be known from the graph because the graph only shows length and width, not vertical height.

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?

Sol.

Bathroom Door

If the bathroom door opens inward from hinge , it may hit the wardrobe if there is not enough space between them.

If the door is made wider, it may create more obstruction.

Suggested change:

Make the door open outward, or

Shift the wardrobe slightly away.

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.

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.

Think and Reflect

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

Sol.

The x-coordinate of every point on the y-axis is 0.

Example: (0, 5), (0,–3), (0, 10)

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

Sol.

Yes.

The y-coordinate of every point on the x-axis is 0.

Example: (4, 0), (–7, 0), (12, 0)

3. Does point Q (y, x) ever coincide with point P (x, y)? Justify your answer.

Sol.

Yes, the two points coincide only when:

x = y

For example:

P(3,3)\ \text{and}\ Q(3,3)

Both are the same point.

But if , then the points are different.

Example: P(2, 5), Q(5, 2)

These are different points.

4. If , then ; and if and only if . Is this claim true?

Yes, this claim is true.

Because two ordered pairs are equal only when their corresponding coordinates are equal.

So:

(x, y) = (y, x)

means

x=y and y=x

Therefore, the statement is correct.

Exercise Set 1.2

3. Reiaan’s Bathroom (The exact coordinates depend on Fig. 1.5, which is not visible here. So only the method can be explained.)

(i) Coordinates of corners and

Read the coordinates directly from the graph where these points are marked.

(ii) Shape of shower area SHWR

If opposite sides are equal and angles are right angles, then SHWR is a rectangle.

Write the coordinates of points and from the graph.

(iii) Washbasin and Toilet Spaces

rectangle can represent the washbasin area.

rectangle can represent the toilet area.

Mark suitable rectangles inside the bathroom and note their corner coordinates.

4. Other Rooms in the House

(i) Dining Room

Length = ft

Width = ft

Suppose point is one corner and represents the length.

Then the dining room will be a rectangle.

Example coordinates (depending on the graph):

P(0,0), A(18,0) B(18,15), C(0,15)

(ii) Dining Table at the Centre

Table size:

5 ft × 3 ft

The centre of the dining room:

Left(18/2, 15/2) = (9,7.5)

Now place the rectangle equally around the centre.

Possible coordinates of table feet:

(6.5,6), (11.5,6), (11.5,9), (6.5,9)

These form a rectangle at the centre.

Exercise Set 1.2 (Using Fig. 1.5)

1. Study Table

The three feet of the rectangular study table are at:

(8,9), (11,9), (11,7)

(i) Where will the fourth foot be?

Since opposite sides of a rectangle are parallel and equal:

x-coordinate from the left side =

y-coordinate from the lower side =

So, the fourth foot is at:

(8,7)

(ii) Is this a good spot for the table?

Yes.

The table is placed properly near the wall and does not block the door or wardrobe.

(iii) What is the width and length of the table?

Horizontal distance:

11 – 8 = 3 ft

Vertical distance:

9 – 7 = 2 ft

So,

Length = ft

Width = ft

The height cannot be determined from the graph because the graph only shows length and width.

2. Bathroom Door

The bathroom door has hinge at and opens into the bedroom.

Yes, if the door opens too wide, it may hit the wardrobe.

Suggested change:

Make the door open outward, or

Move the wardrobe slightly away from the door.

3. Reiaan’s Bathroom

(i) Coordinates of corners and

From the figure:

O = (0, 0)

F = (0, 9)

R = (– 6, 9)

P = (–6, 0)

(ii) Shape of showering area

The showering area is a trapezium.

Coordinates are:

S = (–6, 6)

H = (–3, 6)

W = (–2, 9)

R = (–6, 9)

(iii) Washbasin and Toilet Spaces

One possible arrangement is:

Washbasin (3 ft × 2 ft)

Corners:

(–6, 0), (–3, 0), (–3, 2), (–6, 2)

Toilet (2 ft × 3 ft)

Corners:

(–6, 3), (–4, 3), (–4, 6), (–6, 6)

4. Other Rooms in the House

(i) Dining Room

The dining room extends from point to point

Given:

Length = ft

Width = ft

Coordinates of corners are:

P = (–6, 0)

A = (12, 0)

B = (12, 15)

C = (–6, 15)

(ii) Dining Table at the Centre

Dining table size:

5 ft × 3 ft

The centre of the dining room is:

left(–6 + 12)/2, (0+15)/2 right

= (3,7.5)

A possible placement of the table exactly at the centre is:

(0.5,6), (5.5,6), (5.5,9), (0.5,9)

These are the coordinates of the four feet of the table.

1.4 Distance Between Two Points in the 2-D Plane

We know how to find the distance between two points if they are on the axes or if they form a line segment parallel to the axes. For example, in Fig. 1.5 we can find the distances W₁W₂ and S₁S₂. What should we do if the segment joining the points is not parallel to either axis? We can use the Baudhāyana–Pythagoras theorem which you have studied in Grade 8.

We can use this result to find the distance between any two points in the xy-plane.

Look at triangle ADM in Figure.

Triangle ADM is an acute angled triangle in the first quadrant. How do we find the lengths of its sides AD, DM and MA?

Think and Reflect

1. In moving from A (3, 4) to D (7, 1), what distance has been covered along the x-axis? What about the distance along the y-axis?

2. Can these distances help you find the distance AD? Figure gives us a clue.

The distance moved along the x-axis is given by CD.

CD = x-coordinate of D – x-coordinate of A

= 7 – 3

= 4.

The distance moved along the y-axis is given by AC.

AC = y-coordinate of A – y-coordinate of D

= 4 – 1

= 3.

Using the Baudhāyana–Pythagoras Theorem, we get the distance

AD = √4² + 3² = √16 + 9 = √25 = 5 units.

We similarly find the distances DM and MA:

DM = √2² + 5² = √4 + 25 = √29 units.

MA = √6² + 4² = √36 + 16 = √50 units.

In general, the distance between the points A(x₁ , y₁) and B(x₂ , y₂) is given by

AB = √(x₂ – x₁)² + (y₂ – y₁)²

and is calculated as shown in Figure.

It makes no difference whether (x₂ – x₁) and (y₂ – y₁) are positive or negative, as we are simply measuring the shifts along the two axes.

What if, x₁, x₂, y₁, y₂ take negative values? In Fig. 1.9, triangle AMD is reflected in the y-axis. What are the coordinates of the images of points A, M, and D?

C'D' = x-coordinate of A' – x-coordinate of D'

= –3 – (–7)

= 4.

A'C' = y-coordinate of A' – y-coordinate of D'

= 4 – 1

= 3.

Using the Baudhāyana–Pythagoras Theorem, we get

AD = √4² + 3² = √16 + 9 = √25 = 5 units.

You can similarly calculate both D'M' and M'A':

We similarly find the distances DM and MA:

D'M' = √(–2)² + 5² = √4 + 25 = √29 units.

M'A' = √(–6)² + 2² = √36 + 4 = √40 units.

We see that reflection has preserved the lengths of the sides of the triangles.

Think and Reflect

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

2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?

Think and Reflect

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

In reflection across the y-axis:

Same:

Shape and size of the figure

Lengths of sides

Angles

Changed:

The x-coordinates change sign.

The figure appears reversed left-to-right.

Example:

(3, 2) –> (–3, 2)

2. Would these observations be the same if ΔADM is reflected in the x-axis?

Yes, similar observations would hold.

In reflection across the x-axis:

Shape and size remain the same.

y-coordinates change sign.

Example:

(3, 2) –> (–3, 2)

So the figure flips upside-down.

End-of-Chapter Exercises

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

Sol.

The x-axis and y-axis intersect at the origin.

Coordinates: (0,0)

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?

Sol.

Point W has x-coordinate –5.

A line parallel to the y-axis keeps x-coordinate constant.

So point will have coordinates: (–5, y)

where can be any number.

Which quadrants can H lie in?

Quadrant II if y > 0 (+ve)

Quadrant III if y < 0 (–ve)

It can also lie on the x-axis if y = 0

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.

Sol. Points R, A, M and P

Points: R (3, 0), A(0, –2), M(–5, –2),\ P(–5, 2)

(i) Two perpendicular sides

AM is horizontal.

MP is vertical.

Therefore:

AM ⊥ MP

(ii) One side parallel to an axis

AM is parallel to the x-axis.

Also,

MP is parallel to the y-axis.

(iii) Mirror image points

Points: M (–5, –2) and P (–5, 2)

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.)

Sol. Plot point Z(5, –6)

Choose:

I(5, 0), N (0, –6)

Then triangle IZN is right-angled.

Lengths of sides

IZ Vertical distance: 6

ZN Horizontal distance: 5

IN Using Pythagoras theorem:

IN = √(5² + 6²)

= √(25+36)

=√61

So sides triangle IZN are: 5, 6, √61

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?

Sol. Coordinate system without negative numbers

Without negative numbers:

★ Points could only lie in the first quadrant.

★ Left side and lower side points could not be represented.

So, we would not be able to locate all points on the plane.

*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.

Sol.

6. Are M, A and G on one straight line?

Points:M(–3, –4), A(0, 0), G(6, 8)

Check slopes.

S = (y₂ – y₁) / (x₂ – x₁)

Slope of MA

= {0–(–4)} / {0 –(–3)} ={4} / {3}

Slope of AG

{8–0} {6–0} = {8} / {6 }= {4} / {3}

Since slopes are equal, the 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.

Sol. Check points R, B and C

Points:

R(–5,–1), B(–2,–5), C(4,–12)

Slope of RB

{–5–(–1)} {–2–(–5)}

= {-4} / {3}

Slope of BC

{–12–(–5)} / {4–(–2)}

= {–7} / {6}

Slopes are not equal.

Therefore, the points are not on the same straight line.

The points are or are not on the same straight line.

🌺 Since slopes are equal, the points lie on the same straight line.

🌺 Therefore, the 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.

Sol. Plotting triangles

(i) Right-angled isosceles triangle

One example:

O(0, 0), A(4, 0), B(0, 4)

OA = OB, and angle at O is .

(ii) Isosceles triangle with vertices in Quadrants III and IV

Example:

A(–3, –2), B(3, –2), O(0, 0)

This forms an isosceles triangle.

*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.

When M is the mid-point of ST, can you find any connection between the coordinates of M, S and T?

Sol. Connection for midpoint

If is midpoint of , then:

M Left ({x₁+ x₂} / {2}, / {{y₁+ y₂}/{2}) Right

🌺 Cordinate of Midpoint

M (x, y) = (x₁ + x₂) , (y₁ + y₂)

2 2

In Separate Form

M (x) = (x₁ + x₂)

M (y) = (y₁ + y₂)

Cordinate of Midpoint

M (x, y)

So midpoint coordinates are averages of endpoint coordinates.

*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).

Sol. Find coordinates of B

Given:

B(–7, 1) is midpoint of A(3, –4) and B(x, y)

Using midpoint formula:

B (x) = (x₁ + x₂)

(–7) = (3 + x)

3 + x = –14

x = –17

Now:

B (y) = (y₁ + y₂)

(1) = (–4 + y)

–4 + y = 2

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).

Sol. Coordinates of trisection points

Given:

A(4,7), B(16,-2)

Difference in x-coordinates:

16 – 4 = 12

Difference in y-coordinates:

–2 –7 = – 9

One-third changes:

{12} / {3} = 4

{–9} / {3} = –3

Point P

(4 + 4, 7 – 3 )= (8, 4)

Point Q

(8+4, 4 –3) = (12,1)

Therefore

P(8, 4), Q(12, 1)

🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺

🌺 Distance Formula between two points A(x₁ , y₁) and B(x₂ , y₂) :

AB = √(x₂ – x₁)² + (y₂ – y₁)²

If one point is origin O then

Distance Formula between two points A A(x₁ , y₁) and O (0,0) :

OA = √(0 – x₁)² + (0 – y₁)²

OA = √(–x₁)² + (– y₁)²

OA = √(x₁)² + (y₁)²

🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺

*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.

Sol. In Circle K

(i) Show A, B, C lie on circle

Points:

A(1, –8), B(–4, 7), C(–7,–4) and O (0, 0)

Now the distance from origin.

AO = √(x₁)² + (y₁)²

AO = √{1² + (–8)²} = √{1+64} = √{65}

BO = √{–4)² + (7)²} = √{16+49} = √{65}

CO √{(–7)² + (–4)²} = √{49+16} = √{65}

AO = BO = CO = √65

SO, All distances are equal.

Therefore, all points lie on the same circle with Radius: √{65}

(ii) Points D and E

D (–5, 6) and O(0, 0)

Distance from origin:

DO = √(x₁)² + (y₁)²

DO = √{(–5)² + (–8)²}

= √{25+36}=√61

Since:

√{61} < √{65}

D lies inside the circle.

E (0, 9)

Distance from origin: 9

EO = √(x₁)² + (y₁)²

EO = √(0)² + (9)²

EO = √(81)

EO = 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.

Sol. Find coordinates of A, B and C

Midpoints:

D(5, 1), E(6, 5), F(0, 3)

Let the coordinates of triangle ABC be:

A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

Given midpoints:

D(5,1) is the midpoint of BC

E(6,5) is the midpoint of CA

F(0,3) is the midpoint of AB

Using the midpoint formula:

M (x, y) = (x₁ + x₂) , (y₁ + y₂)

2 2

Step 1: Form equations from the given midpoints

Since is midpoint of

{x₂+ x₃}/{2} = 5

x₂+ x₃= 10 .........(1)

{y₂+ y₃}/ {2} = 1

_y2+ y₃= 2 ............(2)

Since is midpoint of

{x₃+ x₁} / {2} = 6

x₃+ x₁= 12 .........(3)

{y₃+ y₁} / {2} = 5

y₃+ y₁= 10 .........(4)

Since is midpoint of

{x₁+ x₂} / {2} = 0

x₁+ x₂= 0 ............(5)

{y₁+ y₂} / {2} = 3

y₁+ y₂= 6 ..........(6)

Step 2: Find the x-coordinates

We have:

x₂+ x₃= 10 .........(1)

x₃+ x₁= 12 .........(3)

x₁+ x₂= 0 ..........(5)

Add all three equations:

(x₂+ x₃)+(x₃+ x₁)+(x₁+ x₂) = 10 + 12 + 0

2x₁+ 2x_{2 +} 2x₃= 22. ........(7)

x₁+ x_{2 +} x₃= 11

Now subtract equation (5) from equation (7):

x₁+ x_{2 +} x₃= 11

x₁+ x₂= 0

So,

x₃= 11

Now subtract equation (1) from equation (7):

x₁+ x_{2 +} x₃= 11

x₂+ x₃= 10

x₁= 1

Now subtract equation (3) from equation (7):

x₁+ x_{2 +} x₃= 11

x₃+ x₁= 12

x₂= –1

Therefore,

x₁= 1, x₂= –1 and x₃= 11

Step 3: Find the y-coordinates

We have:

{y₂+ y₃} / {2} = 1

Y₂+ y₃= 2 ............(2)

{y₃+ y₁} / {2} = 5

y₃+ y₁= 10 .........(4)

{y₁+ y₂} / {2} = 3

y₁+ y₂= 6 ..........(6)

Add all three equations:

y₂+ y₃+ y₃+ y₁+ y₁+ y₂= 2 + 10 + 6

2y₁+ 2y₂+ 2y₃= 18

y₁+ y₂+ y₃= 9 ........(8)

Now subtract equation (2) from equation (8):

y₁+ y₂+ y₃= 9

_y2+ y₃= 2

_y1= 9 – 2

_y1= 7

Now subtract equation (4) from equation (8):

y₁+ y₂+ y₃= 9

y₃+ y₁= 10

y₂ = 9 – 10

y₂ = – 1

Now subtract equation (6) from equation (8):

y₁+ y₂+ y₃= 9

y₁+ y₂= 6

y₃= 9 – 6

y₃= 3

Therefore,

_y1= 7, y₂= –1 and y₃= 3

For Final Answer

x₁= 1, x₂= –1 and x₃= 11

_y1= 7, y₂= –1 and y₃= 3

A(1,7), B(-1,-1), C(11,3)

So, the coordinates of the vertices are:

A(1,7), B(-1,-1), C(11,3)

2nd Method

Find coordinates of A, B and C

Midpoints:

D(5, 1), E(6, 5), F(0, 3)

where

D(5,1) is the midpoint of BC

E(6,5) is the midpoint of CA

F(0,3) is the midpoint of AB

Let

A(x₁+ y₁), B(x₂+ y₂), C(x₃+ y₃)

Step 1: Form equations from the given midpoints

Since is midpoint of

{x₂+ x₃}/{2} = 5

x₂+ x₃= 10 .........(1)

{y₂+ y₃}/ {2} = 1

_y2+ y₃= 2 ............(2)

Since is midpoint of

{x₃+ x₁} / {2} = 6

x₃+ x₁= 12 .........(3)

{y₃+ y₁} / {2} = 5

y₃+ y₁= 10 .........(4)

Since is midpoint of

{x₁+ x₂} / {2} = 0

x₁+ x₂= 0 ............(5)

{y₁+ y₂} / {2} = 3

y₁+ y₂= 6 ..........(6)

Find x-coordinates quickly

Add (1) and (5):

(x₁+ x₂) + (x₃+ x₁) = 0 + 12

2x₁+ x₂+ x₃ = 12 .........(7)

Now subtract equation (1) from equation (7):

2x₁+ x₂+ x₃ = 12

x₂+ x₃= 10

2x₁ = 12 –10

2x₁ = 2

x₁ = 1

Placing x₁ = 1 equation (3):

x₃+ x₁= 12

x₃+ 1= 12

x₃= 11

Placing x₁ = 1 equation (1):

x₂+ x₃= 10

x₂+ 11= 10

x₂= 10 –11

x₂= –1

Now

x₁= 1, x₂= –1 and x₃= 11

Find y-coordinates quickly

Add (4) and (6):

y₃+ y₁ + y₁+ y₂= 10 + 6

2y₁+ y₂ + y₃= 16 ........(8)

Now subtract equation (2) from equation (8):

2y₁+ y₂ + y₃= 16

_y2+ y₃= 2

2y₁+ 2= 16

2y₁+ 2= 16 –2

2y₁= 14

2y₁= 14/2

y₁= 7

Placing y₁ = 7 equation (4):

y₃+ y₁= 10

y₃+ 7= 10

y₃= 10 – 7

y₃= 3

Placing y₁ = 7 equation (6):

y₁+ y₂= 6

7+ y₂= 6

y₂= 6 –7

y₂= –1

For Final Answer

x₁= 1, x₂= –1 and x₃= 11

_y1= 7, y₂= –1 and y₃= 3

A(1,7), B(-1,-1), C(11,3)

So, the coordinates of the vertices are:

A(1,7), B(-1,-1), C(11,3)

This method is easier because we solve step-by-step without adding all six equations together.

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).

Sol.

Streets and Intersections

(i) Drawing the Model

Scale: 1 { cm} = 200 {m}

Steps:

★ Draw one vertical line for the main North–South road.

★ Draw one horizontal line for the main East–West road crossing at the centre.

★ Draw 10 parallel vertical streets and 10 parallel horizontal streets.

★ Keep each street 1 cm apart because 1 cm represents 200 m.

This gives a grid-like model of the city.

(ii) Street Intersections

Each intersection is named by:

({N–S street number},{E–W street number})

(a) How many intersections can be referred to as ?

Only one intersection.

Because the:

★ 4th N–S street and

★ 3rd E–W street

meet at exactly one point.

(b) How many intersections can be referred to as ?

Again, only one intersection.

Because the:

★ 3rd N–S street and

★ 4th E–W street

meet at one unique point.

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.

Sol. Computer Graphics Circles

Screen size:

800 { pixels wide}, 600 { pixels high}

Circle A

Centre: A(100, 150)

Radius: 80

Check boundaries

Left edge: 100 – 80 = 20

Right edge: 100 + 80 = 180

Bottom edge: 150 – 80 = 70

Top edge: 150 + 80 = 230

All values are within the screen limits.

So, Circle A lies completely inside the screen.

Circle B

Centre: B(250, 230)

Radius: 100

Check boundaries

Left edge: 250 – 100 = 150

Right edge: 250 + 100 = 350

Bottom edge: 230 – 100 = 130

Top edge: 230 + 100 = 330

All values are within the screen.

So, Circle B also lies completely inside the screen.

(i) Answer

No part of either circle lies outside the screen.

(ii) Do the circles intersect?

Distance between centres:

d= √{(250-100)² + (230–150)²}

= √ {150² + 80²}

= √{22500+6400}

= √{28900}

= 170

Sum of radii: 80 + 100 = 180

Since:

170 < 180

the 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?

Sol. Is ABCD a Square?

Points:

A(2, 1), B(–1, 2), C(–2, –1), D(1, –2)

Step 1: Find lengths of sides

Using distance formula: AB

AB = √{(–1–2)² + (2 –1)²}

= √{(–3)²+ (1)²}

= √{9+1}

= √{10}

Similarly,

BC = CD = DA = √{10}

So all four sides are equal.

Step 2: Check diagonals

AC = √{(–2–2)²+(–1–1)²}

= √{(–4)²+(–2)²}

= √{16+4}

= √{20}

Similarly,

BD = √{20}

Both diagonals are equal.

Therefore, is a square.

Area of the square

Side: √{10}

Area: (side)² = (√{10})² = 10

So, the area of the square is: 10 { square units}

Chapter Summary

★ To locate the position of an object or a point in a plane, we require two perpendicular lines — One of them is horizontal, and the other is vertical.

★ The plane is called the cartesian plane, the coordinate plane or the xy-plane and the lines are called the coordinate axes.

★ The horizontal line is called the x-axis and the vertical line is called the y-axis.

★ The coordinate axes divide the plane into four parts called 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 and the distance of the point from the x-axis is its y-coordinate. If the x-coordinate of a point is x, and the y-coordinate is y, then (x, y) are called the coordinates of the point.

The coordinates of a point on the x-axis are of the form (x, 0), and those of the points on the y-axis are of the form (0, y).

★ The coordinates of the origin are (0, 0).

★ The coordinates of a point are of the form (+, +) in the first quadrant, (–, +) in the second quadrant, (–, –) in the third quadrant and (+, –) in the fourth quadrant.

★ If x = y, then (x, y) = (y, x). If x ≠ y, then (x, y) ≠ (y, x).

★ The distance between points (x₁, y) and (x₂, y) is the absolute value |x₂–x₁|of the difference between x₁ and x₂.

★ The distance between points (x, y₁) and (x, y₂) is the absolute value |y₂–y₁|of the difference between y₁ and y₂.

★ By the Baudhāyana–Pythagoras Theorem, the distance between points (x₁, y₁) and (x₂

, y₂) is

C₀C₁C₂C₃

🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺🌺

02: Write the coordinates of the points marked on the axes in Fig.

(आकृति में अक्षों पर अंकित बिन्दुओं के निर्देशांक लिखिएः)

Sol. : You can see that :	हल: आप यहाँ देख सकते हैं कि :
(i) The point A is at a distance of + 4 units from the y - axis and at a distance zero from the x - axis. Therefore, the x - coordinate of A is 4 and the y - coordinate is 0. Hence, the coordinates of A are (4, 0).	(i) बिन्दु A, y-अक्ष +4 एकक की दूरी पर है और x-अक्ष से दूरी 0 पर है। अतः बिन्दु A का x-निर्देशांक 4 है और -निर्देशांक 0 है। इसलिए A के निर्देशांक (4,0) हैं।
(ii) The coordinates of B are (0, 3). Why?	(ii) B के निर्देशांक (0, 3) हैं। क्यों?
(iii) The coordinates of C are (– 5, 0). Why?	(iii) C के निर्देशांक (5,0) हैं। क्यों?
(iv) The coordinates of D are (0, – 4). Why?	(iv) D के निर्देशांक (0,4) हैं। क्यों?
(v) The coordinates of E are ( ⅔, 0) . Why?	(v) E के निर्देशांक (⅔, 0) है। क्यों ?

Solution : Taking 1cm = 1unit, we draw the x - axis and the y - axis. The positions of the points are shown by dots in Fig.

11. In which quadrants or on which axis do each of the point lies

12. Pilot the Following points and write the name of figure obtained. Which point lies in III Quadrant?

13. Pilot the points given in the following table on the plane, choosing suitable units of distance on the axis.

Sol. 15. : The pairs of numbers given in the table can be represented by the points (– 3, 7), (0, –3.5), (– 1, – 3), (4, 4) and (2, – 3). The locations of the points are shown by dots in Fig.

14. Pilot the points given in the following table on the plane, choosing suitable units of distance on the axis.

15. Pilot the points given in the following table on the plane, choosing suitable units of distance on the axis.

16. Plot the points O (0,0), B (16,0) C (16,12) on a graph paper. Join Points O, B and C name the figure so formed.

17. If a point is on negative side of at a distance of 3 units from origin, then find the co-ordinates of the point.

18. Write the coordinates of the vertices of a rectangle whose length and breadth are 4 units and 3 units respectively has one vertex at origin, the longer side is on the and one of the vertices likes in the 4^th quadrant. Also find the area.

19. Find the coordinates of the vertices of an equilateral triangle of side of length ‘8 Unit’ with one of its sides on x-axis and one vertex at origin.

20. From the given graph, Write:

(a) The coordinates of the Points B and F.

(b) The abscissa of points D and H.

(d) The perpendicular distance of the points G from the x-axis.

22. Write the coordinates of points A, B, C, and D from figures.

Plot the following points:

P	Q	R	S	T	U
– 1	0	6	3	– 3	6
3	3	3	0	– 2	– 3

23. (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are about 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.

There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East – West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at somecrossing, then we will call this cross-street (2, 5). Using this convention, find:

(i) how many cross - streets can be referred to as (4, 3).

(ii) how many cross - streets can be referred to as (3, 4).

C₀C₁C₂C₃C₄

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