NCERT Solutions for Class 9 Maths Chapter 1 Orienting Yourself: The Use of Coordinates - Ganita Manjari
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NCERT Solutions for Chapter 1 Orienting Yourself: The Use of Coordinates Class 9 Maths
Page No. 4
Exercise Set 1.1
Fig. 1.3 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.
(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?
(ii) What are the coordinates of D1?
(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?
(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?
Answer
(i) The door D1R1 lies on the x-axis.
The left end of the door is at D1(7.5, 0) and the right end is at R1(11.5, 0).
Therefore, the door is 7.5 ft from the left wall (y-axis).
Since the door lies on the x-axis, its distance from the x-axis is 0 ft.
- Distance from y-axis = 7.5 ft
- Distance from x-axis = 0 ft
(ii) From the figure, D1 lies on the x-axis midway between 7 and 8.
Hence, the coordinates of D1 are:
\((7.5,0)\)
(iii) D1 = \((7.5,0)\)
R1 = \((11.5,0)\)
Width of the door = \(11.5 - 7.5\)
= \(4\) ft
Therefore, the width of the room door is 4 ft.
A 4 ft wide door is quite comfortable for normal use.
Since wheelchairs generally require less than 4 ft of clearance, a person using a wheelchair can enter the room comfortably and easily.
(iv) Coordinates of bathroom door ends:
B1\((0,1.5)\)
B2\((0,4)\)
Width of bathroom door
= \(4 - 1.5\)
= \(2.5\) ft
Width of room door = \(4\) ft
Since \(2.5\ \text{ft} < 4\ \text{ft}\)
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.
Answer
The standard width of a room door is usually between 3 ft and 4 ft. Wider doors are preferred for easy movement and accessibility.
2. Are the doors in your school suitable for people in wheelchairs?
Answer
This depends on the school. Doors wider than about 3 ft and provided with ramps are generally suitable for wheelchair users.
Think and Reflect
1. What is the x-coordinate of a point on the y-axis?
Answer
Every point on the y-axis has x-coordinate equal to \(0\).
Therefore, the x-coordinate of a point on the y-axis is \(0\).
2. Is there a similar generalisation for a point on the x-axis?
Answer
Yes.
Every point on the x-axis has y-coordinate equal to \(0\).
Therefore, the y-coordinate of a point on the x-axis is \(0\).
3. Does point Q \((y,x)\) ever coincide with point P \((x,y)\)? Justify your answer.
Answer
Point Q \((y,x)\) will coincide with point P \((x,y)\) only when \(x=y\).
For example, if \(x=3\) and \(y=3\), then
P = \((3,3)\)
Q = \((3,3)\)
Hence, both points coincide.
If \(x \ne y\), the coordinates are different and the points do not coincide.
4. If \(x \ne y\), then \((x,y) \ne (y,x)\); 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,
\((x,y)=(y,x)\)
implies
\(x=y\) and \(y=x\).
Hence, if \(x \ne y\), then
\((x,y)\ne(y,x)\).
Thus, the statement is true.
Page No. 7
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?
Answer
(i) The given three feet are:
(8, 9), (11, 9) and (11, 7)
Since the table is rectangular, the fourth foot must have the x-coordinate of (8,9) and the y-coordinate of (11,7).
Therefore, the fourth foot will be at: (8,7)
(ii) Yes, this is a good spot because the table is placed in an open area of the room and does not obstruct the bed, wardrobe, or doors.
(iii) Horizontal side: 11-8=3 ft
Vertical side: 9-7=2 ft
Therefore,
Length = 3 ft
Width = 2 ft
The height of the table cannot be determined because the figure shows only a top view (2-D view) of the room.
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?
Answer
The bathroom door extends from B1(0,1.5) to B2(0,4).
The wardrobe is located between x = 3 and x = 7.
Since the wardrobe is far from the bathroom door, the door will not hit the wardrobe when opened.
If the door is made much wider, care should be taken that it still does not obstruct movement in the room.
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.
Answer
(i) From the figure:
O = \((0,0)\)
F = \((0,9)\)
R = \((-6,9)\)
P = \((-6,0)\)
(ii) The showering area SHWR has one pair of parallel sides.
Therefore, it is a trapezium.
Coordinates are:
S = \((-6,6)\)
H = \((-3,6)\)
W = \((-2,9)\)
R = \((-6,9)\)
(iii) This is an activity-based question.
One possible answer is:
Washbasin (3 ft × 2 ft)
\((-6,0)\), \((-3,0)\), \((-3,2)\), \((-6,2)\)
Toilet (2 ft × 3 ft)
\((-6,2)\), \((-4,2)\), \((-4,5)\), \((-6,5)\)
Any other correct placement within the bathroom having the required dimensions is acceptable.
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.
Answer
(i) P = \((-6,0)\)
A = \((12,0)\)
Length PA = \(18\) ft
Width = \(15\) ft
Therefore, the remaining two corners are:
\((-6,-15)\) and \((12,-15)\)
Coordinates of dining room corners:
\((-6,0)\), \((12,0)\), \((12,-15)\), \((-6,-15)\)
(ii) The centre of the dining room is:
\(\left(\frac{-6+12}{2},\frac{0+(-15)}{2}\right)\)
= \((3,-7.5)\)
A table of dimensions \(5\) ft × \(3\) ft centred at \((3,-7.5)\) will have corners at:
\((0.5,-6)\)
\((5.5,-6)\)
\((5.5,-9)\)
\((0.5,-9)\)
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?
Answer
Coordinates of A = \((3,4)\)
Coordinates of D = \((7,1)\)
Distance covered along the x-axis
= \(7-3\)
= \(4\) units
Distance covered along the y-axis
= \(4-1\)
= \(3\) units
- Distance along x-axis = \(4\) units
- Distance along y-axis = \(3\) units
2. Can these distances help you find the distance AD?
Answer
Yes.
The distances along the x-axis and y-axis form the two perpendicular sides of a right-angled triangle.
Using the Baudhāyana–Pythagoras Theorem,
\[ AD=\sqrt{4^2+3^2} \]
\[ =\sqrt{16+9} \]
\[ =\sqrt{25} \]
\[ =5 \]
Therefore, the distance AD is 5 units.
Think and Reflect
1. What has remained the same and what has changed with this reflection?
Answer
When the triangle is reflected in the y-axis:
What remains the same:
• The lengths of all sides remain unchanged.
• The shape and size of the triangle remain unchanged.
• The y-coordinates of all points remain the same.
What changes:
• The x-coordinates change their signs.
• The position of the triangle shifts from one side of the y-axis to the other.
For example:
A\((3,4)\) becomes A′\((-3,4)\)
D\((7,1)\) becomes D′\((-7,1)\)
M\((9,6)\) becomes M′\((-9,6)\)
The shape, size and side lengths remain the same, while the x-coordinates change sign and the figure appears on the opposite side of the y-axis.
2. Would these observations be the same if ΔADM is reflected in the x-axis (instead of the y-axis)?
Answer
Yes, the observations about shape, size and side lengths would remain the same because reflection preserves distances.
However, when reflected in the x-axis:
- The x-coordinates remain unchanged.
- The y-coordinates change their signs.
For example:
A\((3,4)\) becomes \((3,-4)\)
D\((7,1)\) becomes \((7,-1)\)
M\((9,6)\) becomes \((9,-6)\)
Yes. The shape, size and side lengths would remain unchanged. The only difference is that the y-coordinates would change sign instead of the x-coordinates.
Page No. 12
End-of-Chapter Exercises
1. What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Answer
The x-axis and y-axis intersect at the origin.
The coordinates of the origin are \((0,0)\).
- x-coordinate = \(0\)
- 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
A line parallel to the y-axis has the same x-coordinate for all points.
Therefore, if W has x-coordinate \(-5\), then H must also have x-coordinate \(-5\).
Hence, the coordinates of H can be written as:
\((-5,y)\)
where \(y\) can be any real number.
If \(y>0\), H lies in Quadrant II.
If \(y<0\), H lies in Quadrant III.
If \(y=0\), H lies on the x-axis.
Coordinates of H = \((-5,y)\)
H can lie in Quadrant II or Quadrant III.
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?
Answer
(i) AM is horizontal because both points have the same y-coordinate \((-2)\).
MP is vertical because both points have the same x-coordinate \((-5)\).
A horizontal line and a vertical line are perpendicular.
AM and MP are perpendicular to each other.
(ii) AM joins \((0,-2)\) and \((-5,-2)\).
Since the y-coordinate remains constant, AM is parallel to the x-axis.
AM is parallel to the x-axis.
(iii) M = \((-5,-2)\)
P = \((-5,2)\)
These points have the same x-coordinate and opposite y-coordinates.
Therefore, they are mirror images of each other in the x-axis.
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.
Answer
Take:
I = \((5,0)\)
N = \((0,0)\)
Z = \((5,-6)\)
Then triangle IZN is right-angled at I.
Length IZ
= \(6\) units
Length IN
= \(5\) units
Using the Baudhāyana–Pythagoras Theorem,
\[ ZN=\sqrt{5^2+6^2} \]
\[ =\sqrt{25+36} \]
\[ =\sqrt{61} \] units.
IZ = \(6\) units
IN = \(5\) units
ZN = \(\sqrt{61}\) 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 only have non-negative values.
As a result, only the first quadrant and the positive parts of the axes could be represented.
Points lying to the left of the y-axis or below the x-axis could not be located.
Therefore, such a system would not allow us to locate all points on a 2-D plane.
No. Without negative numbers, only part of the plane could be represented, so all points could not be located.
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.
Answer
Method: Compare the slopes.
Slope of MA
\[ =\frac{0-(-4)}{0-(-3)} \]
\[ =\frac{4}{3} \]
Slope of AG
\[ =\frac{8-0}{6-0} \]
\[ =\frac{8}{6} =\frac{4}{3} \]
Since the slopes are equal, the three points lie on the same straight line.
Yes, M, A and G are collinear.
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.
Answer
Slope of RB
\[ =\frac{-5-(-1)}{-2-(-5)} \]
\[ =\frac{-4}{3} \]
Slope of BC
\[ =\frac{-12-(-5)}{4-(-2)} \]
\[ =\frac{-7}{6} \]
Since
\[ -\frac{4}{3}\ne-\frac{7}{6} \]
the slopes are not equal.
Therefore, the points do not lie on the same straight line.
R, B and C are not collinear.
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.
Answer
(i) One possible set of vertices is:
\((0,0)\), \((4,0)\), \((0,4)\)
This forms a right-angled isosceles triangle because the two perpendicular sides are equal.
(ii) One possible set of vertices is:
\((0,0)\), \((-4,-3)\), \((4,-3)\)
The two equal sides are each \(5\) units long.
Therefore, the triangle 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.
When M is the midpoint of ST, can you find any connection between the coordinates of M, S and T?
Answer
| S | M | T | Is M the Midpoint? | Reason |
|---|---|---|---|---|
| (−3, 0) | (0, 0) | (3, 0) | Yes | Midpoint = (0, 0) \[ \left(\frac{-3+3}{2},\frac{0+0}{2}\right)=(0,0) \] |
| (2, 3) | (3, 4) | (4, 5) | Yes | Midpoint = (3, 4) \[ \left(\frac{2+4}{2},\frac{3+5}{2}\right)=(3,4) \] |
| (0, 0) | (0, 5) | (0, -10) | No | Midpoint = (0, -5) \[ \left(\frac{0+0}{2},\frac{0+(-10)}{2}\right)=(0,-5)\neq(0,5) \] |
| (−8, 7) | (0, −2) | (6, −3) | No | Midpoint = (−1, 2) \[ \left(\frac{-8+6}{2},\frac{7+(-3)}{2}\right)=(-1,2)\neq(0,-2) \] |
If S = \((x_1,y_1)\) and T = \((x_2,y_2)\), then the midpoint M is
\[ M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \]
Thus, the coordinates of the midpoint are obtained by taking the average of the corresponding coordinates of the endpoints.
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).
Answer
Given:
A = \((3,-4)\)
M = \((-7,1)\)
Let B = \((x,y)\)
Using the midpoint formula,
\[ \left(\frac{3+x}{2},\frac{-4+y}{2}\right)=(-7,1) \]
Comparing x-coordinates:
\[ \frac{3+x}{2}=-7 \]
\[ 3+x=-14 \]
\[ x=-17 \]
Comparing y-coordinates:
\[ \frac{-4+y}{2}=1 \]
\[ -4+y=2 \]
\[ y=6 \]
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).
Answer
Given:
A = \((4,7)\)
B = \((16,-2)\)
The change in x-coordinate is
\[ 16-4=12 \]
The change in y-coordinate is
\[ -2-7=-9 \]
Each third of AB corresponds to:
\[ \left(\frac{12}{3},\frac{-9}{3}\right)=(4,-3) \]
Coordinates of P:
\[ (4+4,\;7-3) \]
\[ =(8,4) \]
Coordinates of Q:
\[ (8+4,\;4-3) \]
\[ =(12,1) \]
P = \((8,4)\)
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 centre 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.
Answer
(i) Distance OA:
\[ OA=\sqrt{1^2+(-8)^2} \]
\[ =\sqrt{65} \]
Distance OB:
\[ OB=\sqrt{(-4)^2+7^2} \]
\[ =\sqrt{65} \]
Distance OC:
\[ OC=\sqrt{(-7)^2+(-4)^2} \]
\[ =\sqrt{65} \]
Since
\[ OA=OB=OC=\sqrt{65} \]
all three points are at the same distance from the origin.
Therefore, A, B and C lie on a circle centred at O.
Radius of circle K:
\[ \sqrt{65}\text{ units} \]
(ii) Radius of circle K = \(\sqrt{65}\)
Distance OD:
\[ OD=\sqrt{(-5)^2+6^2} \]
\[ =\sqrt{61} \]
Since
\[ \sqrt{61}<\sqrt{65} \]
D lies inside the circle.
Distance OE:
\[ OE=\sqrt{0^2+9^2} \]
\[ =9 \]
Since
\[ 9>\sqrt{65} \]
E lies outside the circle.
- D lies inside the circle.
- 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.
Answer
Let
D, E and F be the midpoints of BC, CA and AB respectively.
Using the midpoint relations:
\[ D=\left(\frac{B+C}{2}\right)=(5,1) \]
\[ E=\left(\frac{C+A}{2}\right)=(6,5) \]
\[ F=\left(\frac{A+B}{2}\right)=(0,3) \]
Using
\[ A=E+F-D \]
\[ =(6,5)+(0,3)-(5,1) \]
\[ =(1,7) \]
Using
\[ B=D+F-E \]
\[ =(5,1)+(0,3)-(6,5) \]
\[ =(-1,-1) \]
Using
\[ C=D+E-F \]
\[ =(5,1)+(6,5)-(0,3) \]
\[ =(11,3) \]
- A = \((1,7)\)
- B = \((-1,-1)\)
- 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) Each street intersection is formed by two streets—one running in the N–S direction and another in the E–W direction. If the second street running in the N–S direction and the 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).
Answer
(i) This is an activity-based question.
Draw two perpendicular roads passing through the centre of the page.
Draw 10 equally spaced parallel streets in the North–South direction and 10 equally spaced parallel streets in the East–West direction.
Use the scale:
\(1\text{ cm} = 200\text{ m}\)
(ii) (a) The notation \((4,3)\) represents a unique intersection of:
• 4th North–South street
• 3rd East–West street
Only one crossing is formed by these two streets.
(b) The notation \((3,4)\) also represents a unique intersection of:
• 3rd North–South street
• 4th East–West street
Only one such crossing exists.
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.
Answer
(i) Circle A
Centre = \((100,150)\)
Radius = \(80\) pixels
Left edge = \(100-80=20\)
Right edge = \(100+80=180\)
Bottom edge = \(150-80=70\)
Top edge = \(150+80=230\)
All these values lie within the screen dimensions \(0 \le x \le 800\) and \(0 \le y \le 600\).
Therefore, Circle A lies completely inside the screen.
Circle B
Centre = \((250,230)\)
Radius = \(100\) pixels
Left edge = \(250-100=150\)
Right edge = \(250+100=350\)
Bottom edge = \(230-100=130\)
Top edge = \(230+100=330\)
All these values also lie within the screen dimensions.
Therefore, Circle B also lies completely inside the screen.
No part of either circle lies outside the screen.
(ii) Distance between centres A and B:
\[ AB=\sqrt{(250-100)^2+(230-150)^2} \]
\[ =\sqrt{150^2+80^2} \]
\[ =\sqrt{22500+6400} \]
\[ =\sqrt{28900} \]
\[ =170 \]
pixels
Sum of radii:
\[ 80+100=180 \]
pixels
Since
\[ 170<180 \]
the two circles intersect each other.
Yes, the two circles intersect.
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?
Answer
Using the distance formula:
\[ AB=\sqrt{(-1-2)^2+(2-1)^2} \]
\[ =\sqrt{9+1} \]
\[ =\sqrt{10} \]
Similarly,
\[ BC=CD=DA=\sqrt{10} \]
Hence all four sides are equal.
Diagonal AC:
\[ AC=\sqrt{(-2-2)^2+(-1-1)^2} \]
\[ =\sqrt{16+4} \]
\[ =\sqrt{20} \]
Diagonal BD:
\[ BD=\sqrt{(1+1)^2+(-2-2)^2} \]
\[ =\sqrt{4+16} \]
\[ =\sqrt{20} \]
Both diagonals are equal.
Also, adjacent sides are perpendicular.
Therefore, ABCD is a square.
Area of square
\[ =(\text{side})^2 \]
\[ =(\sqrt{10})^2 \]
\[ =10 \]
square units.
ABCD is a square.
Side length = \(\sqrt{10}\) units
Area = \(10\) square units.
