Class 9 Maths Chapter 1 End of Chapter Exercise: Orienting Yourself The Use of Coordinates

Class 9 Maths Ganita Manjari Chapter 1 End of Chapter Exercises cover the main ideas from Orienting Yourself: The Use of Coordinates. These exercises help students revise the x-axis, y-axis, origin, quadrants, coordinates of points, midpoint, distance between two points and real-life use of coordinate geometry.

Chapter 1 introduces the Cartesian coordinate system Class 9 through maps, room layouts and points on a plane. The end-of-chapter questions take this learning further by asking students to predict positions, plot points, check collinearity, use midpoint ideas and solve real-life coordinate geometry problems. These Class 9 Maths Ganita Manjari Chapter 1 Solutions are written step-by-step so students can practise the complete chapter in one place.

Class 9 Ganita Manjari Maths Chapter 1 - End of Chapter Exercise

End of Chapter Exercise (Page 12 – 14)

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

Answer:

The point where the x-axis and y-axis intersect is called the origin.

Its coordinates are:

 

$$\boxed{(0,0)}$$

 

So, the x-coordinate = 0 and the 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:

$H$

lies in Quadrant II.

• If
  $y<0$ 

  ,

 

$H$

lies in Quadrant III.

• If
  $y=0$ 

  ,

 

$H$

lies on the negative x-axis.

Therefore,

$H$

can lie in:

 

$$\boxed{\text{Quadrant II or Quadrant III}}$$

$$\boxed{AM\text{ is parallel to the x-axis}}$$

(Also,

$MP$

is parallel to the y-axis.)

(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.

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.)

Answer:

One possible construction is to choose

 

$$Z(5,-6),\qquad I(5,0),\qquad N(0,-6).$$

 

Here,

$IZ$

is vertical and

$ZN$

is horizontal, so they are perpendicular. Therefore,

$\triangle IZN$

is right-angled at

$Z$

.

The side lengths are:

 

$$IZ=|0-(-6)|=\boxed{6\text{ units}}$$

 

$$ZN=|5-0|=\boxed{5\text{ units}}$$

 

Using the Pythagorean theorem,

 

$$IN=\sqrt{IZ^2+ZN^2} =\sqrt{6^2+5^2} =\sqrt{61}$$

 

$$\boxed{IN=\sqrt{61}\text{ units}\approx7.81\text{ units}}$$

 

IZ = 6 units, NZ = 5 units, IN =

$\sqrt{61}$

​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:

If we did not have negative numbers, the coordinates of points could only be zero or positive.

So we could locate points only where

 

$$x \ge 0 \quad \text{and} \quad y \ge 0$$

 

This would cover only the first quadrant and the positive parts of the two axes.

Therefore, such a coordinate system would not allow us to locate all the points on a 2-D plane, because points lying to the left of the y-axis or below the x-axis require negative coordinates.

Negative numbers are necessary to represent all points on a 2-D 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.

Answer:

Yes, the points

$M(-3,-4)$

,

$A(0,0)$

, and

$G(6,8)$

lie on the same straight line.

A method to check this without plotting is to 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 two slopes are equal,

 

$$\boxed{\text{M, A and G are collinear.}}$$

 

So, comparing the slopes of the line segments is one way to check whether 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.

Answer:

Given points: R(−5, −1), B(−2, −5), C(4, −12)
Using the distance formula:

$$RB=\sqrt{(-2-(-5){)}^2+(-5-(-1){)}^2}=\sqrt{3^2+(-4{)}^2}=\sqrt{9+16}=5$$ $$BC=\sqrt{(4-(-2){)}^2+(-12-(-5){)}^2}=\sqrt{6^2+(-7{)}^2}=\sqrt{36+49}=\sqrt{85}$$ $$RC=\sqrt{(4-(-5){)}^2+(-12-(-1){)}^2}=\sqrt{9^2+(-11{)}^2}=\sqrt{81+121}=\sqrt{202}$$

RB + BC = 5 +

$\sqrt{85}$

​ ≠

$\sqrt{202}$

​ = RC
Therefore, the points R, B, and C are not collinear (do not lie on the same straight line).
Verification: On plotting, the three points will not lie on a single straight line

8. Using the origin as one vertex, plot the vertices of:
(i)  A right-angled isosceles triangle.

Answer:

A right-angled isosceles triangle

One possible set of vertices is:

$$\boxed{O(0,0),\ A(4,0),\ B(0,4)}$$

Here,

$$OA=OB=4$$

and

$OA\perp OB$

. Therefore,

$\triangle OAB$

is a right-angled isosceles triangle.

(ii)  An isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.

Answer:

One possible set of vertices is:

 

$$\boxed{O(0,0),\ P(-3,-4),\ Q(3,-4)}$$

 

Here,

$P(-3,-4)$

lies in Quadrant III and

$Q(3,-4)$

lies in Quadrant IV.

Also,

 

$$OP=OQ=5$$

 

so

$\triangle OPQ$

is 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?

Answer:

Conclusion

When

$M$

is the midpoint of

$ST$

, it lies on

$ST$

and divides it into two equal parts:

 

$$\boxed{SM=MT}$$

 

If

$S(x_1,y_1)$

and

$T(x_2,y_2)$

, then the coordinates of the midpoint

$M$

are:

M(2x1​+x2​​,2y1​+y2​​)​

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:

$$OA=\sqrt{(1-0)^2+(-8-0)^2} =\sqrt{1+64} =\sqrt{65}$$

For

$B(-4,7)$

:

$$OB=\sqrt{(-4)^2+7^2} =\sqrt{16+49} =\sqrt{65}$$

For

$C(-7,-4)$

:

$$OC=\sqrt{(-7)^2+(-4)^2} =\sqrt{49+16} =\sqrt{65}$$

Thus,

$$OA=OB=OC=\sqrt{65}$$

Therefore,

$A$

,

$B$

, and

$C$

lie on the same circle with centre

$O(0,0)$

.

$$\boxed{\text{Radius of circle }K=\sqrt{65}\text{ units}}$$

(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.

 

Let

 

$$A(x_1,y_1),\quad B(x_2,y_2),\quad C(x_3,y_3)$$

 

Given that

$D(5,1)$

,

$E(6,5)$

, and

$F(0,3)$

are the midpoints of

$BC$

,

$CA$

, and

$AB$

, respectively.

Using the midpoint formula:

 

$$D=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$$

 

Therefore,

 

$$x_2+x_3=10,\qquad y_2+y_3=2 \tag{1}$$

 

Similarly, from

$E(6,5)$

,

 

$$x_1+x_3=12,\qquad y_1+y_3=10 \tag{2}$$

 

and from

$F(0,3)$

,

 

$$x_1+x_2=0,\qquad y_1+y_2=6 \tag{3}$$

 

For the x-coordinates, adding (2) and (3):

 

$$2x_1+x_2+x_3=12$$

 

Using

$x_2+x_3=10$

,

 

$$2x_1+10=12$$

 

$$x_1=1$$

 

Then,

 

$$x_2=-1,\qquad x_3=11$$

 

For the y-coordinates, adding (2) and (3):

 

$$2y_1+y_2+y_3=16$$

 

Using

$y_2+y_3=2$

,

 

$$2y_1+2=16$$

 

$$y_1=7$$

 

Then,

 

$$y_2=-1,\qquad y_3=3$$

 

Therefore,

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.

Answer:

Take the point where the two main roads cross as the centre

$O$

.

Using the scale

$$\boxed{1\text{ cm}=200\text{ m}}$$

draw one horizontal line for the E–W main road and one vertical line for the N–S main road, intersecting at

$O$

.

Then draw:

• 5 parallel streets on each side of the N–S road, each
  $1$ 

  cm apart.

• 5 parallel streets on each side of the E–W road, each
  $1$ 

  cm apart.

Thus, there are 10 streets in each direction, all

$200$

m apart.

(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).

Answer:

Because the streets are numbered according to their distance from the main roads, there is a street with a given number on both sides of each main road.

(a) Intersections referred to as

$(4,3)$

There are two 4th N–S streets and two 3rd E–W streets.

Therefore,

 

$$2\times2=4$$

4 intersections​

 

(b) Intersections referred to as

$(3,4)$

Similarly, there are two 3rd N–S streets and two 4th E–W streets.

 

$$2\times2=4$$

 

$$\boxed{4\text{ intersections}}$$

 

Final Answer:

(a) 4(b) 4

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.

Answer:

The screen extends from

$$0\le x\le 800,\qquad 0\le y\le 600$$

For the circle centred at

$A(100,150)$

with radius

$80$

:

$$x:100-80=20\text{ to }100+80=180$$

 

$$y:150-80=70\text{ to }150+80=230$$

All these values lie within the screen. Hence, the first circle is completely inside the screen.

For the circle centred at

$B(250,230)$

with radius

$100$

:

$$x:250-100=150\text{ to }250+100=350$$

 

$$y:230-100=130\text{ to }230+100=330$$

This circle is also completely inside the screen.

No part of either circle lies outside the screen.
(ii)  whether the two circles intersect each other.

Answer: