Ganita Manjari Class 9 Chapter 1 Solutions Orienting Yourself The Use of Coordinates

Students often refer to Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 1 Orienting Yourself The Use of Coordinates Solutions to verify their answers.

Orienting Yourself The Use of Coordinates Class 9 Solutions

Class 9 Ganita Manjari Chapter 1 Solutions

Class 9 Maths Ganita Manjari Chapter 1 Solutions Orienting Yourself The Use of Coordinates

Think and Reflect (NCERT Textbook Page No. 5)

Question 1.
What are the standard widths for a room door? Look around your home and in school.
Solution:
Do it yourself

Question 2.
Are the doors in your school suitable for people in wheelchairs?
Solution:
Do it yourself

Think and Reflect (NCERT Textbook Page No. 7)

Question 1.
What is the x-coordinate of a point on the Y-axis?
Solution:
The x-coordinate of any point on the Y-axis is always 0.

Question 2.
Is there a similar generalisation for a point on the X-axis?
Solution:
Yes, they-coordinate of any point on theX-axis is always 0.

Question 3.
Does point Q(y, x) ever coincide with point P(x, y)? Justify your answer.
Solution:i
Point Q(y, x) coincides with point P(x,y) only, when x =y.
Otherwise, the two points are different.

Question 4.
If x ≠ y then (x, y) ≠ (y, x); and (x, y) = (y, x) if and only if x = y. Is this claim true?
Solution:
Yes, the claim is true.
Two ordered pairs are equal only, when their corresponding coordinates are equal.
So, (x, y) = (y, x) if and only if x = y. Otherwise, they are not equal.

Think and Reflect (NCERT Textbook Page No. 11)

Question 1.
According to the figure given below, answer the following questions.

(i) What has remained the same and what has changed with this reflection?
(ii) Would these observations be the same if ∆ADM is reflected in the X-axis (instead of the Y-axis)?
Solution:
(i) In this reflection about the 7-axis, the x-coordinates of all points remain the same, while the x-coordinates change their signs (positive becomes negative and vice-versa), because reflection in the Y-axis only changes the horizontal position and not the vertical position. The shape and size of the figure remain unchanged, as reflection is a rigid transformation.

(ii) No, the observations would not be exactly the same because in reflection about the X-axis, the x-coordinates remain the same, while the y-coordinates change their signs, as the reflection changes the vertical position but not the horizontal position. However, the shape and size of the figure would still remain unchanged.

Ex 1.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 1.1 Solutions

Exercise 1.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 1.1 Solutions

Question 1.
Figure shows Reiaan’s room with points OABC marking its corners. TheX-and Y-axes are marked in the figure. Point O is the origin.

Answer the following questions.
(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:
The room door lies on the X-axis so, its distance from theX-axis is 0 units.
From the figure, D₁ = (8, 0) and R₁ =(11.5, 0)
So, the door begins 8 units from the Y-axis.

(ii) What are the coordinates of D₁?
Solution:
The coordinates of D₁ are (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:
We have, coordinates of D₁ =(8, 0) and R₁ = (11.5, 0).
So, the width of the door
= Distance between D₁ and R₁ = 11.5 – 8 = 3.5 units
If the unit represents feet, then 3.5 ft« 42 inches, which is quite comfortable for a room door.
Standard residential doors are usually 30-36 inches wide, so this is actually slightly wider than average, which is good.
For wheelchair
A wheelchair typically needs atleast 32 inches (~2.7 ft) clear width.
Since, 3.5 ft > 2.7 ft, the door is wide enough.
So, we can say that yes, the width of this is comfortable and a person using a wheelchair should be able to enter easily without difficulty.

(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:
We have, coordinates of B₁ =(0,1.5) and B₂ =(0,4).
So, the width of the bathroom door
= Distance between B₁ and B₂.
= 4 – 1.5 = 2.5 units.
Since, 2.5 < 3.5, the bathroom door is narrower than the room door.

Ex 1.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 1.2 Solutions

Exercise 1.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 1.2 Solutions

Question 1.
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 given figure, answer the given questions.

(i) Place Reiaan’s rectangular study table with three of its feet at the points (8, 9), (11, 9) and (11, 7).
(a) Where will the fourth foot of the table be?
(b) Is this a good spot for the table?
(c) What is the width of the table? The length? Can you make out the height of the table?
Solution:
Given, three vertices (feet) of the rectangular table are (8,9),(11,9),(11,7).
The X-axis, Y-axis, origin and the given points are plotted on the graph sheet as shown below.

(a) Since, the table is rectangular, so opposite sides are parallel and equal.
Points (8,9) and (11,9) lie on the same horizontal line. Points (11,9) and (11,7) lie on the same vertical line.

So, the fourth point will align with
I. x-coordinate of (8, 9), which is 8.
II. y-coordinate of (11, 7), which is 7.
Therefore, the fourth foot is at (8, 7).

(b) Yes, this is a good spot because
I. the table is properly placed within the room.
II. it does not obstruct any doors or pathways.
III. it is located near the wall, making it convenient for study.

(c) Length = Distance between (8, 9) and (11, 9)
= 11 – 8 = 3 units
Width = Distance between (11, 9) and (11, 7)
= 9 – 7 = 2 units
The height of the table cannot be determined from the given diagram, as it represents only a top view (2D) and does not show vertical dimensions.

(ii) If the bathroom door has a hinge at B, and opens into the bedroom, will it hit the wardrobe? Are there any changes you would suggest if the door is made wider?
Solution:
From the given figure, B,(0,1.5),B2(0, 4)
∴ Width of the bathroom door
= B₁B₂ = 4 – 1.5 = 2.5 units
If the door is hinged atB[ and opens into the bedroom, it will sweep an arc of radius 2.5 units.
The wardrobe starts at the points W₁ = (3, 0) and W₄ = (3, 2).
The nearest part of the wardrobe from B₁ (0,1.5) lies along x = 3, which is at a distance greater than the door width of 2.5 units.
Therefore, the bathroom door will not hit the wardrobe. If the door is made wider, the arc of opening will increase and the chances of collision with the wardrobe will also increase.
There are some suggestions given below.
(a) The door can be made to open outwards instead of opening into the bedroom.
(b) The wardrobe may be repositioned.
(c) The hinge position may be changed.

(iii) Look at Reiaan’s bathroom.
(a) What are the coordinates of the four corners O, F, R and P of the bathroom?
(b) What is the shape of the showering area SHWR in Reiaan’s bathroom? Write the coordinates of the four corners.
(c) Mark off a 3 ft x 2 ft space for the washbasin and a 2 ft x 3 ft space for the toilet. Write the coordinates of the corners of these spaces.
Solution:
(a) From the given figure,
The coordinates of the four corners O, F, R and P of the bathroom are
O (0, 0), F (0,9), R (-6,9) and P (-6, 0).

(b) From the given figure,
S (-6, 6), H (-3,6), W (-2,9) and R (-6,9).
Since, one pair of opposite sides is parallel, so SHWR is a trapezium.

(c) A rectangular space of 3 ft × 2 ft is marked for the washbasin.
Taking the rectangle at the bottom-left corner of the bathroom, the coordinates of its corners are (-6, 0), (-3, 0), (-3,2) and (-6,2).
A rectangular space of 2 ft × 3 ft is marked for the toilet.
Taking the rectangle above the washbasin, the coordinates of its corners are
(-6,2), (-4,2), (-4,5) and (-6,5).
The rectangular spaces for the washbasin and toilet are plotted on the coordinate plane as shown below.

(iv) Other rooms in the house.
(a) Reian’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.
(b) Place a rectangular 5 ft x 3 ft dining table precisely in the centre of the dining room. Write down the coordinates of the feet of the table.
Solution:
(a) From the given figure,
Coordinates of P = (—6, 0)
Coordinates of A = (12, 0)
So, the length of PA = 12 – (-6) = 18 ft
This matches the given length of the dining room.
Since, the width of the dining room is 15 ft and it lies below BA it extends 15 ft downward.
Hence, the coordinates of the four corners are (-6, 0), (12,0), (12,-15), (-6,-15).
The dining room is sketched on the coordinate plane and its corners are marked as shown below.

(b) The dining room extends:
I. From x = -6 to x = 12
II. From y = 0 to y = -15
For centre of the dining room :
x-coordinate of centre = \(\frac{-6+12}{2}\) = 3
and y-coordinate of centre = \(\frac{0+(-15)}{2}\) = -7.5
So, the centre is (3, -7.5).

A rectangular table of size 5 ft x 3 ft is placed at the , centre.
Half-length =2.5 and half-width = 1.5
Coordinates of the four corners (feet) of the table are.
(3-2.5,-7.5-15) =(0.5, -9),
(3+ 2.5,-7.5-1.5) =(5.5, -9),
(3+ 2.5,-7.5+ 1.5) =(5.5, -6)
and (3 – 2.5, – 75 + 1.5) = (0.5, – 6).
Hence, the coordinates of the four feet of the dining table ar (0.5, -9),(5.5, -9),(5.5, -6),(0.5, -6).
The rectangular dining table placed at the centre of the dining room is shown on the coordinate plane as shown below.

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

Orienting Yourself The Use of Coordinates End of Chapter Exercise Solutions

Question 1.
What are the x-coordinate and y-coordinate of the point of intersection of the two axes?
Solution:
The X-axis and 7-axis intersect at the origin.
Therefore, x-coordinate = 0 and y-coordinate = 0.
So, the point of intersection is (0,0).

Question 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 7-axis? Which quadrants can H lie in?
Solution:
Point W has X-coordinate equal to -5. Since, pointHlies on the line through W parallel to the y-axis, its x-coordinate will also be -5, because all points on a line parallel to the Y-axis have the same x-coordinate.
Thus, the coordinates of H can be written as (-5, y), where y can be any real number.

Now, depending on the value of y If y > 0,
H lies in the second quadrant, if y < 0,
H lies in the third quadrant.

If y = 0, H lies on the negative X-axis.
Hence, point H can lie in the second quadrant, third quadrant or on the negative X-axis.

Question 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.
Solution:
(i) Points A(0, -2) and M(-5, -2) have the same
y-coordinate, so AM is parallel to the X-axis (horizontal). Points M(-5, -2) and P(-5,2) have the same x-coordinate, so MP is parallel to the 7-axis (vertical).
Since, a horizontal line is perpendicular to a vertical line.
AM ⊥ MP
Hence, the two perpendicular sides are AM and MP.

(ii) Since, points A and M have the same y-coordinate, so AM is parallel to the X-axis. Also, since points M and P have the same x-coordinate, MP is parallel to the Y-axis.
Hence, one side parallel to an axis is AM (parallel to X-axis) or MP (parallel to Y-axis).

(iii) Points M(-5, -2) andP(-5, 2) have the same x-coordinate and their y-coordinates are equal in magnitude but opposite in sign.
Hence, they are mirror images of each other in the X-axis.
The points R(3, 0), A(0, -2), M(-5, -2) and P(-5, 2) are plotted on the coordinate plane as shown below to verily the results.

Verification
On plotting the points, we observe that
(i) AM is horizontal and MP is vertical.
(ii) AM ⊥ MP
(iii) M and P are symmetric about the X-axis.
Thus, the predictions are verified.

Question 4.
Plot point Z(5, -6) on the cartesian plane. Construct a right angled triangle IZN, find the lengths of the three sides.
Solution:
Given point is Z (5, – 6).
To form a right angled triangle, take point 1(5, 0) on the X-axis and point N(0, -6) on the 7-axis.
The coordinates of the points are 1(5, 0), Z(5 – 6) and MO,-6).
The point Z(5, – 6) and the required triangle IZN are plotted on the coordinate plane as shown below.

Points I and Z have the same x-coordinate, so IZ is vertical.
Points Z and N have the same y-coordinate, so ZN is horizontal.
Since, a vertical line is perpendicular to a horizontal line, so, ∠IZN = 90°.
Therefore, ∆IZN is right-angled at Z.
Now, the lengths of the sides are IZ = |0 – (-6) | = 6 units
and ZN = |5 – 0| = 5 units
∴ IN = \(\sqrt{(5-0)^2+(0-(-6))^2}=\sqrt{25+36}=\sqrt{61}\) units
Hence, the lengths of the sides of the triangle are IZ = 6 units, ZN = 5 units and IN = \(\sqrt{61}\) units.

Question 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?
Solution:
If there were no negative numbers, the coordinate system would exist only in the first quadrant, where both x and y-coordinates are positive.
In such a system, we could represent only those points that lie to the right of the 7-axis and above the X-axis. Points lying to the left of the 7-axis or below the X-axis could not be represented, because they require negative coordinates.
Therefore, this system would not allow us to locate all points on a two-dimensional plane, as it would cover only a part of the plane and not the entire plane.

Question 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:
To check whether the pointsM(-3, -4), A(0, 0) and G(6,8) lie on the same straight line, we use the distance formula,
d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} .\)
Now, MA = \(\sqrt{(0+3)^2+(0+4)^2}\)
= \(\sqrt{3^2+4^2}\)
= \(\sqrt{9+16}\)
= \(\sqrt{25}\)
= 5 units

AG = \(\sqrt{(6-0)^2+(8-0)^2}\)
= \(\sqrt{6^2+8^2}\)
= \(\sqrt{36+64}=\sqrt{100}\)
= 10 units

and MG = \(\sqrt{(6+3)^2+(8+4)^2}\)
= \(\sqrt{9^2+12^2}\)
= \(\sqrt{81+144}=\sqrt{225}\)
= 15 units

Now, MA + AG = 5 + 10 = 15 = MG
Since, the sum of two distances is equal to the third distance, so the points M, A and G lie on the same straight line.
Hence, the given points are collinear.

Question 7.
Use your method (from Problem 6) to check if the points R(-5, -l),B(-2, -5) andC(4, -12) are on the same straight line. Now, plot both sets of points and check your answers.
Solution:
To check whether the points R{-5, -1), B(-2, -5) and C(4, -12) lie on the same straight line, we use the distance formula,
d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} .\)
Now, RB = \(\sqrt{(-2+5)^2+(-5+1)^2}=\sqrt{3^2+(-4)^2}\)
\(\sqrt{9+16}=\sqrt{25}\) = 5 units

BC = \(\sqrt{(4+2)^2+(-12+5)^2}=\sqrt{6^2+(-7)^2}\)
= \(\sqrt{36+49}=\sqrt{85}\)

and RC = \(\sqrt{(4+5)^2+(-12+1)^2}=\sqrt{9^2+(-11)^2}\)
= \(\sqrt{81+121}=\sqrt{202}\)
Now, RB + BC = 5 + \(\sqrt{85} \neq \sqrt{202}\) units = RC

Since, the sum of two distances is not equal to the third distance, so the points R, B and C do not lie on the same straight line.
Hence, the given points are not collinear.
The given points are plotted on the coordinate plane as shown below

Verification
On plotting the points on the cartesian plane, we observe that they do not lie on a single straight line, which verifies the result.

Question 8.
Using the origin as one vertex, plot the vertices of
(i) a right angled isosceles triangle.
Solution:
One possible set of vertices is 0(0, 0), A(4, 0) and B(0, 4).
Here, OA = 4 units and OB = 4 units.
The vertices of a right angled isosceles triangle with one vertex at the origin are plotted below.

Points A and B lie on the X-axis andY-axis, respectively,
so OA is perpendicular to OB.
Since, OA = OB and ∠AOB =90°.
So, ΔOAB is a right angled isosceles triangle.

(ii) an isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV.
Solution:
One possible set of vertices is 0(0, 0), P(-3, -4) and Q(3, -4).
OP = \(\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\) = 5 units
OQ = \((3)^2+(-4)^2=\sqrt{9+16}\) = 5 units
Since, OP =OQ.
So, ΔOPQ an isosceles triangle.
The vertices of an isosceles triangle with one vertex in Quadrant III and the other in Quadrant IV are plotted below

Question 9.
The following table shows the coordinates of points S, M and T. In each case, state whether M is the mid-point of segment ST. Justify your answer.

When M is the mid-point of ST, can you find any connection between the coordinates ofM, S and T?
Solution:
To check whether M is the mid-point of ST, we use the distance formula,
d = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

(i) Given, S(-3,0),M(0, 0) and T(3, 0).
Now, SM = \(\sqrt{(0+3)^2+(0-0)^2}=\sqrt{9}\) = 3 units
and MT = \(\sqrt{(3-0)^2+(0-0)^2}=\sqrt{9}\) = 3 units.
Since, SM = MT,M is the mid-point of ST.

(ii) Given, S(2, 3), M(3, 4) and T(4, 5)
Now, SM = \(\sqrt{(3-2)^2+(4-3)^2}=\sqrt{1+1}=\sqrt{2}\) units
and MT = \(\sqrt{(4-3)^2+(5-4)^2}=\sqrt{1+1}=\sqrt{2}\) units
Since, SM = MT, M is the mid-point of ST.

(iii) Given, S(0, 0), M(0, 5) and 7’(0, -10).
Now, SM = \(\sqrt{(0-0)^2+(5-0)^2}=\sqrt{25}\) = 5 units
and MT = \(\sqrt{(0-0)^2+(-10-5)^2}=\sqrt{225}\) = 15 units
Since, SM ≠ MT
So, M is not the mid-point of ST.

(iv) Given, S(-8,7), M(0, -2) and T(6,-3).
Now, SM = \(\sqrt{(0+8)^2+(-2-7)^2}=\sqrt{64+81}\)
= \(\sqrt{145}\) units
and MT = \(\sqrt{(6-0)^2+(-3+2)^2}=\sqrt{36+1}=\sqrt{37}\) units
Since, SM ≠ MT
So,M is not the mid-point of ST.
Therefore, M is the mid-point of ST only in Cases (i) and (ii).
When M is the mid-point of ST, where S(x₁, y₁) and T(x₂, y₂) then the coordinates of M are given by
M = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Question 10.
Use the connection you found to find the coordinates of/l given thatM(-7,1) is the mid-point of A(3, – 4) and B(x, y).
Solution:
Given, A(3, -4),B(x, y) and mid-point, M(-7, 1)
Using the mid-point formula,
M = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
We have, — 7 = \(\frac{3+x}{2}\)
⇒ 3 + x = -14
⇒ x = -17
and 1 = \(\frac{-4+y}{2}\)
⇒ -4 + y = 2
⇒ y = 6
Therefore, the coordinates of point B are (-17, 6).

Question 11.
Let P and Q be the 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 mid-point of a segment, how would you find the coordinates ofP and Q? Do this for the case, when the points are A(4, 7) and B(16, -2).
Solution:
Given, A(4,7) and B(16, -2)
Let P(x₁, y₁) and Q(x₂, y₂) be the points of trisection.

On putting the value of xl in Eq. (iii), we get

Since, P and Q divide AB into three equal parts, P is the mid-point of A and Q, and Q is the mid-point ofPandB. Using mid-point formula,

On putting the value of xl in Eq. (iii), we get

From Eq.(i),
⇒ 4x₂ = x₂ + 36 ⇒ 3x₂ = 36
⇒ x₂ = 12
From Eq. (i),
x₁ = \(\frac{4+12}{2}\)
⇒ x₁ = 8

On putting the value of y, in Eq. (iv), we get

⇒ 4y₂ = y₂ + 3 ⇒ 3y₂ = 3
⇒ y₂ = 1

From Eq. (ii), y₁ = \(\frac{1}{2}\) ⇒ y₁ = 4
Therefore, P = (8, 4) and Q = (12, 1).

Question 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 0(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.
Solution:
(i) Given, A(1, -8), B(-4, 7), C(-7, -4) and centre 0(0, 0).
Using the distance formula,
OA = \(\sqrt{(1)^2+(-8)^2}=\sqrt{1+64}=\sqrt{65}\) units
OB = \(\sqrt{(-4)^2+(7)^2}=\sqrt{16+49}=\sqrt{65}\) units
and OC = \(\sqrt{(-7)^2+(-4)^2}=\sqrt{49+16}=\sqrt{65}\) units
Since, OA = OB = OC = \(\sqrt{65}\), so the points AB and C are at the same distance from the origin.
Hence, they lie on a circle with centre 0(0,0) and radius \(\sqrt{65}\).

(ii) Given, D (-5,6)andE(0,9).
∴ OD = \(\sqrt{(-5)^2+(6)^2}=\sqrt{25+36}=\sqrt{61}\)
and OE = \(\sqrt{(0)^2+(9)^2}\) = 9
Here, radius of the circle = \(\sqrt{65}\).
Since, \(\sqrt{61}\) < \(\sqrt{65}\), point D lies inside the circle. Since, 9 > \(\sqrt{65}\), point £ lies outside the circle.

Question 13.
The mid-points of the sides of ΔABC 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 the coordinates of ΔABC be A(x₁, y₁), B(x₂, y₂) andC(x₃, y₃).
Given mid-points are D(5,1), E(6,5) and F(0,3).

Using the mid-point formula,
Coordinates of mid-point of(x₁, y₁) and (x₂, y₂)
= \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Since, D is the mid-point of BC.
\(\frac{x_2+x_3}{2}\) = 5
⇒ x₂ + x₃ = 10 ………..(i)

and \(\frac{y_2+y_3}{2}\) = 1
⇒ y₂ + y₃ = 2 ……….(ii)

Since, E is the mid-point of CA.
∴ \(\frac{x_3+x_1}{2}\) = 6
⇒ x₃ + x₁ = 12 …(iii)
and \(\frac{y_3+y_1}{2}\) = 5

⇒ y₃ + y₁ =10 …(iv)
Since, F is the mid-point of AB.
∴ \(\frac{x_1+x_2}{2}\) = 0
⇒ x₁ + x₂ = 0 …(v)
and \(\frac{y_1+y_2}{2}\) = 3

⇒ y₁ + y₂ = 6 ………………….(vi)
From Eq. (v), x₂ = -x₁

On putting the value of x₂ in Eq.(i), we get
-x₁ + x₃ = 10
⇒ x₃ = 10 + x₁

On putting the value of x₃ in Eq. (iii), we get
(10 + x₁) + x₁ = 12
⇒ 2x₁ + 10 = 12
⇒ 2x₁ = 2
⇒ x₁ = 1

On substituting x₃ = 1 in Eq.(v), we get x₂ = -1
On substituting xl = lin Eq.(vii), we get
x₃ = 11
From Eq. (vi), y₂ = 6 – y₁

On putting the value of y₂ in Eq.(ii), we get
(6 – y₁) + y₃ =2
⇒ y₃ = y₁ – 4

On putting the value of 7 , in Eq. (iv), we get (y₁ – 4) + y₁ = 10
⇒ 2y₁ – 4 = 10
⇒ 2y₁ = 14
⇒ y₁ = 7

On substituting y₁ = 7 in Eq. (vi), we get
⇒ y₃ = -1
On substituting y₁ = 7 in Eq.(viii), we get y₃ = 3
Therefore, the coordinates of the points are A(1, 7), B(-1, -1) and C(11, 3).

Question 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)?
Solution:
(i) Given, scale 1 cm =200 m
Since, the streets are 200 m apart, so the distance between two consecutive streets in the model will be 1cm.

There are 10 streets in the N-S direction and 10 streets in the E-W direction.
Therefore, draw 10 equally spaced parallel lines in the N-S direction and 10 equally spaced parallel lines in the E-W direction, each 1 cm apart. The two main roads intersect at the centre and the model forms a square grid.

The model of the city showing N-S and E-W streets drawn to scale is given below.

(ii) Each street intersection is represented by an ordered pair (m,n), where m denotes the street number in the N-S direction and n denotes the street number in the E-W direction.
(a) The ordered pair (4,3) represents the intersection of the 4th N-S street and the 3rd E-W street.
Since, one N-S street and one E-W street intersect at only one point, there is exactly one such intersection.
(b) Similarly, the ordered pair (3, 4) represents the intersection of the 3rd N-S street and the 4th E-W street. This is also a unique intersection.

The street intersections such as (2, 5), (3, 4) and (4, 3) are shown in the model below.

Question 15.
A computer graphics program displays images on a t 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 5(250, 230). Determine:
(i) Whether any part of either circle lies outside the screen.
(ii) Whether the two circles intersect each other.
Solution:
Given
Screen size = 800 × 600 pixels
Circle 1 Centre A (100, 150) and radius = 80 pixels
Circle 2 Centre 5 (250, 230) and radius = 100 pixels

(i) For circle with centre A(100,150) and radius 80 pixels 100 – 80 = 20 > 0,100 + 80 = 180 < 800 150-80 = 70 > 0, 150 + 80 = 230 < 600 Hence, the circle lies completely inside the screen. For circle with centre 5(250,230) and radius 100: 250 -100 = 150 > 0,250 + 100 =350 <800 230 -100 = 130 > 0,230 + 100 =330 < 600
Hence, this circle also lies completely inside the screen.

(ii) Distance between centres,
AS = \(\sqrt{(250-100)^2+(230-150)^2}\)
= \(\sqrt{150^2+80^2}\)
= \(\sqrt{22500+6400}=\sqrt{28900}\) = 170 units
Sum of radii = 80 + 100 = 180 units
Since, AS < Sum of radii
Hence, the two circles intersect each other.
The positions of the two circular icons on the screen are shown in the coordinate plane below.

Question 16.
Plot the points A (2, 1), 5 (-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:
Given points are A (2, 1), 5 (-1, 2), C (-2, -1) and D(1, -2).
The points A (2, 1), 5 (-1, 2), C (-2, -1) andD(1, -2) are plotted on the coordinate plane and joined as shown below.

We check whether ABCD is a square by finding the lengths of its sides and diagonals.
Lengths of the sides,

Thus, all four sides are equal. Lengths of the diagonals,

Thus, the diagonals are equal.
Since, all sides are equal and diagonals are equal. ABCD is a square.
∵ Side of square = \(\sqrt{10}\) units
∴ Area of square = (Side)² = (\(\sqrt{10}\))² = 10 sq units.
Hence, ABCD is a square and its area is 10 sq units.