class 9 ganit manjari exercise 1.3 solution

Class 9 Maths · Ganita Manjari · End-of-Chapter Exercises

What are the x‑coordinate and y‑coordinate of the point of intersection of the two axes?

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

Therefore:

The x‑coordinate = 0
The y‑coordinate = 0

✅ So, the point of intersection is (0, 0).

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?

If point W has x‑coordinate –5, then any point on the line through W parallel to the y‑axis will also have x‑coordinate –5.

Therefore, the coordinates of H will be of the form:

H = (-5, y), where y can be any real number.

Now, depending on the value of y:

If y > 0, then H lies in Quadrant II.
If y < 0, then H lies in Quadrant III.
If y = 0, then H lies on the x‑axis.

📍 So, H can lie in: Quadrant II, Quadrant III, or on the x‑axis.

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.

Let us observe:

AM joins A(0, –2) to M(–5, –2), so it is horizontal.
MP joins M(–5, –2) to P(–5, 2), so it is vertical.

A horizontal line and a vertical line are perpendicular.

📐 Therefore: AM ⟂ MP — the two perpendicular sides are AM and MP.

(ii)

One side of RAMP that is parallel to one of the axes.

AM is parallel to the x‑axis because both points A and M have the same y‑coordinate (–2).
MP is parallel to the y‑axis because both points M and P have the same x‑coordinate (–5).

📏 One side parallel to an axis is: AM, which is parallel to the x‑axis, or MP, which 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.

Comparing M(–5, –2) and P(–5, 2):

They have the same x‑coordinate.
Their y‑coordinates are equal in magnitude but opposite in sign.

🪞 The points M and P are mirror images of each other. The axis is the x‑axis.

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.)

Point Z is (5, –6).

To form a right‑angled triangle easily, take:

I = (5, 0) on the x‑axis

N = (0, –6) on the y‑axis

Then triangle IZN is right‑angled at Z? Let’s check:

IZ is vertical
ZN is horizontal
So, triangle IZN is right‑angled at Z.

Coordinates of the points:

I = (5, 0)

Z = (5, –6)

N = (0, –6)

Now find the lengths of the sides:

IZ: Distance between (5, 0) and (5, –6) = 0 – (–6) = 6 units
ZN: Distance between (5, –6) and (0, –6) = 5 – 0 = 5 units
IN: Using distance formula: $IN = \sqrt[(5 - 0)² + (0 - (-6))²] = \sqrt(5² + 6²) = \sqrt(25 + 36) = \sqrt61 units$

✅ One possible right‑angled triangle is formed by: I = (5, 0), Z = (5, –6), N = (0, –6)
Lengths: IZ = 6 units, ZN = 5 units, IN = √61 units

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?

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

So:

On the x‑axis, we could mark only the points to the right of the origin.
On the y‑axis, we could mark only the points above the origin.

This means we could locate points only in:

Quadrant I
the positive part of the x‑axis
the positive part of the y‑axis
and the origin

We would not be able to locate:

points in Quadrant II
points in Quadrant III
points in Quadrant IV
points on the negative parts of the axes

⚠️ Therefore, such a system would not allow us to locate all the points on a 2‑D plane.

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.

To check whether the points M (–3, –4), A (0, 0), and G (6, 8) lie on the same straight line using the Distance formula:

d = \sqrt[(x₂ - x₁)² + (y₂ - y₁)²]

MA = \sqrt[(0 + 3)² + (0 + 4)²] = \sqrt(3² + 4²) = \sqrt(9 + 16) = \sqrt25 = 5

AG = \sqrt[(6 - 0)² + (8 - 0)²] = \sqrt(36 + 64) = \sqrt100 = 10

MG = \sqrt[(6 + 3)² + (8 + 4)²] = \sqrt(9² + 12²) = \sqrt(81 + 144) = \sqrt225 = 15

Now checking: MA + AG = 5 + 10 = 15 = MG

✅ Since the sum of two distances equals the third, the points M, A and G lie on the same straight line.

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.

To check whether the points R (–5, –1), B (–2, –5) and C (4, –12) lie on the same straight line using the distance formula:

RB = \sqrt[(-2 + 5)² + (-5 + 1)²] = \sqrt(3² + (-4)²) = \sqrt(9 + 16) = \sqrt25 = 5

BC = \sqrt[(4 + 2)² + (-12 + 5)²] = \sqrt(6² + (-7)²) = \sqrt(36 + 49) = \sqrt85

RC = \sqrt[(4 + 5)² + (-12 + 1)²] = \sqrt(9² + (-11)²) = \sqrt(81 + 121) = \sqrt202

Now: RB + BC = 5 + √85 ≠ √202

❌ Since the sum of two distances is not equal to the third, the points R, B and C do not lie on the same straight line.

Using the origin as one vertex, plot the vertices of:

(i)

A right‑angled isosceles triangle.

One possible set of vertices of triangle OAB is:

O = (0, 0)

A = (4, 0)

B = (0, 4)

Because:

OA = 4 units
OB = 4 units
OA is perpendicular to OB

✅ So triangle OAB is a right‑angled isosceles triangle.

(ii)

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

One possible set of vertices is:

O = (0, 0)

P = (–3, –4)

Q = (3, –4)

Explanation:

P lies in Quadrant III
Q lies in Quadrant IV
OP = OQ = 5 units

✅ So triangle OPQ is an isosceles triangle.

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?

Using Distance Formula: d = √[(x₂−x₁)² + (y₂−y₁)²]

Row	S	M	T	SM	MT	M midpoint?
(i)	(−3, 0)	(0, 0)	(3, 0)	3	3	Yes
(ii)	(2, 3)	(3, 4)	(4, 5)	√2	√2	Yes
(iii)	(0, 0)	(0, 5)	(0, −10)	5	15	No
(iv)	(−8, 7)	(0, −2)	(6, −3)	√145	√37	No

Detailed calculations:

(i) S(−3, 0), M(0, 0), T(3, 0)

SM = \sqrt[(0-(-3))² + (0-0)²] = \sqrt[9 + 0] = 3 MT = \sqrt[(3-0)² + (0-0)²] = \sqrt[9 + 0] = 3 SM = MT = 3 \to Yes, M is the midpoint

(ii) S(2, 3), M(3, 4), T(4, 5)

SM = \sqrt[(3-2)² + (4-3)²] = \sqrt[1 + 1] = \sqrt2 MT = \sqrt[(4-3)² + (5-4)²] = \sqrt[1 + 1] = \sqrt2 SM = MT = \sqrt2 \to Yes, M is the midpoint

(iii) S(0, 0), M(0, 5), T(0, −10)

SM = \sqrt[(0-0)² + (5-0)²] = \sqrt[0 + 25] = 5 MT = \sqrt[(0-0)² + (-10-5)²] = \sqrt[0 + 225] = 15 SM \neq MT (5 \neq 15) \to No, M is NOT the midpoint

(iv) S(−8, 7), M(0, −2), T(6, −3)

SM = \sqrt[(0-(-8))² + (-2-7)²] = \sqrt[64 + 81] = \sqrt145 MT = \sqrt[(6-0)² + (-3-(-2))²] = \sqrt[36 + 1] = \sqrt37 SM \neq MT (\sqrt145 \neq \sqrt37) \to No, M is NOT the midpoint

🔗 Connection: When M is the midpoint of ST, the coordinates of M are the averages of the corresponding coordinates of S and T: M = ((x₁+x₂)/2, (y₁+y₂)/2)

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).

Method 1: Using distance formula

Given: M (–7, 1) is the midpoint of A (3, –4) and B (x, y).

Using distance formula: AM = √[(–7 − 3)² + (1 − (–4))²]

AM = \sqrt[(-10)² + (5)²] = \sqrt(100 + 25) = \sqrt125

Now, MB = √[(x + 7)² + (y − 1)²]

Since Distance AM = Distance MB:

\sqrt[(x + 7)² + (y - 1)²] = \sqrt125 Squaring both sides: (x + 7)² + (y - 1)² = 125 \dots(1)

Also, since M is the midpoint, it lies between A and B, so coordinates of B will be such that M divides AB into two equal parts. From symmetry (or balancing coordinates):

From x‑coordinates: Distance from A to M is –7 − 3 = –10. So from M to B must also be –10.
x = –7 − 10 = –17
From y‑coordinates: Distance from A to M is 1 − (–4) = 5. So from M to B must also be 5.
y = 1 + 5 = 6

Method 2: Using midpoint formula

Given: A = (3, –4), B = (x, y), M = (–7, 1)

Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

(3 + x)/2 = -7 \Rightarrow 3 + x = -14 \Rightarrow x = -17 (-4 + y)/2 = 1 \Rightarrow -4 + y = 2 \Rightarrow y = 6

✅ Therefore, coordinates of B: B = (–17, 6)

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).

To trisect AB, P and Q divide the segment into 3 equal parts.

Given: A (4, 7), B (16, –2)

Let P (x₁, y₁) and Q (x₂, y₂)

Here, P is midpoint of A and Q, so:

x₁ = (4 + x₂)/2 \dots(1) y₁ = (7 + y₂)/2 \dots(2)

And Q is midpoint of P and B, so:

x₂ = (x₁ + 16)/2 \dots(3) y₂ = (y₁ - 2)/2 \dots(4)

Solving for x‑coordinates:

From (1): x₁ = (4 + x₂)/2 and from (3): x₂ = (x₁ + 16)/2

Substituting (1) into (3):

x₂ = ((4 + x₂)/2 + 16)/2 = (4 + x₂ + 32)/4 = (x₂ + 36)/4 So, 4x₂ = x₂ + 36 \Rightarrow 3x₂ = 36 \Rightarrow x₂ = 12 Then from (1): x₁ = (4 + 12)/2 = 8

Solving for y‑coordinates:

From (2): y₁ = (7 + y₂)/2 and from (4): y₂ = (y₁ − 2)/2

Substituting (2) into (4):

y₂ = ((7 + y₂)/2 - 2)/2 = (7 + y₂ - 4)/4 = (y₂ + 3)/4 So, 4y₂ = y₂ + 3 \Rightarrow 3y₂ = 3 \Rightarrow y₂ = 1 Then from (2): y₁ = (7 + 1)/2 = 4

✅ Therefore, P = (8, 4) and Q = (12, 1).

(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?

Distance OA: = \sqrt(1² + (-8)²) = \sqrt(1 + 64) = \sqrt65

Distance OB: = \sqrt((-4)² + 7²) = \sqrt(16 + 49) = \sqrt65

Distance OC: = \sqrt((-7)² + (-4)²) = \sqrt(49 + 16) = \sqrt65

Since all three points are at the same distance from origin, they lie on a circle centred at (0, 0).

⭕ Radius = √65
Points A, B and C lie on circle K with center (0, 0) and radius √65.

(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.

Distance OD: = \sqrt((-5)² + 6²) = \sqrt(25 + 36) = \sqrt61

Distance OE: = \sqrt(0² + 9²) = 9

Compare with radius √65 ≈ 8.06:

√61 ≈ 7.81 < √65 → D lies inside the circle
9 > √65 → E lies outside the circle

📍 D lies inside the circle.
E lies outside the circle.

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.

Let the vertices of triangle ABC be A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

Given midpoints: D(5, 1), E(6, 5), F(0, 3)

D is midpoint of BC, so using midpoint formula:

x₂ + x₃ = 10 \dots(1) y₂ + y₃ = 2 \dots(2)

E is midpoint of CA, therefore:

x₃ + x₁ = 12 \dots(3) y₃ + y₁ = 10 \dots(4)

F is midpoint of AB, therefore:

x₁ + x₂ = 0 \dots(5) y₁ + y₂ = 6 \dots(6)

Solving for x‑coordinates:

From (5): x₂ = –x₁

Substitute into (1): –x₁ + x₃ = 10 ⇒ x₃ = 10 + x₁ …(7)

Substituting into (3): (10 + x₁) + x₁ = 12

\Rightarrow 2x₁ + 10 = 12 \Rightarrow 2x₁ = 2 \Rightarrow x₁ = 1 Then: x₂ = -1 \Rightarrow x₃ = 10 + 1 = 11

Solving y‑coordinates:

From (6): y₂ = 6 – y₁

Substituting into (2): (6 – y₁) + y₃ = 2 ⇒ y₃ = y₁ – 4 …(8)

Substituting into (4): (y₁ – 4) + y₁ = 10

\Rightarrow 2y₁ - 4 = 10 \Rightarrow 2y₁ = 14 \Rightarrow y₁ = 7 Then: y₂ = 6 - 7 = -1 \Rightarrow y₃ = 7 - 4 = 3

✅ Therefore, the coordinates of the points are:
A (1, 7), B (–1, –1) and C (11, 3).

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.

Drawing the model of the city

Since:

Scale is 1 cm = 200 m
Each pair of adjacent streets is 200 m apart

Given that:

10 streets in the N–S direction
10 streets in the E–W direction

Hence, the model will consist of:

10 vertical parallel lines (for N–S streets)
10 horizontal parallel lines (for E–W streets)

Each consecutive line 1 cm apart, which forms a square grid.

(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).

A street intersection is named by: (first number, second number)

= (N–S street number, E–W street number)

So:

(4, 3) means the intersection of the 4th N–S street and the 3rd E–W street
(3, 4) means the intersection of the 3rd N–S street and the 4th E–W street

Now, one N–S street and one E–W street can meet at only one point.

Therefore:

✅ (a) Only one intersection can be called (4, 3).
(b) Only one intersection can be called (3, 4).

A computer graphics program displays images on a rectangular screen whose coordinate system has the origin at the bottom-left corner. The screen is 800 pixels wide and 600 pixels high. A circular icon of radius 80 pixels is drawn with its centre at the point A (100, 150). Another circular icon of radius 100 pixels is drawn with its centre at the point B (250, 230). Determine:
(i) whether any part of either circle lies outside the screen.
(ii) whether the two circles intersect each other.

🖥️ (i) No. Each of the two circles lies inside the screen.

🔵 (ii) Yes. The two circles intersect each other as shown in the picture.

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?

Yes, ABCD is a square.

We can check it by finding the lengths of sides and diagonals.

Finding the lengths of all sides:

AB = \sqrt[(-1 - 2)² + (2 - 1)²] = \sqrt[(-3)² + 1²] = \sqrt(9 + 1) = \sqrt10

BC = \sqrt[(-2 - (-1))² + (-1 - 2)²] = \sqrt[(-1)² + (-3)²] = \sqrt(1 + 9) = \sqrt10

CD = \sqrt[(1 - (-2))² + (-2 - (-1))²] = \sqrt[3² + (-1)²] = \sqrt(9 + 1) = \sqrt10

DA = \sqrt[(2 - 1)² + (1 - (-2))²] = \sqrt[1² + 3²] = \sqrt(1 + 9) = \sqrt10

All four sides are equal.

Now finding diagonals:

AC = \sqrt[(-2 - 2)² + (-1 - 1)²] = \sqrt[(-4)² + (-2)²] = \sqrt(16 + 4) = \sqrt20

BD = \sqrt[(1 - (-1))² + (-2 - 2)²] = \sqrt[2² + (-4)²] = \sqrt(4 + 16) = \sqrt20

Diagonals are equal.

✅ All sides are equal and diagonals are equal, so ABCD is a square.

Area of ABCD:

Area = (side)² = (\sqrt10)² = 10 square units

📐 Area of the square = 10 square units.

class 9 ganit manjari exercise 1.3 solution

End-of-Chapter Exercises Class 9 · Ganita Manjari

Leave a Comment Cancel Reply