Class 9 Maths Chapter 1 End-of-Chapter Exercises Solutions

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

At the origin, both coordinates are zero:

O(0, 0)

Coordinates of point H:

Since the line through W is parallel to the y-axis, it is a vertical line.

All points on a vertical line have the same x-coordinate.

where y can be any real number.

Quadrants in which H can lie:

Quadrant II or Quadrant III

Join the points in the order:

R → A → M → P → R

The four sides of quadrilateral RAMP are RA, AM, MP, and PR.

Consider side AM with endpoints:

A = (0, -2), M = (-5, -2)

Both points have the same y-coordinate (y = −2), so segment AM is horizontal.

Now consider side MP with endpoints:

M = (-5, -2), P = (-5, 2)

Both points have the same x-coordinate (x = −5), so segment MP is vertical.

Since a horizontal line and a vertical line are perpendicular:

AM ⟂ MP

Since A(0, -2) and M(-5, -2) share the same y-coordinate:

AM ∥ x-axis

Also, since M(-5, -2) and P(-5, 2) share the same x-coordinate, MP ∥ y-axis.

AM ∥ x-axis

Consider the vertices:

M = (-5, -2), P = (-5, 2)

Notice that:

Reflection of any point (x, y) across the x-axis gives (x, −y):

(-5, -2) → (-5, 2)

x-axis

Verification by plotting:

Plot the four points on the Cartesian coordinate plane:

R(3, 0), A(0, -2), M(-5, -2), P(-5, 2)

and join them in order to form quadrilateral RAMP.

From the plot:

Position of point Z:

Point Z has coordinates:

Since its x-coordinate is positive and y-coordinate is negative (+, −), Z lies in Quadrant IV.

Construction of right-angled triangle IZN:

There are many possible choices for points I and N. We choose a convenient construction with the right angle at vertex Z that gives simple whole-number side lengths.

Choose point I vertically above Z:

Choose point N horizontally to the left of Z:

Since I and Z have the same x-coordinate (x = 5), segment IZ is vertical.

Since N and Z have the same y-coordinate (y = −6), segment ZN is horizontal.

A vertical segment and a horizontal segment meet at a right angle:

Therefore, ΔIZN is a right-angled triangle with the right angle at Z.

Length of vertical side IZ:

IZ = |-2 - (-6)| = |-2 + 6| = 4 units

Length of horizontal side ZN:

ZN = |5 - 2| = 3 units

Length of hypotenuse IN:

By the Pythagorean theorem:

Taking the square root:

IZ = 4 units, ZN = 3 units, IN = 5 units

This forms a classic 3–4–5 right-angled triangle.

Understanding a coordinate system without negative numbers:

In the standard Cartesian coordinate system, the coordinate axes extend infinitely in both directions from the origin O(0, 0), using positive and negative real numbers to cover all four quadrants.

If negative numbers did not exist:

Can this system locate all points on a 2-D plane?

No, this system would not allow us to locate all points on a 2-D plane.

Reason:

Suggested Method (The Distance Formula Method):

Three points M, A, and G lie on the same straight line (are collinear) if and only if the sum of the distances between two pairs of points is equal to the distance between the third pair.

We use the Distance Formula derived from the Pythagorean theorem:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

If MA + AG = MG, then the three points lie on the same straight line, with point A lying between M and G.

Given points:

M = (-3, -4), A = (0, 0), G = (6, 8)

1. Distance MA:

MA = √[(0 − (−3))² + (0 − (−4))²]

MA = √[9 + 16] = √25 = 5 units

2. Distance AG:

AG = √[(6 − 0)² + (8 − 0)²]

AG = √[36 + 64] = √100 = 10 units

3. Distance MG:

MG = √[(6 − (−3))² + (8 − (−4))²]

MG = √[(6 + 3)² + (8 + 4)²]

MG = √[81 + 144] = √225 = 15 units

Checking the collinearity condition:

MA + AG = 5 + 10 = 15 units

Alternative Method (Slope / Ratio of Coordinates):

The ratio of vertical change to horizontal change from M to A is 4/3, and from A to G is 8/6 = 4/3. Since both segments share point A and have the same ratio, they form a single straight line.

Given points:

R = (-5, -1), B = (-2, -5), C = (4, -12)

Using the Distance Formula:

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

1. Distance RB:

RB = √[(−2 − (−5))² + (−5 − (−1))²]

RB = √[(−2 + 5)² + (−5 + 1)²]

RB = √[9 + 16] = √25 = 5 units

2. Distance BC:

BC = √[(4 − (−2))² + (−12 − (−5))²]

BC = √[(4 + 2)² + (−12 + 5)²]

BC = √[36 + 49] = √85 units ≈ 9.22 units

3. Distance RC:

RC = √[(4 − (−5))² + (−12 − (−1))²]

RC = √[(4 + 5)² + (−12 + 1)²]

RC = √[81 + 121] = √202 units ≈ 14.21 units

Checking the collinearity condition:

RB + BC = 5 + √85 ≈ 5 + 9.22 = 14.22 units

Notice that:

(Exact check: (5 + √85)² = 25 + 10√85 + 85 = 110 + 10√85 ≈ 202.196 ≠ 202)

Also, comparing slopes:

Slope of RB = (−5 − (−1)) / (−2 − (−5)) = −4/3

Slope of BC = (−12 − (−5)) / (4 − (−2)) = −7/6

Since −4/3 ≠ −7/6, the directions of segments RB and BC differ.

Plotting both sets of points and checking the answers:

Refer to Fig. 1.11:

An isosceles right-angled triangle has two perpendicular sides of equal length.

Let the origin O(0, 0) be the vertex containing the right angle (90°).

Choose the second vertex A along the positive x-axis:

Side length OA = |4 − 0| = 4 units.

Choose the third vertex B along the positive y-axis at the same distance:

Side length OB = |4 − 0| = 4 units.

Since the x-axis and y-axis are mutually perpendicular:

And since OA = OB = 4 units, ΔOAB is an isosceles right-angled triangle.

Length of hypotenuse AB:

AB = √[4² + 4²] = √[16 + 16] = √32 = 4√2 units

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

(Note: Any non-zero distance a along perpendicular axes yields a valid triangle, such as (a, 0) and (0, a).)

Recall the sign rules for the lower quadrants:

We take origin O(0, 0) as the shared apex vertex, and choose two points P and Q with equal distances from the origin (OP = OQ):

Choose vertex P in Quadrant III:

Distance from origin OP:

OP = √[(−3)² + (−4)²] = √[9 + 16] = √25 = 5 units

Choose vertex Q in Quadrant IV as the reflection of P across the y-axis:

Distance from origin OQ:

OQ = √[3² + (−4)²] = √[9 + 16] = √25 = 5 units

Since OP = OQ = 5 units, triangle OPQ is isosceles.

PQ = |3 − (−3)| = 6 units

Since 5 + 5 = 10 > 6 (triangle inequality holds) and OP = OQ, ΔOPQ is a genuine isosceles triangle with P in Quadrant III and Q in Quadrant IV.

O(0, 0), P(-3, -4), Q(3, -4)

A point M is the midpoint of segment ST when its coordinates are halfway between endpoints S(x₁, y₁) and T(x₂, y₂):

M = ((x_S + x_T)/2, (y_S + y_T)/2)

For any two points S(x₁, y₁) and T(x₂, y₂), the coordinates of midpoint M are the averages of the coordinates of S and T:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Representative calculation (Row 2):

For S(2, 3) and T(4, 5):

Midpoint = ((2 + 4)/2, (3 + 5)/2)

Since the given coordinates are M(3, 4), which matches the calculated midpoint, M is indeed the midpoint of ST.

The same midpoint calculation is used for all four rows in the table.

When M is the midpoint of segment ST, each coordinate of M is the arithmetic average (mean) of the corresponding coordinates of S and T:

M = ((x_S + x_T)/2, (y_S + y_T)/2)

This fundamental coordinate connection directly leads to Question 10, where it is used to determine an unknown endpoint.

Using the midpoint relationship:

Since M(-7, 1) is the midpoint of segment AB:

M = ((x_A + x)/2, (y_A + y)/2)

((3 + x)/2, (-4 + y)/2) = (-7, 1)

Equating corresponding coordinates:

For the x-coordinate:

For the y-coordinate:

Verification:

Calculate the midpoint of A(3, -4) and B(-17, 6):

Midpoint = ((3 + (-17))/2, (-4 + 6)/2) = (-14/2, 2/2) = (-7, 1)

This is exactly M(-7, 1), confirming the solution.

Understanding the trisection relationship:

Points P and Q trisect segment AB into three equal parts, with P closer to A and Q closer to B:

Using our knowledge of midpoints:

Given endpoints:

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

Let the coordinates of the trisection points be:

P = (x₁, y₁), Q = (x₂, y₂)

Setting up midpoint equations:

1. Since P is the midpoint of AQ:

x₁ = (4 + x₂)/2, y₁ = (7 + y₂)/2 ... (1)

2. Since Q is the midpoint of PB:

x₂ = (x₁ + 16)/2, y₂ = (y₁ + (-2))/2 ... (2)

From (1), we have x₁ = (4 + x₂)/2.

Substitute this expression for x₁ into equation (2):

x₂ = (((4 + x₂)/2) + 16) / 2

Multiply both sides by 2:

Multiply both sides by 2 again:

Substitute x₂ = 12 into (1) to determine x₁:

x₁ = (4 + 12)/2 = 16/2 = 8

From (1), we have y₁ = (7 + y₂)/2.

Substitute this expression for y₁ into equation (2):

y₂ = (((7 + y₂)/2) - 2) / 2

Multiply both sides by 2:

Multiply both sides by 2 again:

Substitute y₂ = 1 into (1) to determine y₁:

y₁ = (7 + 1)/2 = 8/2 = 4

Coordinates of the trisection points:

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

Verification using midpoints:

Length verification:

AP = √[(8 − 4)² + (4 − 7)²] = √[16 + 9] = 5 units

PQ = √[(12 − 8)² + (1 − 4)²] = √[16 + 9] = 5 units

QB = √[(16 − 12)² + (−2 − 1)²] = √[16 + 9] = 5 units

A circle is the set of all points in a plane equidistant from a fixed centre point.

Here, the centre of circle K is the origin O(0, 0).

The squared distance of any point (x, y) from the origin is:

For point A(1, -8):

For point B(-4, 7):

For point C(-7, -4):

Taking the square root:

OA = OB = OC = √65 units

Since points A, B, and C are all at the exact same distance from the centre O(0, 0), they all lie on the same circle K.

From part (i), the squared radius of circle K is:

A point (x, y) lies:

For point D(-5, 6):

61 < 65 ⟹ OD² < r² ⟹ OD < r

For point E(0, 9):

81 > 65 ⟹ OE² > r² ⟹ OE > r

Conclusion:

Point D lies within the circle, and point E lies outside circle K.

Let the coordinates of the vertices of triangle ABC be:

Given coordinates of the midpoints:

where:

Using the midpoint formula:

Since D is the midpoint of BC:

((x₂ + x₃)/2, (y₂ + y₃)/2) = (5, 1)

Since E is the midpoint of CA:

((x₃ + x₁)/2, (y₃ + y₁)/2) = (6, 5)

Since F is the midpoint of AB:

((x₁ + x₂)/2, (y₁ + y₂)/2) = (0, 3)

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

x₁ + x₂ + x₃ = 11 ... (4)

Now, solve for each x-coordinate using equation (4):

x₁ = 11 − (x₂ + x₃) = 11 − 10 = 1

x₂ = 11 − (x₁ + x₃) = 11 − 12 = −1

x₃ = 11 − (x₁ + x₂) = 11 − 0 = 11

(y₂ + y₃) + (y₃ + y₁) + (y₁ + y₂) = 2 + 10 + 6

y₁ + y₂ + y₃ = 9 ... (5)

Now, solve for each y-coordinate using equation (5):

y₁ = 9 − (y₂ + y₃) = 9 − 2 = 7

y₂ = 9 − (y₃ + y₁) = 9 − 10 = −1

y₃ = 9 − (y₁ + y₂) = 9 − 6 = 3

Checking the midpoints of the calculated vertices:

A = (1, 7), B = (−1, −1), C = (11, 3)

Given scale:

Since adjacent streets are 200 m apart in reality, they should be drawn:

1 cm apart

in the model.

To construct the model:

According to the convention given in the problem, an intersection is written as:

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

The ordered pair (4, 3) refers to:

Since two straight lines that are perpendicular intersect at exactly one unique point:

These two particular streets meet at only one point.

street intersection can be referred to as (4, 3).

The ordered pair (3, 4) refers to:

Again, these two particular straight lines intersect at exactly one unique point.

street intersection can be referred to as (3, 4).

Key idea:

The order of the numbers in an ordered pair specifies which street runs N–S and which runs E–W, identifying two completely different crossings.

The rectangular screen has its origin (0, 0) at the bottom-left corner.

Given screen dimensions:

For any circle with centre (x, y) and radius r, the circle spans:

x − r to x + r

y − r to y + r

to

to

Comparing with the screen boundaries:

All these values lie completely within the screen boundaries [0, 800] and [0, 600].

to

to

Comparing with the screen boundaries:

These values also lie completely within the screen boundaries.

No part of either circle lies outside the screen.

First find the distance between their centres A(100, 150) and B(250, 230):

AB = √[(250 − 100)² + (230 − 150)²]

The radii of the two circles are:

For two circles to intersect at two distinct points, the distance between their centres must satisfy the triangle inequality:

|r₁ − r₂| < AB < r₁ + r₂

Here,

|r₁ − r₂| = |80 − 100| = 20

r₁ + r₂ = 80 + 100 = 180

Comparing the values:

Since 20 < 170 < 180 is strictly satisfied, the two circles intersect each other at two distinct points.

Given coordinates of the four points:

First, plot the four points on the Cartesian coordinate plane and join them in order to form quadrilateral ABCD.

We compare the squares of the side lengths using the distance formula, which avoids unnecessary square-root calculations:

AB² = (−1 − 2)² + (2 − 1)²

BC² = (−2 − (−1))² + (−1 − 2)²

CD² = (1 − (−2))² + (−2 − (−1))²

DA² = (2 − 1)² + (1 − (−2))²

AB² = BC² = CD² = DA² = 10

Taking square roots,

AB = BC = CD = DA = √10 units

We already know:

Now calculate the square of diagonal AC:

AC² = (−2 − 2)² + (−1 − 1)²

Checking the Pythagorean relation:

AB² + BC² = 10 + 10 = 20 = AC²

Since AB² + BC² = AC², by the converse of the Pythagorean theorem:

ABCD is a square.

Since the side length is √10 units (and AB² = 10):

ABCD is a square and its area is 10 square units.