Coordinate System & Linear Graphs

Draw the coordinate axes X'OX and Y'OY on a graph paper and choose the units such that 1 cm represents 1 unit on both the axes.

Steps:

1. To plot A(3, 5), start from the origin O, move 3 units to the right along the x-axis and then move 5 units upwards parallel to the y-axis. Mark a dot at this place.

2. To plot B(2, 6), start from the origin O, move 2 units to the right along the x-axis and then move 6 units upwards parallel to the y-axis. Mark a dot at this place.

3. To plot C(0, 3), start from the origin O and move 3 units upwards along the y-axis, as its abscissa is zero. Mark a dot at this place.

4. To plot D(5, 0), start from the origin O and move 5 units to the right along the x-axis, as its ordinate is zero. Mark a dot at this place.

5. To plot E(2, -4), start from the origin O, move 2 units to the right along the x-axis and then move 4 units downwards parallel to the y-axis. Mark a dot at this place.

6. To plot F(-3, 5), start from the origin O, move 3 units to the left along the x-axis and then move 5 units upwards parallel to the y-axis. Mark a dot at this place.

7. To plot G(1, -3), start from the origin O, move 1 unit to the right along the x-axis and then move 3 units downwards parallel to the y-axis. Mark a dot at this place.

8. To plot H(-2, -3), start from the origin O, move 2 units to the left along the x-axis and then move 3 units downwards parallel to the y-axis. Mark a dot at this place.

9. To plot I(-4, -1), start from the origin O, move 4 units to the left along the x-axis and then move 1 unit downwards parallel to the y-axis. Mark a dot at this place.

10. To plot J(-3, 0), start from the origin O and move 3 units to the left along the x-axis, as its ordinate is zero. Mark a dot at this place.

11. To plot K(0, -4), start from the origin O and move 4 units downwards along the y-axis, as its abscissa is zero. Mark a dot at this place.

12. To plot L(-1, -1), start from the origin O, move 1 unit to the left along the x-axis and then move 1 unit downwards parallel to the y-axis. Mark a dot at this place.

The required points are plotted on the graph paper as shown below:

The point P lies on the x-axis at a distance of 3 units to the right of the origin.

So, the abscissa of P is 3 and its ordinate is 0.

∴ P = (3, 0)

The point Q is 4 units to the right of the origin and 3 units below it.

So, the abscissa of Q is 4 and its ordinate is -3.

∴ Q = (4, -3)

The point R is 4 units to the left of the origin and 2 units below it.

So, the abscissa of R is -4 and its ordinate is -2.

∴ R = (-4, -2)

The point S is 6 units to the left of the origin and 3 units above it.

So, the abscissa of S is -6 and its ordinate is 3.

∴ S = (-6, 3)

The point T is 3 units to the left of the origin and 2 units above it.

So, the abscissa of T is -3 and its ordinate is 2.

∴ T = (-3, 2)

Hence, the coordinates are P(3, 0), Q(4, -3), R(-4, -2), S(-6, 3) and T(-3, 2).

The signs of the coordinates of a point in the four quadrants are shown in the table given below:

Quadrants ↓	x (abscissa)	y (ordinate)
1st XOY	+ve	+ve
2nd YOX'	-ve	+ve
3rd X'OY'	-ve	-ve
4th Y'OX	+ve	-ve

(i) In the point (3, -7), the abscissa is 3, which is positive, and the ordinate is -7, which is negative.

The abscissa is positive and the ordinate is negative in the fourth quadrant.

Hence, the point (3, -7) lies in the IV quadrant.

(ii) In the point (-5, 1), the abscissa is -5, which is negative, and the ordinate is 1, which is positive.

The abscissa is negative and the ordinate is positive in the second quadrant.

Hence, the point (-5, 1) lies in the II quadrant.

(iii) In the point (-4, -6), the abscissa is -4 and the ordinate is -6, and both of them are negative.

Both the abscissa and the ordinate are negative in the third quadrant.

Hence, the point (-4, -6) lies in the III quadrant.

(i) In the point (0, -2), the abscissa is zero.

We know that the abscissa of every point lying on the y-axis is zero.

So, the point (0, -2) lies on the y-axis at a distance of 2 units below the origin.

Hence, the point (0, -2) lies on the y-axis.

(ii) In the point (7, 0), the ordinate is zero.

We know that the ordinate of every point lying on the x-axis is zero.

So, the point (7, 0) lies on the x-axis at a distance of 7 units to the right of the origin.

Hence, the point (7, 0) lies on the x-axis.

(iii) In the point (0, 6), the abscissa is zero.

We know that the abscissa of every point lying on the y-axis is zero.

So, the point (0, 6) lies on the y-axis at a distance of 6 units above the origin.

Hence, the point (0, 6) lies on the y-axis.

Draw the coordinate axes X'OX and Y'OY on a graph paper and plot the points A(4, 1), B(-2, 1), C(-3, -2) and D(3, -2) on it.

Join AB, BC, CD and DA.

The points A(4, 1) and B(-2, 1) have the same ordinate 1, so AB is parallel to the x-axis and AB = 6 units.

The points D(3, -2) and C(-3, -2) have the same ordinate -2, so DC is parallel to the x-axis and DC = 6 units.

So, AB and DC are equal and parallel to each other.

From the graph, we also find that AD and BC are equal and parallel to each other.

So, ABCD is a quadrilateral in which both the pairs of opposite sides are equal and parallel, but no angle is a right angle.

Hence, the figure ABCD is a parallelogram.

Draw the coordinate axes X'OX and Y'OY on a graph paper and plot the points A(5, 1), B(-2, 2) and C(4, 2) on it.

Join AB, BC and CA.

The three given points are not collinear, so on joining them we get a closed figure bounded by three line segments.

The points B(-2, 2) and C(4, 2) have the same ordinate 2, so BC is parallel to the x-axis and BC = 6 units.

From the graph, we find that the sides AB, BC and CA are of different lengths.

Hence, the figure ABC is a triangle.

(i) The point A is 1 unit to the right of the origin and 2 units below it.

So, the abscissa of A is 1 and its ordinate is -2.

The point B is 6 units to the right of the origin and 4 units above it.

So, the abscissa of B is 6 and its ordinate is 4.

Hence, the end points of the line segment AB are A(1, -2) and B(6, 4).

(ii) The centre C of the circle is 1 unit to the left of the origin and 4 units above it.

So, the abscissa of C is -1 and its ordinate is 4.

Hence, the coordinates of the centre of the circle are C(-1, 4).

(iii) The point P is 7 units to the left of the origin and 4 units below it, so its coordinates are (-7, -4).

The point Q is 5 units to the left of the origin and 2 units above it, so its coordinates are (-5, 2).

The point R is 4 units to the left of the origin and 6 units below it, so its coordinates are (-4, -6).

Hence, the vertices of the triangle PQR are P(-7, -4), Q(-5, 2) and R(-4, -6).

(iv) The point K is 7 units to the right of the origin and 6 units below it, so its coordinates are (7, -6).

The point L is 3 units to the right of the origin and 9 units below it, so its coordinates are (3, -9).

The point M is 3 units to the right of the origin and 5 units below it, so its coordinates are (3, -5).

The point N is 7 units to the right of the origin and 2 units below it, so its coordinates are (7, -2).

Hence, the vertices of the parallelogram KLMN are K(7, -6), L(3, -9), M(3, -5) and N(7, -2).

(i) From the figure, the coordinates of the end points of the line segment PQ are P(5, 2) and Q(5, 9).

Since P and Q have the same abscissa 5, the line segment PQ is parallel to the y-axis.

So, its length is the difference of the ordinates of Q and P.

PQ = 9 units - 2 units = 7 units

Hence, the length of the line segment PQ is 7 units.

(ii) In the right triangle KLM, the right angle is at L, so KM is the hypotenuse.

From the figure, the coordinates of K and M are K(1, -5) and M(8, -5).

Since K and M have the same ordinate -5, the line segment KM is parallel to the x-axis.

So, its length is the difference of the abscissae of M and K.

KM = 8 units - 1 unit = 7 units

Hence, the length of the hypotenuse of the right ΔKLM is 7 units.

(iii) From the figure, the coordinates of the centre of the circle are C(-4, 4).

Draw CD ∥ X'OX, meeting the circle at the point D.

Since D lies on the circle and has the same ordinate as C, from the figure the coordinates of D are (-1.5, 4).

Radius (CD) = -1.5 units - (-4 units) = 2.5 units

Hence, the radius of the circle with centre C is 2.5 units.

(iv) From the figure, the coordinates of the end points of the diagonal SU of the rectangle STUV are S(-6, -9) and U(-6, -2).

Since S and U have the same abscissa -6, the diagonal SU is parallel to the y-axis.

So, its length is the difference of the ordinates of U and S.

SU = -2 units - (-9 units) = 7 units

Hence, the length of a diagonal of the rectangle STUV is 7 units.

Draw the coordinate axes X'OX and Y'OY on a graph paper and plot the points A(2, -5), B(7, -5) and C(7, -2) on it.

Join AB and BC.

The points A(2, -5) and B(7, -5) have the same ordinate -5, so AB is parallel to the x-axis and AB = 5 units.

The points B(7, -5) and C(7, -2) have the same abscissa 7, so BC is parallel to the y-axis and BC = 3 units.

To complete the rectangle ABCD, draw a line through C parallel to BA and a line through A parallel to BC.

These two lines meet at the point D, as shown below:

The point D is 2 units to the right of the origin and 2 units below it.

So, the abscissa of D is 2, which is the same as that of A, and its ordinate is -2, which is the same as that of C.

Hence, the coordinates of the fourth vertex are D(2, -2).

Draw the coordinate axes X'OX and Y'OY on a graph paper and plot the points A(5, -2), B(6, 2) and C(0, 2) on it.

Join AB and BC.

The points B(6, 2) and C(0, 2) have the same ordinate 2, so BC is parallel to the x-axis and BC = 6 units.

In the parallelogram ABCD, the side AD is opposite to the side BC.

So, AD must be parallel to BC and AD = BC = 6 units.

Starting from the point A(5, -2), move 6 units to the left parallel to the x-axis and mark the point D, as shown below:

The point D is 1 unit to the left of the origin and 2 units below it.

So, the abscissa of D is -1 and its ordinate is -2.

Hence, the coordinates of the fourth vertex are D(-1, -2).

We know that :

Area of a circle = $\pi r^2$, where $r$ = radius of the circle.

Taking π = $\frac{22}{7}$, we get

Area of the circle of radius 3.5 cm = $\frac{22}{7}$ × 3.5 × 3.5 = 38.5 cm²

Area of the circle of radius 7 cm = $\frac{22}{7}$ × 7 × 7 = 154 cm²

Area of the circle of radius 10.5 cm = $\frac{22}{7}$ × 10.5 × 10.5 = 346.5 cm²

Area of the circle of radius 14 cm = $\frac{22}{7}$ × 14 × 14 = 616 cm²

Thus, we have the following table :

Radius (r) in cm	Area (πr2) in cm2
3.5	38.5
7	154
10.5	346.5
14	616

Represent the radius of the circle (in cm) along the x-axis and its area (in cm²) along the y-axis.

Scale:

Take,

1 unit on x-axis = 3.5 cm,

1 unit on y-axis = 50 cm².

Steps:

1. Plot the points A(3.5, 38.5), B(7, 154), C(10.5, 346.5) and D(14, 616) on the graph paper.

2. Since the points A, B, C and D are non-collinear, we join them by a free hand curve to obtain the required graph.

Hence, ABCD is the required graph.

We know that :

The multiples of 5 are : (5 × 1), (5 × 2), (5 × 3), (5 × 4), .....

First multiple of 5 = 5 × 1 = 5

Second multiple of 5 = 5 × 2 = 10

Third multiple of 5 = 5 × 3 = 15

Fourth multiple of 5 = 5 × 4 = 20

Thus, we have the following table :

n	5n
1	5
2	10
3	15
4	20

Represent the number n along the x-axis and its multiple 5n along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 (number),

1 unit on y-axis = 5 (number).

Steps:

1. Plot the points P(1, 5), Q(2, 10), R(3, 15) and S(4, 20) on the graph paper.

2. Since the points P, Q, R and S are collinear, we join them by a straight line to obtain the required graph.

Hence, PQRS is the required graph.

We know that :

Perimeter of an equilateral triangle of side a = 3a

So, semi-perimeter, $s = \frac{3a}{2}$

Semi-perimeter of the equilateral triangle of side 2 cm = $\frac{3 \times 2}{2}$ = 3 cm

Semi-perimeter of the equilateral triangle of side 3 cm = $\frac{3 \times 3}{2}$ = 4.5 cm

Semi-perimeter of the equilateral triangle of side 4 cm = $\frac{3 \times 4}{2}$ = 6 cm

Semi-perimeter of the equilateral triangle of side 5 cm = $\frac{3 \times 5}{2}$ = 7.5 cm

Thus, we have the following table :

Length of side (a) in cm	Semi-perimeter (s) in cm
2	3
3	4.5
4	6
5	7.5

Represent the length of the side (in cm) along the x-axis and the semi-perimeter (in cm) along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 cm,

1 unit on y-axis = 1 cm.

Steps:

1. Plot the points A(2, 3), B(3, 4.5), C(4, 6) and D(5, 7.5) on the graph paper.

2. Since the points A, B, C and D are collinear, we join them by a straight line to obtain the required graph.

Hence, ABCD is the required graph.

We know that :

S.I. = $\frac{P \times R \times T}{100}$, where P = principal, R = rate % p.a. and T = time in years.

Here, P = ₹ 5000 and R = 5% p.a.

S.I. for 1 year = ₹ $\Big(\frac{5000 \times 5 \times 1}{100}\Big)$ = ₹ 250

S.I. for 2 years = ₹ $\Big(\frac{5000 \times 5 \times 2}{100}\Big)$ = ₹ 500

S.I. for 3 years = ₹ $\Big(\frac{5000 \times 5 \times 3}{100}\Big)$ = ₹ 750

S.I. for 4 years = ₹ $\Big(\frac{5000 \times 5 \times 4}{100}\Big)$ = ₹ 1000

Thus, we have the following table :

Time (in years)	Simple interest (in ₹)
1	250
2	500
3	750
4	1000

Represent the time (in years) along the x-axis and the simple interest (in ₹) along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 year,

1 unit on y-axis = ₹ 250.

Steps:

1. Plot the points K(1, 250), L(2, 500), M(3, 750) and N(4, 1000) on the graph paper.

2. Since the points K, L, M and N are collinear, we join them by a straight line to obtain the required graph.

Hence, KLMN is the required graph.

We know that :

Perimeter of a square, P = 4a, where a = side of the square.

For different values of a, we get the corresponding values of P.

⇒ a = 1

⇒ P = 4 × 1 = 4;

⇒ a = 2

⇒ P = 4 × 2 = 8;

⇒ a = 3

⇒ P = 4 × 3 = 12;

⇒ a = 4

⇒ P = 4 × 4 = 16;

⇒ a = 6

⇒ P = 4 × 6 = 24.

Thus, we have the following table :

Length of side (a)	Perimeter (4a)
1	4
2	8
3	12
4	16
6	24

Represent the length of the side (a) along the x-axis and the perimeter (P) along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 unit of side,

1 unit on y-axis = 4 units of perimeter.

Steps:

Plot the points A(1, 4), B(2, 8), C(3, 12), D(4, 16) and E(6, 24) on the graph paper and join them successively to obtain the required graph.

(i) Corresponding to a = 2.5, locate a point M on the x-axis and draw a vertical line through M to meet the graph at P.

From P, draw a horizontal line to meet the y-axis at the point N.

The point N represents 10.

Hence, when a = 2.5, the value of P is 10 units.

(ii) Corresponding to a = 5, locate a point R on the x-axis and draw a vertical line through R to meet the graph at S.

From S, draw a horizontal line to meet the y-axis at the point T.

The point T represents 20.

Hence, when a = 5, the value of P is 20 units.

The given function is y = 2x.

For different values of x, we get the corresponding values of y.

⇒ x = 0

⇒ y = 2 × 0 = 0;

⇒ x = 1

⇒ y = 2 × 1 = 2;

⇒ x = 2

⇒ y = 2 × 2 = 4;

⇒ x = 3

⇒ y = 2 × 3 = 6;

⇒ x = 5

⇒ y = 2 × 5 = 10.

Thus, we have the following table :

x	y
0	0
1	2
2	4
3	6
5	10

Represent the values of x along the x-axis and the corresponding values of y along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 unit,

1 unit on y-axis = 1 unit.

Steps:

Plot the points O(0, 0), A(1, 2), B(2, 4), C(3, 6) and D(5, 10) on the graph paper and join them successively to obtain the required graph.

(i) Corresponding to x = 1.5, locate a point L on the x-axis and draw a vertical line through L to meet the graph at P.

From P, draw a horizontal line to meet the y-axis at the point N.

The point N represents 3.

Hence, when x = 1.5, the value of y is 3.

(ii) Corresponding to x = 4, locate a point M on the x-axis and draw a vertical line through M to meet the graph at Q.

From Q, draw a horizontal line to meet the y-axis at the point R.

The point R represents 8.

Hence, when x = 4, the value of y is 8.

We know that :

Distance = Speed × Time

Here, the speed of the object is 5 metres per second.

So, the distance (d) covered in time (t) is given by the relation d = 5t.

For different values of t, we get the corresponding values of d.

⇒ t = 0

⇒ d = 5 × 0 = 0;

⇒ t = 1

⇒ d = 5 × 1 = 5;

⇒ t = 2

⇒ d = 5 × 2 = 10;

⇒ t = 4

⇒ d = 5 × 4 = 20;

⇒ t = 7

⇒ d = 5 × 7 = 35.

Thus, we have the following table :

Time (t) in seconds	Distance (d) in metres
0	0
1	5
2	10
4	20
7	35

Represent the time (in seconds) along the x-axis and the distance (in metres) along the y-axis.

Scale:

Take,

1 unit on x-axis = 1 second,

1 unit on y-axis = 5 metres.

Steps:

Plot the points O(0, 0), A(1, 5), B(2, 10), C(4, 20) and D(7, 35) on the graph paper and join them successively to obtain the required graph.

(i) Corresponding to 3 seconds, locate a point L on the x-axis and draw a vertical line through L to meet the graph at P.

From P, draw a horizontal line to meet the y-axis at the point N.

The point N represents 15.

Hence, the distance covered in 3 seconds is 15 metres.

(ii) Corresponding to 5 seconds, locate a point M on the x-axis and draw a vertical line through M to meet the graph at Q.

From Q, draw a horizontal line to meet the y-axis at the point R.

The point R represents 25.

Hence, the distance covered in 5 seconds is 25 metres.

(iii) Corresponding to 30 metres, locate a point S on the y-axis and draw a horizontal line through S to meet the graph at T.

From T, draw a vertical line to meet the x-axis at the point U.

The point U represents 6.

Hence, the body covered 30 metres in 6 seconds.

In the first quadrant, both the abscissa and the ordinate of a point are positive.

The point (0, 0) is the origin, the point (4, 0) lies on the x-axis and the point (0, 3) lies on the y-axis.

Among the given points, only the point (2, 3) has both its abscissa and ordinate positive.

Hence, option 4 is the correct option.

The perpendicular distance of a point from the y-axis is the absolute value of x-coordinate, which is called the abscissa of the point.

The perpendicular distance of a point from the x-axis is the absolute value of y-coordinate, which is called the ordinate of the point.

Hence, option 1 is the correct option.

The point lies on the x-axis, so its ordinate is zero.

The distance of the point from the y-axis is 3 units, so its abscissa can be 3 (if it lies to the right of the y-axis) or -3 (if it lies to the left of the y-axis).

So, the coordinates of the point can be (3, 0) or (-3, 0).

Among the given options, only (3, 0) is present.

Hence, option 2 is the correct option.

The point lies on the y-axis, so its abscissa is zero.

The distance of the point from the x-axis is 6 units, so its ordinate can be 6 (if it lies above the x-axis) or -6 (if it lies below the x-axis).

So, the coordinates of the point can be (0, 6) or (0, -6).

Among the given options, only (0, 6) is present.

Hence, option 1 is the correct option.

The point lies on the y-axis, so its abscissa is zero.

The point is at a distance of 4 units from the x-axis and it lies below the x-axis, so its ordinate is -4.

So, the coordinates of the point are (0, -4).

Hence, option 3 is the correct option.

In the point (-4, 3), the abscissa is -4, which is negative, and the ordinate is 3, which is positive.

The abscissa is negative and the ordinate is positive in the second quadrant.

Hence, option 2 is the correct option.

Every point on a line parallel to the y-axis is at the same distance from the y-axis, that is, each such point has the same abscissa.

Here, each point of the line is at a distance of 4 units to the right of the y-axis, so the abscissa of each of its points is 4.

So, the equation of the line is x = 4.

Hence, option 1 is the correct option.

Every point on a line parallel to the y-axis has the same abscissa.

Here, each point of the line is at a distance of 3 units to the left of the y-axis, so the abscissa of each of its points is -3.

So, the equation of the line is x = -3.

Hence, option 2 is the correct option.

Every point on a line parallel to the x-axis is at the same distance from the x-axis, that is, each such point has the same ordinate.

Here, each point of the line is at a distance of 9 units above the x-axis, so the ordinate of each of its points is 9.

So, the equation of the line is y = 9.

Hence, option 2 is the correct option.

Every point on a line parallel to the x-axis has the same ordinate.

Here, each point of the line is at a distance of 5 units below the x-axis, so the ordinate of each of its points is -5.

So, the equation of the line is y = -5.

Hence, option 3 is the correct option.

The x-coordinate of a point is the perpendicular distance of the point from the y-axis, and it is called the abscissa of the point.

So, Assertion (A) is true.

The distance of any point lying on the x-axis from the x-axis is zero, so its ordinate is zero and its coordinates are of the form (x, 0).

So, Reason (R) is also true.

Reason (R) speaks of the coordinates of a point lying on the x-axis, whereas Assertion (A) speaks of the x-coordinate of any point in the plane.

So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

In the fourth quadrant, a point lies to the right of the origin and below it, so its abscissa is positive and its ordinate is negative.

So, Assertion (A) is true.

The y-coordinate of a point is positive if it is measured above the origin and it is negative if it is measured below the origin.

So, Reason (R) is false.

Hence, option 3 is the correct option.

The distance of any point lying on the y-axis from the y-axis is zero, so the abscissa of every point on the y-axis is 0.

So, Reason (R) is true.

In the point (0, -2), the abscissa is 0, so the point lies on the y-axis at a distance of 2 units below the origin.

So, Assertion (A) is also true, and it follows directly from Reason (R).

So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

The perpendicular distance of a point from the x-axis is its y-coordinate, that is, its ordinate.

In the point A(7, 5), the ordinate is 5.

So, the distance of the point A(7, 5) from the x-axis is 5 units.

Hence, option 2 is the correct option.

In the point (-2, 0), the ordinate is zero, so the point lies on the x-axis.

Also, the abscissa is -2, which is negative, so the point lies to the left of the y-axis.

So, the point (-2, 0) lies on the x-axis and on the left side of the y-axis.

Hence, option 1 is the correct option.

A point lying on the x-axis has its ordinate equal to zero and a point lying on the y-axis has its abscissa equal to zero.

So, a point lying on both the axes has both its abscissa and its ordinate equal to zero, that is, it is the point (0, 0), which is the origin.

Hence, option 4 is the correct option.

From the figure, the book is at a distance of 2 units from the y-axis, so its abscissa is 2.

The book is at a distance of 5 units from the x-axis, so its ordinate is 5.

Since the abscissa precedes the ordinate, the position of the book on the table is given by (2, 5).

Hence, option 1 is the correct option.