Coordinate System and Graphs

(i) A point whose x-coordinate is zero and y-coordinate is non-zero is of the form (0, y), where y is non-zero.

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

So, the given point lies on the y-axis.

Hence, the given statement is True.

(ii) A point whose y-coordinate is zero and x-coordinate is 7 is the point (7, 0).

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 and not on the y-axis.

Hence, the given statement is False.

Correct statement : A point whose y-coordinate is zero and x-coordinate is 7 will lie on x-axis.

(iii) The origin is the point of intersection of the x-axis and the y-axis.

Since the origin lies on both the axes, its abscissa as well as its ordinate is zero.

So, the coordinates of the origin are (0, 0).

Hence, the given statement is True.

(iv) For a point lying on the x-axis, the distance measured above or below the origin is zero.

So, if the point (x, y) lies on the x-axis, then its ordinate is zero and not its abscissa.

Hence, the given statement is False.

Correct statement : If the point (x, y) lies on x-axis, then its ordinate is zero.

(v) For a point lying on the y-axis, the distance measured to the right or to the left of the origin is zero.

So, if the point (x, y) lies on the y-axis, then its abscissa is zero and not its ordinate.

Hence, the given statement is False.

Correct statement : If the point (x, y) lies on y-axis, then its abscissa is zero.

(vi) The abscissa of every point lying on the y-axis is zero.

So, the point (x, y) lies on the y-axis when x = 0.

Hence, the given statement is True.

(vii) In the point (-3, -2), the abscissa is -3 and the ordinate is -2.

Since both the abscissa and the ordinate are negative, the point lies in the third quadrant.

Hence, the given statement is False.

Correct statement : The point (-3, -2) lies in the third quadrant.

(viii) In the point (5, -3), the abscissa is 5 and the ordinate is -3.

Since the abscissa is positive and the ordinate is negative, the point lies in the fourth quadrant.

Hence, the given statement is False.

Correct statement : The point (5, -3) lies in the fourth quadrant.

(i) The only point which lies on the x-axis as well as on the y-axis is the point of intersection of the two axes, that is, the origin.

Hence, the coordinates of the point are (0, 0).

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

The ordinate of the point is -4.

Since the abscissa precedes the ordinate, the coordinates are (0, -4).

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

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

The abscissa of the point is 5.

Since the abscissa precedes the ordinate, the coordinates are (5, 0).

Hence, the coordinates of the point are (5, 0).

(iv) The abscissa of the point is -2 and its ordinate is 3.

Since the abscissa precedes the ordinate, the coordinates are (-2, 3).

Hence, the coordinates of the point are (-2, 3).

(v) The abscissa of the point is 3 and its ordinate is -2.

Since the abscissa precedes the ordinate, the coordinates are (3, -2).

Hence, the coordinates of the point are (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

In the point (-3, 5), the abscissa is negative and the ordinate is positive.

So, the point (-3, 5) lies in the second quadrant.

In the point (4, -1), the abscissa is positive and the ordinate is negative.

So, the point (4, -1) lies in the fourth quadrant.

In the point (2, 0), the ordinate is zero.

So, the point (2, 0) lies on the x-axis.

In the point (2, 2), both the abscissa and the ordinate are positive.

So, the point (2, 2) lies in the first quadrant.

In the point (-3, -6), both the abscissa and the ordinate are negative.

So, the point (-3, -6) lies in the third quadrant.

Hence, the given points lie in the second quadrant, fourth quadrant, on the x-axis, in the first quadrant and in the third quadrant respectively.

(i) The ordinate of every point lying on the x-axis is zero.

Among the given points, the ordinate is zero for C(23, 0), F(-6, 0) and G(3.5, 0).

Hence, the points C, F and G lie on the x-axis.

(ii) The abscissa of every point lying on the y-axis is zero.

Among the given points, the abscissa is zero for A(0, 2), D(0, 23) and E(0, -4).

Hence, the points A, D and E lie on the y-axis.

(i) The point B is 5 units to the left of the origin and 2 units above it.

So, the abscissa of B is -5 and its ordinate is 2.

The point C is 2 units to the left of the origin and 3 units below it.

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

The point E is 3 units to the right of the origin and 1 unit below it.

So, the abscissa of E is 3 and its ordinate is -1.

Hence, the coordinates are B(-5, 2), C(-2, -3) and E(3, -1).

(ii) The point identified by the coordinates (0, -2) lies on the y-axis at a distance of 2 units below the origin.

From the figure, the point lying at this position is F.

Hence, the point identified by the coordinates (0, -2) is F.

(iii) The point H lies 1 unit to the right of the origin.

Hence, the abscissa of the point H is 1.

(iv) The point A lies on the y-axis at a distance of 3 units above the origin.

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

The point D lies on the x-axis at a distance of 4 units to the right of the origin.

So, the abscissa of D is 4 and its ordinate is 0.

Hence, the coordinates are A(0, 3) and D(4, 0).

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 (-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.

2. To plot (4, 2.5), start from the origin O, move 4 units to the right along the x-axis and then move 2.5 units upwards parallel to the y-axis. Mark a dot at this place.

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

4. To plot (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.

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

6. To plot (-4.5, 0), start from the origin O and move 4.5 units to the left along the x-axis, as its ordinate is zero. Mark a dot at this place.

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

(i) Draw the coordinate axes on a graph paper and plot the points A(-1, 0), B(0, 1) and C(2, 3) on it.

Draw a line passing through the points A and B.

We find that the point C(2, 3) also lies on this line.

Hence, the points (-1, 0), (0, 1) and (2, 3) are collinear.

(ii) Draw the coordinate axes on a graph paper and plot the points A(1, 1), B(2, -3) and C(-1, -2) on it.

We note that the line passing through A and B does not pass through the point C(-1, -2).

Thus the three points do not lie on the same straight line.

Hence, the points (1, 1), (2, -3) and (-1, -2) are not collinear.

The rule of multiplication by 5 is given by the relation y = 5x, where x represents the positive numbers 0, 1, 2, 3, ... including 0, and y represents their corresponding multiples of 5.

The table of values is given below:

x	y = 5x
0	0
1	5
2	10
3	15
4	20
...	...

Represent the numbers 0, 1, 2, 3, ... on the x-axis and their respective multiples of 5 on the y-axis.

Scale: Take,

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

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

Plot the points (0, 0), (1, 5), (2, 10), (3, 15), (4, 20), ... on the graph paper and join these points.

To find the product of 5 and 3.5 :

Corresponding to the number 3.5, 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 17.5.

Hence, the product of 5 and 3.5 is 17.5.

Represent the quantity of rice (in kg) along the x-axis and its cost (in ₹) along the y-axis.

Scale: Take,

1 unit on x-axis = 1 kg,

1 unit on y-axis = ₹ 30.

Plot the points (1, 30), (2, 60), (3, 90) and (5, 150) on the graph paper and join these points.

We note that the graph so obtained is a linear graph.

(i) Corresponding to 4 kg of rice, 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 Q.

The point Q represents 120.

Hence, the amount needed to buy 4 kg rice is ₹ 120.

(ii) Corresponding to ₹ 105, locate a point R on the y-axis and draw a horizontal line through R to meet the graph at S.

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

The point T represents 3.5.

Hence, the quantity of rice that can be purchased for ₹ 105 is 3.5 kg.

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

Scale: Take,

1 unit on x-axis = ₹ 500,

1 unit on y-axis = ₹ 30.

Plot the points (1000, 60), (1500, 90), (2000, 120), (2400, 144), (2700, 162) and (3000, 180) on the graph paper and join these points.

We note that the graph so obtained is a linear graph.

(i) Corresponding to the deposit of ₹ 1,800, 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 Q.

The point Q represents 108.

Hence, the simple interest on ₹ 1,800 for one year is ₹ 108.

(ii) Corresponding to the interest of ₹ 156, locate a point R on the y-axis and draw a horizontal line through R to meet the graph at S.

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

The point T represents 2600.

Hence, ₹ 2,600 should be deposited to get an interest of ₹ 156.

Represent the temperature in °C along the x-axis and the temperature in °F along the y-axis.

Scale: Take,

1 unit on x-axis = 5°C,

1 unit on y-axis = 10°F.

Plot the points (0, 32), (5, 41), (10, 50), (15, 59), (20, 68), (30, 86) and (40, 104) on the graph paper and join these points.

We note that the graph so obtained is a linear graph.

(i) Corresponding to 25°C, 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 Q.

The point Q represents 77.

Hence, the temperature in °F corresponding to 25°C is 77°F.

(ii) Corresponding to 95°F, locate a point R on the y-axis and draw a horizontal line through R to meet the graph at S.

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

The point T represents 35.

Hence, the temperature in °C corresponding to 95°F is 35°C.

(i) Corresponding to 6 a.m. on the x-axis, draw a vertical line to meet the graph at a point.

From this point, draw a horizontal line to meet the y-axis. It meets the y-axis at 3.

Hence, the temperature at 6 a.m. is 3°C.

(ii) Corresponding to 10 °C on the y-axis, draw a horizontal line to meet the graph at a point.

From this point, draw a vertical line to meet the x-axis. It meets the x-axis at 11 a.m.

Hence, the temperature is 10 °C at 11 a.m.

(iii) The temperature remains constant during the hour in which the graph is a horizontal line segment.

From the graph, the temperature is 7 °C at 8 a.m. and it is 7 °C at 9 a.m. also.

Hence, there was a constant temperature between 8 a.m. and 9 a.m.

(iv) The temperature falls during the hour in which the graph slopes downwards.

From the graph, the temperature is 4 °C at 5 a.m. and it falls to 3 °C at 6 a.m.

Hence, the temperature fell between 5 a.m. and 6 a.m.

(v) From the graph,

Temperature at 8 a.m. = 7 °C

Temperature at 11 a.m. = 10 °C

Change in temperature = 10 °C - 7 °C = 3 °C

Hence, the change in temperature between 8 a.m. to 11 a.m. is a rise of 3 °C.

(i) The horizontal line X'OX in a coordinate plane is called x-axis.

(ii) The point of intersection of x-axis and y-axis is called origin.

(iii) In a cartesian plane x-coordinate of a point is also called abscissa.

(iv) The two axes divide the plane into four parts called quadrants.

(v) The y-coordinate of a point is negative, if it is measured below the origin.

(vi) Area of circle depends on the radius of circle, then radius of circle is called independent variable.

(i) The horizontal line X'OX and the vertical line Y'OY, taken together, are called the coordinate axes.

Hence, the given statement is True.

(ii) The y-coordinate of a point is called its ordinate, whereas the x-coordinate of a point is called its abscissa.

Hence, the given statement is False.

(iii) The point (0, 0) is the origin, which is the point of intersection of the two axes.

So, it lies on the x-axis as well as on the y-axis.

Hence, the given statement is True.

(iv) The abscissa of the point (0, 2) is zero, so the point lies on the y-axis and not on the x-axis.

Hence, the given statement is False.

(v) The abscissa of each of the points (2, 3), (2, 5) and (2, 7) is 2.

So, each of these points is at a distance of 2 units to the right of the y-axis, and hence the line joining them is parallel to the y-axis.

Hence, the given statement is True.

(vi) In the point (-2, 3), the abscissa is negative and the ordinate is positive.

So, the point (-2, 3) lies in the second quadrant and not in the third quadrant.

Hence, the given statement is False.

In the point (-2, 7), the abscissa is -2, which is negative, and the ordinate is 7, which is positive.

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

Hence, option 2 is the correct option.

In the point (0, -3), the abscissa is zero.

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

So, the point (0, -3) lies on the y-axis at a distance of 3 units below the origin.

Hence, option 3 is the correct option.

In the point (-10, 0), the ordinate is zero, so the point lies on the x-axis.

Also, the abscissa is -10, which is negative, so the point lies to the left of the origin.

So, the point (-10, 0) lies on the negative direction of the x-axis.

Hence, option 1 is the correct option.

In the second quadrant, a point lies to the left of the origin and above it.

So, its abscissa is negative and its ordinate is positive.

Hence, option 3 is the correct option.

The abscissa of a point is positive when it is measured to the right of the origin.

The points lying to the right of the origin are those in the first quadrant and in the fourth quadrant.

Hence, option 2 is the correct option.

In the fourth quadrant, the abscissa of a point is positive and its ordinate is negative.

Among the given points, only the point (4, -7) has a positive abscissa and a negative ordinate.

Hence, option 3 is the correct option.

The abscissa of every point lying on the y-axis is zero.

Among the given points, only the point (0, -7) has its abscissa equal to zero.

Hence, option 1 is the correct option.

In the point (-3, 4), the abscissa is negative and the ordinate is positive, so it lies in the second quadrant.

In the point (4, -3), the abscissa is positive and the ordinate is negative, so it lies in the fourth quadrant.

So, the given points lie in the II and IV quadrants respectively.

Hence, option 3 is the correct option.

(i) The abscissa is -3 and the ordinate is 5, so the point is (-3, 5).

Here, the abscissa is negative and the ordinate is positive.

Hence, the point lies in the II quadrant.

(ii) The abscissa is -5 and the ordinate is -3, so the point is (-5, -3).

Here, both the abscissa and the ordinate are negative.

Hence, the point lies in the III quadrant.

(iii) The abscissa is -5 and the ordinate is 3, so the point is (-5, 3).

Here, the abscissa is negative and the ordinate is positive.

Hence, the point lies in the II quadrant.

(iv) The abscissa is 3 and the ordinate is 5, so the point is (3, 5).

Here, both the abscissa and the ordinate are positive.

Hence, the point lies in the I quadrant.

Draw the coordinate axes on a graph paper.

Start from the origin O, move 3 units to the left along the x-axis and then move 4 units upwards parallel to the y-axis. Mark this point as P(-3, 4).

From P, draw PM perpendicular to the x-axis and PN perpendicular to the y-axis.

The point M is the foot of the perpendicular drawn from P to the x-axis.

So, M lies on the x-axis and its ordinate is 0, while its abscissa is the same as that of P, that is, -3.

The point N is the foot of the perpendicular drawn from P to the y-axis.

So, N lies on the y-axis and its abscissa is 0, while its ordinate is the same as that of P, that is, 4.

Hence, the coordinates of the points are M(-3, 0) and N(0, 4).

Draw the coordinate axes on a graph paper and plot the points A(1, 2), B(2, -1) and C(-1, 4) on it.

We note that the line passing through A and B does not pass through the point C(-1, 4).

Thus the three points do not lie on the same straight line.

Hence, the points A(1, 2), B(2, -1) and C(-1, 4) are non-collinear.

Represent the age (in years) along the x-axis and the height (in cm) along the y-axis.

Scale: Take,

1 unit on x-axis = 2 years,

1 unit on y-axis = 10 cm.

A break is shown near the origin to indicate that the scale on the y-axis starts at 70 cm.

Plot the points (2, 75), (4, 90), (6, 110), (8, 120) and (10, 130) on the graph paper and join the consecutive points by line segments.

(i) Corresponding to the age of 5 years, 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 Q.

The point Q represents 100.

Hence, the height at the age of 5 years is 100 cm.

(ii) From the graph,

Height at the age of 8 years = 120 cm

Height at the age of 4 years = 90 cm

Difference in the heights = 120 cm - 90 cm = 30 cm

Hence, the child was 30 cm taller at the age of 8 than at the age of 4.

(iii) The child grows fastest during the period in which the graph is the steepest.

From the graph, the growth in height during the consecutive periods is as follows:

Between 2 years and 4 years = 90 cm - 75 cm = 15 cm

Between 4 years and 6 years = 110 cm - 90 cm = 20 cm

Between 6 years and 8 years = 120 cm - 110 cm = 10 cm

Between 8 years and 10 years = 130 cm - 120 cm = 10 cm

The greatest growth of 20 cm is between 4 years and 6 years.

Hence, the child grew fastest between 4 years and 6 years.