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#### Question 1:

Write down the coordinates of each of the points A, B, C, D, E shown below:

Draw perpendicular AL, BM, CN, DP and EQ on the X-axis.

(i) Distance of A from the Y-axis = OL = -6 units
Distance of A from the X-axis = AL = 5 units
Hence, the coordinates of A are (-6,5).

(ii) Distance of B from the Y-axis = OM = 5 units
Distance of B from the X-axis = BM = 4 units
Hence, the coordinates of B are (5,4).

(iii) Distance of C from the Y-axis = ON = -3 units
Distance of C from the X-axis = CN = 2 units
Hence, the coordinates of C are (-3,2).

(iv) Distance of D from the Y-axis = OP = 2 units
Distance of D from the X-axis = DP = -2 units
Hence, the coordinates of D are (2,-2).

(v) Distance of E from the Y-axis = OL = -1 units
Distance of E from the X-axis = AL = -4 units
Hence, the coordinates of E are (-1,-4).

#### Question 2:

Draw the lines X' OX and YOY' as the coordinate axes on a paper and plot the following points on it.
(i) P(7, 4)
(ii) Q(−5, 3)
(iii) R(−6, −3)
(iv) S(3, −7)
(v) A(6, 0)
(vi) B(0, 9)
(vii) O(0, 0)
(viii) C(−3, −3)

Let X'OX and YOY' be the coordinate axes.
Fix a convenient unit of length and form point O, mark equal distances on OX, OX', OY and OY'. Use the convention of signs.

(i) Starting from O, take 7 units on the x-axis and then 4 units on the y-axis to obtain the point P(7,4).
(ii) Starting from O, take -5 units on the x-axis and then 3 units on the y-axis to obtain the point Q(-5,3).
(iii) Starting from O, take -6 units on the x-axis and then -3 units on the y-axis to obtain the point R(-6,-3).
(iv) Starting from O, take 3 units on the x-axis and then -7 units on the y-axis to obtain the point S(3,-7).
(v) Starting from O, take 6 units on the x-axis to obtain the point A(6,0).
(vi) Starting from O, take 9 units on the y-axis to obtain the point B(0,9).
(vii) Same as origin.
(viii) ​Starting from O, take -3 units on the x-axis and then -3 units on the y-axis to obtain the point C(-3,-3).

#### Question 3:

On which axis do the following points lie?
(i) (7, 0)
(ii) (0, −5)
(iii) (0, 1)
(iv) (−, 0)

(i) In (7,0), ordinate = 0
∴ (7,0) lies on the x-axis.

(ii) In (0,-5), abscissa = 0
∴ (0,-5) lies on the y-axis.

(iii) In (0,1), abscissa = 0
∴ (0,1) lies on the y-axis.

(iv) In (-4,0), ordinate = 0
∴ (-4,0) lies on the x-axis.

#### Question 4:

In which quadrant do the given points lie?
(i) (−6, 5)
(ii) (−3, −2)
(iii) (2, −9)

(i)  Points of the type (-,+) lie in the second quadrant.
Hence, the point (-6,5) lies in quadrant II.

(ii) Points of the type (-,-) lie in the third quadrant.
Hence, the point (-3,-2) lies in quadrant III.

(iii) Points of the type (+,-) lie in the fourth quadrant.
Hence, the point (-6,5) lies in quadrant IV.

#### Question 5:

Draw the graph of the equation, y = x + 1.

The given equation is y = x + 1.
Putting x = 0, we get y = 0 + 1 = 1
Putting x = 1, we get y = 1 + 1 = 2
Thus, we have the following table:

 x 0 1 y 1 2

On a graph paper, draw the lines X'OX and YOY' as the x-axis and y-axis, respectively.
Now, plot the points A(0,1) and B(1,2) on the graph paper.
Join AB and extend it on both directions.

Thus, line AB is the required graph of the equation, y = x + 1.

#### Question 6:

Draw the graph of the equation, y = 3x + 2.

The given equation is y = 3x + 2.
Putting x = 0, we get y = (3 × 0) + 2 = 2.
Putting x = 1, we get y = (3 × 1) + 2 = 5.
Thus, we have the following table:
 x 0 1 y 2 5

On a graph paper, draw the lines X'OX and YOY' as the x-axis and y-axis, respectively.
Now, plot the points A(0,2) and B(1,5) on the graph paper.
Join AB and extend it on both sides.
Thus, line AB is the required graph of the equation, y = 3x + 2.

#### Question 7:

Draw the graph of the equation, y = 5x 3.

The given equation is y = 5x - 3.
Putting x = 0, we get y = (5 × 0) - 3 = -3
Putting x = 1, we get y = (5 × 1) - 3 = 2
Thus, we have the following table:
 x 0 1 y -3 2

On a graph paper, draw the lines X'OX and YOY' as the x-axis and y-axis, respectively.
Now, plot the points A(0,-3) and B(1,2) on the graph paper.
Join AB and extend it on both sides.
Thus, line AB is the required graph of the equation, y = 5x - 3.

#### Question 8:

Draw the graph of the equation, y = 3x.

The given equation is y = 3x.
Putting x = 0, we get y = (3 × 0) = 0.
Putting x = 1, we get y = (3 × 1) = 3
Thus, we have the following table:

 x 0 1 y 0 3

On a graph paper, draw the lines X'OX and YOY' as the x-axis and y-axis, respectively.
Now, plot the points A(0,0) and B(1,3) on the graph paper.
Join AB and extend it on both sides.
Thus, line AB is the required graph of the equation, y = 3x.

#### Question 9:

Draw the graph of the equation, y = −x.

The given equation is y = -x.
Putting x = 0, we get y = 0.
Putting x = 1, we get y = (-1).
Thus, we have the following table:

 x 0 1 y 0 -1

On a graph paper, draw the lines X'OX and YOY' as the x-axis and y-axis, respectively.
Now, plot the points A(0,0) and B(1,-1) on the graph paper.
Join AB and extend it on both sides.
Thus, line AB is the required graph of the equation,  y = -x.

#### Question 1:

The point P(−5, 3) lies in

Points of the type (-, +) lie in the second quadrant.
Hence, (-5,3) lies in quadrant II.

#### Question 2:

The point Q(4, −6) lies in

Explanation:
The points of the type (+, -) lie in the fourth quadrant.
Hence, (4,-6) lies in quadrant IV.

#### Question 3:

The point A(0, −4) lies
(c) on the x-axis
(d) on the y-axis

(d) on the y- axis

​Explanation:
As the abscissa of the point A(0,-4) is 0, it lies on the y-axis.

#### Question 4:

The point B(8, 0) lies
(c) on the x-axis
(d) on the y-axis

(c) on the x-axis

​Explanation:
As the ordinate of the point B(8,0) is 0, it lies on the x-axis.

#### Question 5:

The point C(−6, 0) lies
(c) on the x-axis
(d) on the y-axis

(c) on the x-axis

​Explanation:
As the ordinate of the point C(-6,0) is 0, it lies on the x-axis.

#### Question 6:

The point at which the two coordinate axes meet is called
(a) the abscissa
(b) the ordinate
(c) the origin

(c) the origin
​Explanation: The point at which two axes meet is called as the origin.

#### Question 7:

If x > 0 and y < 0, then the point (x, y) lies in

​Explanation:
The points of the type (+,-) lie in fourth quadrant.
Hence, the point (x,y), where > 0 and y <0, lies in quadrant IV.

#### Question 8:

The points (other than the origin) for which the abscissa is equal to the ordinate lie in

​Explanation:
If abscissa = ordinate, there could be two possibilities.
Either both are positive or both are negative. So, a point could be either (+,+), which lie in quadrant I or it could be of the type (-,-), which lie in quadrant III.
Hence, the points (other then the origin) for which the abscissae are equal to the ordinates lie in quadrant I and III.

#### Question 9:

The points in which abscissa and ordinate have different signs will lie in

​Explanation:
If the abscissa and ordinate have different signs, there could be two possibilities:
Either the abscissa is positive and the ordinate is negative or the abscissa is positive and the ordinate is negative.
So, a point could be either (+,-), which lie in quadrant IV, or it could be of the type (-,+), which lie in quadrant II.
Hence, points whose abscissae and ordinates have different signs lie in quadrants IV and II.

#### Question 10:

The perpendicular distance of the point A(7, 5) from y-axis is
(a) 7 units
(b) 5 units
(c) 12 units
(d) 2 units

(a) 7 units

​Explanation:
The abscissa is the distance of a point from the y-axis. For point A(7,5), the abscissa is 7.
Hence, the perpendicular distance of the point A from y-axis is 7 units.

#### Question 11:

A point both of whose coordinates are negative lies in

​Explanation:
Points of the type (-,-) lie in the third quadrant.

#### Question 12:

Abscissa of a point is positive in

​Explanation:
If abscissa of a point is positive, then the ordinate could be either positive or negative.
It means that the type of any point can be either (+,+) or (+, -).
Points of the type (+,+) lie in quadrant I, whereas points of the type (+,-) lie in quadrant IV.

#### Question 13:

The coordinates of two points are A(3, 4) and B(−2, 5) then (abscissa of A) − (abscissa of B) = ?
(a) 1
(b) −1
(c) 5
(d) −5

(c) 5

​Explanation:
Abscissa of A = 3
Abscissa of B = -2
Hence, (abscissa of A) - (abscissa of B) = 3 - (-2) = 5

#### Question 14:

The points A(2, −2), B(3, 3), C(4, −4) and D(5, −5) all lie in

​Explanation:
For all the given points, the abscissa is positive and the ordinate is negative.
Such points of the type (+,-) lie in quadrant IV.

#### Question 15:

Which of the points A(0, 6) B(−2, 0), C(0, −5), D(3, 0) and E(1, 2) does not lie on x-axis?
(a) A and C
(b) B and D
(c) A, C and E
(d) E only

(c) A,C and E

​Explanation:
The ordinate of the points lying on the x-axis = 0
So, the points B and D lie on the x-axis. The rest of the points do not lie on the x-axis, as their ordinates are not equal to 0.
Thus, the points A, C and E do not lie on the x-axis.

#### Question 16:

The signs of abscissa and ordinate of a point in quadrant II are respectively
(a) (+, )
(b) (−, +)
(c) (−, −)
(d) (+, +)

(b) (-, +)

In quadrant II, the sign of the abscissa is negative and the sign of the ordinate is positive.

#### Question 17:

Which of the following points does not lie on the line y = 3x + 4?
(a) (1, 7)
(b) (2, 0)
(c) (−1, 1)
(d) (4, 12)

(d) (4,12)

Explanation:
(a) Point (1,7) satisfy the equation y = 3x + 4.                      (∵y = 3 × 1 + 4 = 7)
(b) Point (2,10) satisfy the equation y = 3x + 4.                    (∵y = 3 × 2 + 4 = 10)
(c) Point (-1,1) satisfy the equation y = 3x + 4.                     (∵y = 3 × -1 + 4 = 1)
(d) Point (4,12) does not satisfy the equation y = 3x + 4.    (∵ y = 3 × 4 + 4 = 16 ≠ 12)
Hence, the point (4,12) do not lie on the line y = 3x +4.

#### Question 18:

Which of the following points lies on the line y = 2x + 3?
(a) (2, 8)
(b) (3, 9)
(c) (4, 12)
(d) (5, 15)

(b) (3,9)

Explanation:
Point (2,8) does not satisfy the equation y = 2x + 3.              (​∵ y = 2 × 2 + 8 = 12$\ne$ 8)
Point (3,9) satisfy the equation y = 2x + 3.                             (​∵ y =2 × 3 + 3 = 9)
Point (4,12) does not satisfy the equation y = 2x + 3.    (∵ y = 2 × 4 + 3 = 11$\ne$ 12)
Point (5,15) does not satisfy the equation y = 2x +3.    (∵ y= 2 × 5 + 3 = 13$\ne$15)
Hence, the point (3,9) lies on the line ​y = 2x +3.

#### Question 19:

If a < 0 and b < 0, then the point P(a, b) lies in

Explanation:
Points of the type (-,-) lie in the third quadrant.
Hence, the point P(a,b), where a < 0 and b < 0, lie in quadrant III.

#### Question 20:

The perpendicular distance of the point P(4, 3) from the y-axis is
(a) 3 units
(b) 4 units
(c) 5 units
(d) 7 units

(b) 4 units
Explanation:
The perpendicular distance of the point P(4,3) from the y-axis is 4 units (the abscissa).

#### Question 21:

The area of the OAB with O(0, 0), A(4, 0) and B(0, 6) is
(a) 8 sq units
(b) 12 sq units

(c) 16 sq units
(d) 24 sq units

(b) 12 sq units
Explanation:
On plotting the points on a graph paper, we get ∆OAB as a right angle triangle, where OA = base = 4 units and OB = 6 units
∴ Area of ∆OAB = ½ × OA × OB = ½ × 4 × 6 = 12 sq units

#### Question 22:

The area of the OPQ with O(0, 0), P(1, 0) and Q(0, 1) is
(a) 1 sq unit
(b)
(c)
(d) 2 sq units

(b)  ½​ sq unit
Explanation:
On plotting the points on a graph paper, we get ∆OPQ as a right angle triangle, where OP = base = 1 units and OQ = 1 units
∴ Area of (∆OPQ) = ½ × OP × OQ = ½ × 1 × 1 = ½ sq unit

#### Question 23:

Consider the three statements given below:
I. Any point on x-axis is of the form (a, 0).
II. Any point on y-axis is of the form (0, b).
III. The point P(3, 3) lies on both the axes.
Which is true?
(a) I and II
(b) I and III
(c) II and III
(d) III only

(a) I and II

Explanation:
Ordinates of points lying on the x-axis = 0
Abscissae of points lying on the y-axis = 0
In point P(3,3), neither the abscissa nor the ordinate is 0. Hence, statements I and II are true.

#### Question 24:

Assertion: The point P(−3, 0) lies on x-axis.
Reason: Every point on x-axis is of the form (x, 0).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Explanation:
Assertion (A): The point P(-3,0) lies on the x-axis. This is true, as the ordinate of the point is 0.
Reason (R): Every point on the x- axis is of the form (x,0). This is also a true statement.
Hence, both the assertion and the reason are true and reason (R) is the correct explanation of assertion (A).

#### Question 25:

Assertion: The point O(0, 0) lies in quadrant I.
Reason: The point O(0, 0) lies on both the axes.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(d) Assertion (A) is false and Reason (R) is true.
Explanation:
Assertion (A): The point O(0,0) lies in quadrant I. This is a false statement, as point O is the origin where two axes intersect each other.
Reason (R): The point O(0, 0) lies on both the axes. This is a true statement.
​Hence, assertion (A) is false and reason (R) is true.
So, the correct answer is (d).

#### Question 26:

Assertion: The point P(−6, −4) lies in quadrant III.
Reason: The signs of points in quadrants I, II, III and IV are respectively (+, +), (−, +), (−, −) and (+, ).
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(a)  Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Explanation:
Assertion (A): The point P(-6,-4) lies in quadrant III. This is a true statement, as points of the type (-,-) lie in quadrant III.
Reason (R): The signs of the points in quadrants I, II, III and IV are (+, +), (−,+), (−,−) and (+,), respectively. This is also a true statement.
Clearly, reason ( R) justifies assertion (A), as those points of the type (-,-) lie in quadrant III.
​Hence,  (a).

#### Question 27:

Assertion: If ab, then (a, b) ≠ (b, a).
Reason: (4, −3) lies in quadrant IV.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

(b)  Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
Explanation:
Assertion (A): If a ≠ b, then (ab) ≠ (b, a), which is a true statement.
Reason ( R ): (4, −3) lies in quadrant IV, as points of the type (+,-) lie in the fourth quadrant. So, the reason (R)  is also a true statement.
But, the reason does not justify the assertion.
​Hence, the correct answer is (b).

#### Question 28:

Write whether the following statements are true or false?
(i) The point P(6, 0) lies in the quadrant I.
(ii) The perpendicular distance of the point A(5, 4) for x-axis is 5 units.

(i) False
Explanation:
The ordinate of the point P(6,0) is 0. So, it lies on the x-axis.

(ii) False
Explanation:
The perpendicular distance of the point A( 5,4) from the x-axis will be 4 units, not 5 units.

#### Question 29:

State whether true or false:
(i) The mirror image of the pint A(4, 5) in the x-axis is A'(−4, 5).
(ii) The mirror image of the pint A(4, 5) in the y-axis is A'(−4, 5).

(i) False
Explanation:
The mirror image of the point A(4,5) on the x-axis is A'(4,-5), not A'(-4,5).

(ii) True
Explanation:
The mirror image of the point A(4,5) on the y-axis is A'(-4,5).

#### Question 30:

Write whether the following statements are true or false:
(a) The point (−5, 0) lies on x-axis.
(b) The point (0, −3) lies in quadrant II.

(i)True
Explanation:
The point (−5,0) lies on the x-axis, as any point whose ordinate is 0 lies on the x-axis .Therefore, the given statement is correct.

(ii) False
Explanation:
The point (0,-3) lies on the y-axis. So, the given statement is false.

#### Question 31:

Match the following columns:

 Column I Column II (a) Equation of x-axis is (p) (a, 0) (b) Equation of y-axis is (q) y = 0 (c) Any point on x-axis is of the form (r) (0, b) (d) Any point on y-axis is of the form (s) x = 0
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a)-(q), (b)-(s), (c)-(p) and (d)-(r)

Explanation:
(a) As the points that lie on the x-axis have their ordinates equal to 0, the equation of the x-axis will be y = 0.
(b) As the points that lie on the y-axis have their absiccae equal to 0, the equation of the y-axis will be x = 0.
(c) Any point on the x-axis is of the form (a,0).
(d) Any point on the y-axis is of the form (0,b).

#### Question 32:

Match the following columns:

 Column I Column II (a) The point A(−3, 0) lies on (p) y-axis (b) The point B(−5, −1) lies in quadrant (q) IV (c) The point C(2, −3) lies in quadrant (r) III (d) The point D(0, −6) lies on (s) x-axis
(a) ......,
(b) ......,
(c) ......,
(d) ......,

(a)-(s), (b)-(r), (c)-(q) and (d)-(p)
Explanation:
The points of the type (a,0) lie on the x-axis.
The points of the type (-,-) lie in quadrant III.
The points of the type (+,-) lie in quadrant IV.
The points of the type (
0,b) lie on the y-axis.

#### Question 33:

Without plotting the given points on a graph paper indicate the quadrants in which they lie, it
(a) ordinate = 6, abscissa = −3
(b) ordinate = 6, abscissa = 4
(c) abscissa = −5, ordinate = −7
(d) ordinate = 3, abscissa = 5

(a) Point (-3,6) lie in quadrant II.
(b) Point (4,-6) lie in quadrant IV.
(c) Point (-5,-7) lie in quadrant III.
(d) Point (5,3) lie in quadrant I.​

#### Question 34:

Plot the point P(−6, 3) on a graph paper. Draw PL x-axis and PM ⊥ y-axis. Write the coordinates of L and M.

The required point is shown in the graph given above.

Also, draw PL ​⊥ x-axis and PM ⊥ y-axis.
The coordinates of L and M are (-6,0) and(0,3), respectively.

#### Question 35:

Plot the points A(−5,2), B(3,−2), C(−4,−3) and D(6, 0) on a graph paper.

The points A(-5,2), B(3,-2), C(-4,-3) and D(6,0) are plotted on the graph paper.

#### Question 36:

The three vertices of ABC are A(1, 4), B(−2, 2) and C(3, 2). Plot these points on a graph paper and calculate the area of ∆ABC.

Let A(1,4), B(-2,2) and C(3,2) be the vertices of ∆ABC.
On plotting the points on the graph paper and joining the points, we get ∆ABC as shown above.
Let BC intersect y-axis at D.
Then BC = BD + DC = (2 + 3) units = 5 units                                    ( Abscissa of B  = -2, which indicates that it is on the left side of y-axis. So, for calculating the length of BC, we will consider only the magnitude)
Draw AM ⊥​ x -axis meeting BC at L.
Ordinate of point L = ordinate of point C = 2
So, AL = AM - LM = (4 - 2) units = 2 units
∴ Area of (∆ABC) = ½ × BC × AL = ½ × 5 × 2 = 5 sq units

#### Question 37:

The three vertices of a rectangle ABCD are A(2, 2) B(−3, 2) and C(−3, 5). Plot these points on a graph paper and find the coordinates of D. Also, find the area of rectangle ABCD.

Let A(2,2), B(-3,2) and C(-3,5) be the three vertices of rectangle ABCD.
On plotting the points on the graph paper and joining the points, we see that points B and C lie on quadrant II and point A lies on quadrant I.
Let D be the fourth vertex of the rectangle.
So, abscissa of D = abscissa of A = 2
Also, ordinate of D = ordinate of C = 5
So, coordinates of point D = (2,5)
Let the y-axis cut AB and CD at points L and M, respectively.
Now, AB = (BL + LA) = (3 + 2) units = 5 units               (Abscissa of B = -3, which indicates that it is on the left side of y-axis. So, for calculating the length of AB, we will consider only the magnitude.)
Thus, BC = (5 - 2) units = 3 units
∴ Area of rectangle ABCD = BC × AB = 3 × 5 = 15 sq units

#### Question 38:

The three vertices of a square ABCD are A(3, 2) B(−2, 2) and D(3, −3). Plot these points on a graph paper and hence, find the coordinates of C. Also, find the area of square ABCD.

Let A(3,2), B(-2,2) and D(3,-3) be the three vertices of square ABCD.
On plotting the points on the graph paper and joining the points, we see that A, B and D lie in different quadrants.
Let C be the fourth vertex of the square.
∴ Abscissa of C = abscissa of B = -2
Also, ordinate of C = ordinate of D = -3
So, coordinates of D = (-2,-3)
Let the y-axis cut AB and CD at points L and M, respectively.
Now, AB = (BL + LA) = (2 + 3) units = 5 units              (Abscissa of B = -2, which indicates that it is on the left side of y-axis. So, for calculating the length of AB, we will consider only the magnitude.)
∴ Area of ABCD = AB × AB = 5 × 5 = 25 sq units

#### Question 39:

From the figure given below write each of the following:
(i) The coordinates of point D
(ii) The abscissa of the point A
(iii) The point whose coordinates are (2, −3)
(iv) The point whose coordinates are (−3, −4)
(v) The ordinate of point E
(vi) The coordinates of B
(vii) The coordinates of F
(viii) The coordinates of the origin

(i) As the abscissa of point D is 0 and the ordinate is -5, the coordinates of point D are (0,-5).
(ii) The abscissa of point A is -4.
(iii) The coordinates of point E are (2,-3).

(iv) The coordinates of point C are (-3,-4).
(v) Ordinate of point E = -3
(vi) The point B lies on the x-axis, i.e., abscissa = -2 and ordinate = 0.
So, the coordinates of B are (-2,0).
(vii) Abscissa of point F = 5 and ordinate = -1
​So, coordinates of point F are (5,-1).
(viii) The coordinates of the origin are (0,0).

#### Question 1:

If x < 0 and y > 0, then the point (x, y) lies in

Explanation:
Those points of the type (-,+) lie on the second quadrant. Hence, if  x < 0 and y > 0, then the point (xy) lies in quadrant II.

#### Question 2:

Which point does not lie in any quadrant?
(a) (3, −6)
(b) (−3, 4)
(c) (5, 7)
(d) (0, 3)

(d)  (0,3)

Explanation:
The point (0,3) lies on the y-axis.

#### Question 3:

The area of AOB having vertices A(0, 6), O(0, 0) and B(6, 0) is
(a) 12 sq units
(b) 36 sq units

(c) 18 sq units
(d) 24 sq units

(c)  18​ sq units

Explanation:
On plotting the points on the graph paper, we get the right angle ∆AOB, where OB = base = 6 units and height = OA = 6 units
∴ Area of ∆AOB$\frac{1}{2}$ × OA × OB = $\frac{1}{2}$ × 6 × 6 = 18 sq units

#### Question 4:

I. Any point on x-axis is of the form (x, 0) for all x.
II. Any point on y-axis is of the form (0, y) for all y.
III. Any point on both the axes is of the form (x, y) for all x and y.
Which of the following is true?
(a) I and II
(b) I and III
(c) I only
(d) III only

(a) I and II
The correct statements are:
I: Any point on the x-axis is of the form (x,0) for all x.
II. Any point on the y-axis is of the form (0, y) for all y.

#### Question 5:

Which of the following points does not lie on the line 3y = 2x − 5?
(a) (7, 3)
(b) (1, −1)
(c) (−2, −3)
(d) (−5, 5)

(d) (-5,5)

Explanation:
(-5,5) does not satisfy the equation 3y = 2x - 5
[RHS = 2 x (-5) - 5 = -15; LHS = 3 x 5 = 15 and 15 ≠ (-15)]
So, the point (-5,5) does not lie on the equation.

#### Question 6:

Plot each of the following points on a graph paper:
A(3, −5), B(−5, −2), C(−6, 1) and D(4, 0).

The points A(3,-5), B(-5,-2), C(-6,1) and D(4,0) are plotted on the graph paper.

#### Question 7:

If 2y = 3 − 5x, find the value of y when x = −1.

On putting the value of x = -1 in the equation, 2y =  3 - 5x, we get:
2y = 3 - 5 ×​ (-1)
y = $\frac{1}{2}$ ×​ [3 - 5 ×​ (-1)] = 4
∴ y = 4 when x = -1

#### Question 8:

On which axis does the point A(0, −4) lie?

Abscissa of point A(0,-4) = 0
Hence, A lies on the y-axis.

#### Question 9:

In which quadrant does the point B(−3, −5) lie?

The abscissa and ordinate of point B(-3,-5) are negative and those points of the type (-,-) lie in the third quadrant.
Hence, point B lies in quadrant III.

#### Question 10:

What is the perpendicular distance of the point P(−2, −3) from the y-axis?

Abscissa of point P(-2,-3) = -2

However, distance cannot be negative.
Hence, the perpendicular distance of point P(-2,-3) from the y-axis is 2 units.

#### Question 11:

At what point do the coordinate axes meet?

The coordinate axes (x-axis and y-axis) meet at point O(0,0), known as the origin.

#### Question 12:

For each of the following write true or false
(i) The point (4, 0) lies in quadrant I.
(ii) The ordinate of a point P is −3 and its abscissa is −4. The point is P(−3, −4).
(iii) The points A(1, −1) and B(−1, 1) both lies in quadrant IV.
(iv) A point lies on y-axis at a distance of 3 units from x-axis. Its coordinates are (3, 0).
(v) The point C(0, −5) lies on y-axis.
(vi) The point O(0, 0) lies on x-axis as well as y-axis.

(i) False. It lies on the x-axis.
(ii) False. The point is P(-4,-3).
(iii) False. A(1,-1) lies in quadrant IV and B (-1,1) lies in quadrant II.
(iv) False. The coordinates of the point are (0,3).
(v) True.
(vi) True.

#### Question 13:

Taking a suitable scale, plot the following points on a graph paper:

 x −4 −2 5 0 3 −5 y 6 −7 5 −1 −6 0

The points A(-4, 6), B( -2,-7), C( 5,5), D (0,-1), E( 3, -6) and F(-5,0) are plotted on the graph paper.

#### Question 14:

(i) Write the points whose ordinate is 0.
(ii) Write the points whose abscissa is 0.
(iii) Write the points whose ordinate is −3.
(iv) Write the points whose abscissa is 2.
(v) Write the coordinates of all points in quadrant II.
(vi) Write the coordinates of all those points for which abscissa and ordinate have the same value.

(i) The points G(-3,0), H(-8,0), Q(4,0)and R(9,0) lie on the x-axis. Hence, their ordinates are equal to 0.

(ii) The points L(0,-6), K(0,-2), D(0,3) and C(0,7) lie on the y-axis. Hence, their abscissae are equal to 0.

(iii) The ordinates of points M(1,-3), J(-4,-3) and P(6,-3) are equal to -3.

(iv) B(2,4) and N 2,-1)

(v) The points E and F lie in quadrant II.
Coordinates of E = (-4,4)
Coordinates of F = (-6,2)
(vi) A( 3,3) and I(-2,-2)

#### Question 15:

(i) Write the mirror image of the point (2, 5) in the x-axis.
(ii) Write the mirror image of the point (3, 6) in the y-axis.
(iii) A point (a, b) lies in quadrant II. In which quadrant does (b, a) lie?

(i) The mirror image of the point (2,5) in the x-axis is (2,5).
(ii) The mirror image of the point (3,6) in the y-axis is (3,6).
(iii) If a point (a,b) lies in quadrant II, then a must be a negative number and b must be a positive number. So, the point (b,a) or (+,) lie in quadrant IV.

#### Question 16:

Without plotting the points on a graph paper indicate the quadrant in which they lie:
(i) ordinate = 4, abscissa = −3
(ii) ordinate = −5, abscissa = 4
(iii) abscissa = −1, ordinate = −2
(iv) abscissa = −5, ordinate = 3
(v) abscissa = 2, ordinate = 1
(vi) abscissa = 7, ordinate = −4

(i) Point (3,4) lies in quadrant II.
(ii) Point (4,5) lies in quadrant IV.
(iii) Point (1,2) lies in quadrant III.
(iv) Point (5, 3) lies in quadrant II.
(v) Point (2,1) lies in quadrant I.
(vi) Point (7,4) lies in quadrant IV.

#### Question 17:

Which of the following points do not lie on x-axis?
(i) A(0, 6)
(ii) B(2, 0)
(iii) C(0, −2)
(iv) D(−6, 0)
(v) E(2, 1)
(vi) F(0, 4)

The points B(2,0) and D(6,0) have their ordinates equal to 0. Hence, they lie on the x-axis.
The rest of the points whose ordinate is not equal to zero (i.e., A, C, E and F) do not lie on the  x-axis.
Hence, the points ​A, C, E and F do not lie on the x-axis.

#### Question 18:

Three vertices of a rectangle ABCD are  A(3, 1), B(−3, 1) and C(−3, 3). Plot these points on a graph paper and find the coordinates of the fourth vertex D.

Let A( 3,1), B(-3,1) and C(-3,3) be the three vertices of rectangle ABCD.
On plotting the points on a graph paper and joining the points, we see that A lie in quadrant I and B and C lie in quadrant II.
Let D be the fourth vertex of the rectangle.
i.e., Abscissa of D = abscissa of A = 3
Also, ordinate of D = ordinate of C = 3
∴ Coordinates of the fourth vertex, D = (3,3)

#### Question 19:

Write the coordinates of vertices of a rectangle OABC, where O is the origin, length OA = 5 units lying along x-axis, breadth AB = 3 units and B lying in the fourth quadrant.

Given: OABC is a rectangle. O is the origin, OA = 5 units along the x-axis, AB = 3 units and B lies in quadrant IV.
Solution: Coordinates of origin, i.e., O = (0,0)
Point A lies on the x-axis. So, coordinates of point A = (5,0)
Point B lies in the fourth quadrant. So, ordinate of point B is negative.
As width AB = 3 units, coordinates of point B = ( 5,−3)
Point C and point O lies on the same line.
Hence, abscissa of C = abscissa of O = 0
It means that point C lies on the y-axis.
Similarly, point C and point B lie on the same altitude. So, the ordinates of both points must be equal.
i.e., ordinate of C = ordinate of B = (−3)
i.e., coordinates of C = (0, 3)
Thus, the coordinates of the vertices of rectangle OABC are O(0,0), A( 5,0), B( 5, -3) and C( 0,-3).

#### Question 20:

Plot the points A(2, 5), B(−2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of ABC.

Let A( 2,5), B(-2,2) and C(4,2) be the three vertices of  ∆​ABC.
On plotting the points on a graph paper and joining the points, we see that points A and C lie in quadrant I and point B lie in quadrant II.
Let BC intersects y-axis at point D.

BC = (BD + DC) = (2 + 4) units = 6 units          (Abscissa of B = −2, which indicates that it is on the left side of y-axis. So, for calculating the length of BC, we will consider the magnitude only)
Draw AM x-axis and intersect BC at L.​
Ordinate of point L = ordinate of point B = ordinate of point C
AL = AM − LM = (5 2) units = 3 units
∴ Area of ∆ABC = ½ × BC × AL = ½ × 6 × 3 = 9 sq units

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