RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry


Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry download pdf.





RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry





Exercise 6A





Question 1:
Draw the perpendiculars from the AF, BG, CH, DI and EJ on the x-axis.
(1) The distance of A from the y-axis = OF = -6 units
The distance of A from the x-axis = AF = 5 units
Hence, the coordinate of A are (-6, 5)





(2)The distance of B from the y-axis = OG = 5 units
The distance of B from the x-axis = BG = 4 units
Hence, the coordinate of B are (5, 4)





(3)The distance of C from the y-axis = OH = -3 units
The distance of C from the x-axis = HC = 2 units
Hence, the coordinate of C are (-3, 2)





(4)The distance of D from the y-axis = OI = 2 units
The distance of D from the x-axis = ID = -2 units
Hence, the coordinate of D are (2, -2)





(5)The distance of E from the y-axis = OJ = -1 unit
The distance of E from the x-axis = JE = -4 units
Hence, the coordinate of E are (-1, -4)





Thus, the coordinates of A, B, C, D and E are respectively, A(-6,5), B(5,4), C(-3,2), D(2,-2) and E(-1,-4)





Question 2:
Let X’OX and Y’OY be the coordinate axes.
Fix the side of the small squares as one units.
(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) Mark the point O as O(0, 0)
(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)
These points are shown in the following graph:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q2





Question 3:
(i) In (7, 0), we have the ordinate = 0.
Therefore, (7,0) lies on the x-axis





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





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





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





Question 4:
(i) Points of the type (-, +) lie in the second quadrant. Therefore, the point (-6,5) lies in the II quadrant.





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





(iii) Points of the type (+, -) lie in the fourth quadrant. Therefore, the point  (2,-9) lies in the IV quadrant.





Question 5:
The given equation is y = x + 1
Putting x = 1, we get y = 1 + 1 = 2
Putting x = 2, we get y = 2 + 1 = 3





Thus, we have the following table:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q5-1
On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively.
Then, plot points P (1, 2) and Q (2, 3) on the graph paper. Join PQ and extend it to both sides.
Then, line PQ is the graph of the equation y = x + 1.
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q5-2





Question 6:
The give equation is y = 3x + 2
Putting x = 1, we get y = (3 1) + 2 = 5
Putting x = 2, we get y = (3 2) + 2 = 8





Thus, we have the following table:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q6-1
On the graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively.
Now, plot points P(1,5) and Q(2,8) on the graph paper.
Join PQ and extend it to both sides.
Then, line PQ is the graph of the equation y = 3x + 2.
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q6-2





Question 7:
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 following table:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q7-1
On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively.
Now plot the points P(0,-3) and Q(1,2).
Join PQ and extend it in both the directions.
Then, line PQ is the graph of the equation, y = 5x – 3.
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q7-2





Question 8:
The given equation is y = 3x
Putting x = 1, we get y = (3 1) = 3
Putting x = 2, we get y = (3 2) = 6





Thus, we have the following table:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q8-1
On a graph paper draw the lines X’OX and YOY’ as the x-axis and y-axis respectively.
Now, plot points P(1,3) and Q(2,6).
Join PQ and extend it in both the directions.
Then, line PQ is the graph of the equation y = 3x.
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q8-2





Question 9:
The given equation is y = -x
Putting x = 1, we get y = -1
Putting x = 2, we get y = -2





Thus, we have the following table:
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q9-1
On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively.
Now, plot the points P(1,-1) and Q(2,-2).
Join PQ and extend it in both the directions.
Then, line PQ is the graph of the equation y = -x.
rs-aggarwal-class-9-solutions-coordinate-geometry-2-6a-q9-2





Complete RS Aggarwal Solutions Class 9





If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.





If these solutions have helped you, you can also share alarity.in to your friends.





Best of Luck For Your Future!!


RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry RS Aggarwal Solutions Class 9 Chapter 6 Coordinate Geometry Reviewed by Admin on November 15, 2020 Rating: 5

No comments:

Powered by Blogger.