Contents
CoordinateGeometry
Question 1:
Exercise 6A
Draw the perpendiculars from the AF, BG, CH, DI and EJ on the xaxis.
 The distance of A from the yaxis = OF = 6 units The distance of A from the xaxis = AF = 5 units Hence, the coordinate of A are (6, 5)
 The distance of B from the yaxis = OG = 5 units The distance of B from the xaxis = BG = 4 units Hence, the coordinate of B are (5, 4)
 The distance of C from the yaxis = OH = 3 units The distance of C from the xaxis = HC = 2 units Hence, the coordinate of C are (3, 2)
 The distance of D from the yaxis = OI = 2 units The distance of D from the xaxis = ID = 2 units Hence, the coordinate of D are (2, 2)
 The distance of E from the yaxis = OJ = 1 unit The distance of E from the xaxis = 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.

 Starting from O, take +7 units on the xaxis and then +4 units on the yaxis to obtain the point P(7, 4)
 Starting from O, take 5 units on the xaxis and then +3 units on the yaxis to obtain the point Q(5, 3)
 Starting from O, take 6 units on the xaxis and then 3 units on the yaxis to obtain the point R(6, 3)
 Starting from O, take +3 units on the xaxis and then 7 units on the yaxis to obtain the point S(3, 7)
 Starting from O, take 6 units on the xaxis to obtain the point A(6, 0)
 Starting from O, take 9 units on the yaxis to obtain the point B(0,9)
 Mark the point O as O(0, 0)
 Starting from O, take 3 units on the xaxis and then 3 units on the yaxis to obtain the point C(3, 3)
These points are shown in the following graph:
Question 3:
 In (7, 0), we have the ordinate = 0.
Therefore, (7,0) lies on the xaxis
 In (0, 5), we have the abscissa = 0.
Therefore, (0,5) lies on the yaxis
 In (0,1), we have the abscissa = 0.
Therefore, (0,1) lies on the yaxis
 In (4,0), we have the ordinate = 0.
Therefore, (4,0) lies on the xaxis
Question 4:
 Points of the type (, +) lie in the second quadrant. Therefore, the point (6,5) lies in the II quadrant.
 Points of the type (, ) lie in the third quadrant. Therefore, the point (3,2) lies in the III quadrant.
 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:
On a graph paper, draw the lines X’OX and YOY’ as the xaxis and yaxis 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.
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:
On the graph paper, draw the lines X’OX and YOY’ as the xaxis and yaxis 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.
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:
On a graph paper, draw the lines X’OX and YOY’ as the xaxis and yaxis 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.
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:
On a graph paper draw the lines X’OX and YOY’ as the xaxis and yaxis 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.
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:
On a graph paper, draw the lines X’OX and YOY’ as the xaxis and yaxis 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.