Plot the following points on the graph paper:
(i) (2, 5) (ii) (4, -3)
(iii) (-5, -7) (iv) (7, -4)
(v) (-3, 2) (vi) (7, 0)
(vii) (-4, 0) (viii) (0, 7)
(ix) (0, -4) (x) (0, 0)
Write the coordinates of each of the following points marked in the graph paper:
A (3, 1)
From Point A draw a perpendicular to x-axis we get 3 and perpendicular to y-axis we get 1. Therefore co-ordinates of point A is (3, 1).
B(6, 0)
Since Point B lies on x-axis six places away from origin. Therefore co-ordinates of point B is (6, 0).
C(0, 6)
Since Point C lies on y-axis six places away from origin. Therefore co-ordinates of point C is (0, 6).
D(-3, 0)
Since Point D lies on x-axis three places away from origin on left side. Therefore co-ordinates of point D is (-3, 0).
E(-4, 3)
From Point E draw a perpendicular to x-axis we get-4 and perpendicular to y-axis we get 3. Therefore co-ordinates of point E is (-4, 3).
F(-2, -4)
From Point F draw a perpendicular to x-axis we get-2 and perpendicular to y-axis we get -4. Therefore co-ordinates of point F is (-2, -4).
G(0, -5)
Since Point G lies on y-axis 5 places away from origin in the downward disrection since value of the co-ordinate is negative. Therefore co-ordinates of point G is (0, -5).
H(3, -6)
From Point H draw a perpendicular to x-axis we get 3 and perpendicular to y-axis we get -6. Therefore co-ordinates of point H is (3, -6).
P(7, -3)
From Point P draw a perpendicular to x-axis we get 7 and perpendicular to y-axis we get -3. Therefore co-ordinates of point P is (7, -3).
The point of intersect of the coordinate axes is
A. ordinate
B. abscissa
C. quadrant
D. origin
The point where coordinate axes intersect is known as origin O(0, 0).
The abscissa and ordinate of the origin are
A. (0, 0)
B. (1, 0)
C. (0, 1)
D. (1, 1)
The point where coordinate axes intersect is known as origin The abscissa and the ordinate of Origin are (0, 0).
The measure of the angle between the coordinate axes is
A. 0°
B. 90°
C. 180°
D. 360°
Coordinate axes intersect each other at 90° or coordinate axes are perpendicular to eact other.
A point whose abscissa and ordinate are 2 and -5 respectively lies in
A. First quadrant
B. Second quadrant
C. Third quadrant
D. Fourth quadrant
As we know in the fourth coordinate abscissa is positive and ordinate is negative.
Points (-4, 0) and (7, 0) lie
A. on x-axis
B. y-axis
C. a line parallel to y-axis
D. a line parallel to x-axis
Since the ordinate of both the given points is 0, therefore both the points lie on
The ordinate of any point on x-axis is
A. 0
B. 1
C. -1
D. any number
The ordinate of any point on x-axis is always zero. This means that this point hasn’t covered at any distance on y-axis.
The abscissa of any point on y-axis is
A. 0
B. 1
C. -1
D. any number
The abscissa of any point on y-axis is always zero. This means that this point hasn’t covered at any distance on x-axis.
The abscissa of a point is positive in the
A. First and Second quadrant
B. Second and Third quadrant
C. Third and Fourth quadrant
D. Fourth quadrant
We knw that abscissa is always positive in first and fourth coordinate and ordinate is always positive in first and second coordinate.
A point whose abscissa is -3 and ordinate 2 lies in
A. First quadrant
B. Second quardant
C. Third quadrant
D. Fourth quadrant
As we know that abscissa is negative in second and third coordinate and ordinate is positive in first and second coordinate. Therefore the given point -3, 2 lies in second coordinate.
Two points having same abscissa but different ordinates lie on
A. x-axis
B. y-axis
C. a line parallel to y-axis
D. a line parallel to x-axis
Two points having same abscissa but different ordinate always amke a line which is parallel to y-axis.
The perpendicular distance of the point P(4,3) from x-axis is
A. 4
B. 3
C. 5
D. none of these
The perpendicular distance of any point from x-axis is always equal to the value of ordinate.
The perpendicular distance of the point P(4,3) from y-axis is
A. 4
B. 3
C. 5
D. none of these
The perpendicular distance of any point from y-axis is always equal to the value of abscissa.
The distance of the point P(4,3) from the origin is
A. 4
B. 3
C. 5
D. 7
Using Pythagorous theorem: OP2 = OQ2 + QP2
OP2 = 42 + 32
OP2 = = 5
The area of the triangle formed by the points A(2,0), B(6,0) and C(4,6) is
A. 24 sq. units
B. 12 sq. units
C. 10 sq. units
D. none of these
If (x1, y1), (x2 , y2), (x3, y3) (x1, y1) (x2, y2), (x3, y3) are the vertices of a triangle then its area is given by
Area = |1/2 (x1(y2 −y3)+x2 (y3 −y1 )+x3 (y1 −y2 ))|
Area = [(2(0-6)+6(6-0)+4(0-0)]
⇒ [-12+36+0]
⇒ 12 sq. units
The area of the triangle formed by the points P(0,1), Q(0,5) and R(3,4) is
A. 16 sq. units
B. 8 sq. units
C. 4 sq. units
D. 6 sq. Units
If (x1, y1), (x2, y2), (x3, y3) (x1, y1) (x2, y2), (x3, y3) are the vertices of a triangle then its area is given by
Area = |1/2 (x1(y2 −y3)+x2 (y3 −y1 )+x3 (y1 −y2 ))|
Area = [(0(5-4)+0(4-1)+3(1-5)]
⇒| [-12]|
⇒ 6 sq. units