How to find the coordinates of a point on the coordinate graph paper?
In the adjoining figure, for locating the coordinates of a point draw XOX' and YOY' are coordinate axes.
To locate the position of point P, we draw a perpendicular from P on X'OX, i.e., PT ┴ XOX'
So, the coordinate of point P are (OT, PT).
Example to find the coordinates of a point:
1. In the adjoining figure, XOX' and YOY' are the coordinate axes. Find out the coordinates of point A, B, C and D.
Solution:
To locate the position of point A, draw AQ ┴ X'OX.
Then the coordinate of point A are (OQ, QA) i.e., A (5, 2). These points lie in the I quadrant.
To locate the position of point B, draw BP ┴ X'OX.
Then the coordinate of point B are (OP, PB) i.e., B (3, 4). These points lie in the II quadrant.
To locate the position of point C, draw CS ┴ X'OX.
Then the coordinate of point C are (OS, SC), i.e., C (4, 2). These points lie in the III quadrant.
To locate the position of point D, draw DR ┴ X'OX.
Then the coordinate of point D are (OR, RD) i.e., D (3, 2). These points lie in the IV quadrant.
2. In the
adjoining figure, XOX' and YOY' are the coordinate axes. Find out the
coordinates of point P, Q, R, S, T and U. Also write the abscissa and ordinate
in each case.
To locate the position of point Q:
Point Q is the I quadrant where abscissa and ordinate both are positive.
Perpendicular distance of Q from yaxis is 4 units.
So, xcoordinate of Q is 4.
Perpendicular distance of Q from xaxis is 3 units.
So, ycoordinate of Q is 3.
Therefore, coordinate of Q are (4, 3).
To locate the position of point P:
Point P is the II quadrant where abscissa is negative and ordinate is positive.
Perpendicular distance of P from yaxis is 2 units.
So, xcoordinate of P is 2
Perpendicular distance of P from xaxis is 5 units.
So, ycoordinate of P is 5
Therefore, coordinate of P are (2, 5)
To locate the position of point S:
Point S is the III quadrant where abscissa and ordinate both are negative.
Perpendicular distance of S from yaxis is 4 units.
So, xcoordinate of S is 4.
Perpendicular distance of S from xaxis is 1 unit.
So, ycoordinate of S is 1.
Therefore, coordinate of S are (4, 1)
To locate the position of point R:
Point R is the IV quadrant where abscissa is positive and ordinate is negative.
Perpendicular distance of R from yaxis is 2 units.
So, xcoordinate of R is 2
Perpendicular distance of R from xaxis is 4 units.
So, ycoordinate of R is 4
Therefore, coordinate of R are (2, 4)
To locate the position of point T:
Point T is in the positive xaxis. We know, that the coordinate of a point on xaxis are of the form (x, 0)
Perpendicular distance of T from yaxis is 2 units.
So, xcoordinate of T is 2
Perpendicular distance of T from xaxis is 0 unit.
So, ycoordinate of T is 0
Therefore, coordinate of T are (2, 0)
To locate the position of point U:
Point U is in the negative yaxis. We know, that the coordinate of a point on yaxis are of the form (0, y)
Perpendicular distance of U from yaxis is 0 units.
So, xcoordinate of U is 0
Perpendicular distance of U from xaxis is 4 units.
So, ycoordinate of U is 4
Therefore, coordinate of U are (0, 4)
The above workedout problems will help us to find the coordinates of a point on the graph paper.
Related Concepts:
● Ordered pair of a Coordinate System
● Find the Coordinates of a Point
● Coordinates of a Point in a Plane
● Plot Points on Coordinate Graph
● Simultaneous Equations Graphically
● Graph of Perimeter vs. Length of the Side of a Square
● Graph of Area vs. Side of a Square
● Graph of Simple Interest vs. Number of Years
7th Grade Math Problems
8th Grade Math Practice
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