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 co-ordinate axes.
To locate the position of point P, we draw a perpendicular from P on X'OX, i.e., PT ┴ XOX'
So, the co-ordinate of point P are (OT, PT).
Example to find the coordinates of a point:
1. In the adjoining figure, XOX' and YOY' are the co-ordinate 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 co-ordinate 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 co-ordinate 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 co-ordinate 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 co-ordinate 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 co-ordinate axes. Find out the
coordinates of point P, Q, R, S, T and U. Also write the abscissa and ordinate
in each case.
Solution:
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 y-axis is 4 units.
So, x-co-ordinate of Q is 4.
Perpendicular distance of Q from x-axis is 3 units.
So, y-co-ordinate of Q is 3.
Therefore, co-ordinate 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 y-axis is 2 units.
So, x-co-ordinate of P is -2
Perpendicular distance of P from x-axis is 5 units.
So, y-co-ordinate of P is 5
Therefore, co-ordinate 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 y-axis is 4 units.
So, x-co-ordinate of S is -4.
Perpendicular distance of S from x-axis is 1 unit.
So, y-co-ordinate of S is -1.
Therefore, co-ordinate 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 y-axis is 2 units.
So, x-co-ordinate of R is 2
Perpendicular distance of R from x-axis is 4 units.
So, y-co-ordinate of R is -4
Therefore, co-ordinate of R are (2, -4)
To locate the position of point T:
Point T is in the positive x-axis. We know, that the co-ordinate of a point on x-axis are of the form (x, 0)
Perpendicular distance of T from y-axis is 2 units.
So, x-co-ordinate of T is 2
Perpendicular distance of T from x-axis is 0 unit.
So, y-co-ordinate of T is 0
Therefore, co-ordinate of T are (2, 0)
To locate the position of point U:
Point U is in the negative y-axis. We know, that the co-ordinate of a point on y-axis are of the form (0, y)
Perpendicular distance of U from y-axis is 0 units.
So, x-co-ordinate of U is 0
Perpendicular distance of U from x-axis is 4 units.
So, y-co-ordinate of U is -4
Therefore, co-ordinate of U are (0, -4)
The above worked-out 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 Co-ordinate 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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