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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