CoordinateGrid &

DistanceGeometry

Formula

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Grid Grid

A pattern of horizontal and vertical lines, usually forming squares.

Coordinate grid A grid used to locate a point

by its distances from 2 intersecting straight lines

A

B

C

D

E

1 2 3 4 5

What are the coordinatesfor the foot ball?

The Coordinate Plane In coordinate geometry, points are placed on the

"coordinate plane" as shown below. It has two scales:

X axis – A horizontal number line on a coordinate grid.

Y axis - A vertical number line on a coordinate grid.

1 2 3 4 50 6

x1

2

34

5

6y

Coordinates Coordinates

An ordered pair of numbers that give the location of a point on a grid.

1

2

3

4

5

0

6

1 2 3 4 50 6

(3,4)

How to Plot Ordered Pairs Step 1 – Always find the x value first, moving

horizontally either right (positive) or left (negative). Step 2 – Starting from your new position find the y

value by moving vertically either up (positive) or down (negative).

(3, 4)

1

3

2

45

0 1 2 3 4 5 6

y 6

1

3

2

45

0 1 2 3 4 5 6

y 6

Step 1 Step 2

x x

(3, 4)

Four Quadrants of Coordinate Grid

Origin – The point where the axes cross is called the origin and is where both x and y are zero.

On the x-axis, values to the right are positive and those to the left are negative. 

On the y-axis, values above the origin are positive and those below are negative.

Four Quadrants of Coordinate Grid When the number lines are extended into the

negative number lines you add 3 more quadrants to the coordinate grid.

-2 -1

1

2

-3

3

-2

-1

0 1 2

-3

3

y

x(+ , +)( -, +)

( -, -) (+ , - )

1st Quadrant2nd Quadrant

3rd Quadrant 4th Quadrant

Four Quadrants The following relationship between the signs of

the coordinates of a point and the quadrant of a point in which it lies.

1) If a point is in the 1st quadrant, then the point will be in the form (+, +), since the 1st quadrant is enclosed by the positive x - axis and the positive y- axis.

2) If a point is in the 2nd quadrant, then the point will be in the form (–, +), since the 2nd quadrant is enclosed by the negative x - axis and the positive y - axis.

Four Quadrants3) If a point is in the 3rd quadrant, then the point

will be in the form (–, –), since the 3rd quadrant is enclosed by the negative x - axis and the negative y – axis.

4) If a point is in the 4th quadrant, then the point will be in the form (+, –), since the 4th quadrant is enclosed by the positive x - axis and the negative y - axis

x

y(+, +) (–, +)

(–, –) (+, –)

III

III IV

Coordinate Geometry

A system of geometry where the position of points on the plane is described using an ordered pair of numbers.

The method of describing the location of points in this way was proposed by the French mathematician René Descartes .

He proposed further that curves and lines could be described by equations using this technique, thus being the first to link algebra and geometry. 

In honor of his work, the coordinates of a point are often referred to as its Cartesian coordinates, and the coordinate plane as the Cartesian Coordinate Plane.

René Déscartes (1596 -1650)

Distance Formula The distance of a point from the y-axis is

called its x-coordinate, or abscissa.

The distance of a point from the x-axis is called its y-coordinate, or ordinate.

The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y).

Distance Formula Let us now find the distance between any two

points P(x1, y1) and Q(x1, y2)

Draw PR and QS x-axis. A perpendicular from the point P on QS is drawn to meet it at the point T

So, OR = x1 , OS = x2 , PR = PS = y1 , QS = y2

Then , PT = x2 – x1 ,

QT = y2 – y1

x

Y

P (x1 , y1)

Q(x2 , y2)

T

R SO

Distance Formula Now, applying the Pythagoras theorem in ΔPTQ,

we get

Therefore

222 QTPTPQ

212

212 yyxx

212

212 yyxxPQ

which is called the distance formula.

Section Formula Consider any two points A(x1 , y1) and B(x1 ,y2)

and assume that P (x, y) divides AB internally in the ratio m1: m2 i.e.

Draw AR, PS and BT x-axis. Draw AQ and PC parallel to the x-axis.

Then, by the AA similarity criterion, x

Y

A (x1 , y1)

B(x2 , y2)

P (x , y)

R SO T

2

1

m

m

PB

PA

m1

m2

Q

C

Section FormulaΔPAQ ~ ΔBPC

---------------- (1)

Now,AQ = RS = OS – OR = x– x1

PC = ST = OT – OS = x2– x

PQ = PS – QS = PS – AR = y– y1

BC = BT– CT = BT – PS = y2– y

Substituting these values in (1), we get

BC

PQ

PC

AQ

BP

PA

yy

yy

xx

xx

m

m

2

1

2

1

2

1

Section FormulaFor x - coordinate Taking

or

xx

xx

m

m

2

1

2

1

1221 xxmxxm

122121 xmxmxmxm or

or 121221 mmxxmxm

12

1221

mm

xmxmx

Section Formula

For y – coordinateTaking

yy

yy

m

m

2

1

2

1

1221 yymyym

122121 ymymymym

121221 mmyymym

12

1221

mm

ymymy

or

or

or

Section FormulaSo, the coordinates of the point P(x, y) which

divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1: m2 are

12

1221

12

1221 ,mm

ymym

mm

xmxm

This is known as the section formula.

Mid- Point• The mid-point of a line segment divides the line

segment in the ratio 1 : 1. Therefore,

the coordinates of the mid-point P of the join of the points A(x1, y1) and B(x2, y2) is

From section formula

11

11,

11

11 1212 yyxx

2,

21212 yyxx

Area of a Triangle Let ABC be any triangle whose vertices are A(x1 ,

y1), B(x2 , y2) and C(x3 , y3). Draw AP, BQ and CR

perpendiculars from A, B and C, respectively, to the x-axis.

Clearly ABQP, APRC and BQRC are all trapezium, Now, from figureQP = (x2 – x1)

PR = (x3 – x1)

QR = (x3 – x2) x

Y

A (x1 , y1)

B(x2 , y2)

C (x3 , y3)

P QO R

Area of a Triangle Area of Δ ABC = Area of trapezium ABQP + Area of

trapezium BQRC– Area of trapezium APRC.

We also know that , Area of trapezium =

Therefore, Area of Δ ABC =

embetween th distancesides parallel of sum2

1

PRCR + AP2

1CR + BQ

2

1QPAP + BQ

2

1 QR

133123321212 2

1

2

1

2

1xxyyxxyyxxyy

1333113123332232112112222

1xyxyxyxyxyxyxyxyxyxyxyxy

1233122312

1yyxyyxyyx

Area of Δ ABC

The End

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