Coordinategeometry1 1

Rene DescartesA French mathematician who

discovered the co-ordinate geometry.

He was the first man who unified Algebra and Geometry.

According to him every point of the plane can be represented uniquely by

two numbers

co-ordinate geometry

It is that branch of geometry in which two numbers called co-

ordinates , are used to locate the position of a point in a plane.

It is also called as Cartesian-geometry

Axes of reference• The whole plane is divided in to the four

parts by two straight lines , which are perpendicular to each other.

• The line which is parallel to the horizontal line is called as the X - AXIS

• The line which is parallel to the vertical line is called as the Y - AXIS

• the point of intersection is called as ORIGIN

CO-ORDINATES OF A POINT

• X-CO-ORDINATE -the distance of the point from the origin along X - axis is called X- co-ordinate ( abscissa)

• Y-CO-ORDINATE -the distance of the point from the origin along Y - axis is called Y- co-ordinate ( ordinate)

REPRESENTATION OF A POINT

THE CO-ORDINATES OF A POINT IS ALWAYS REPRESENTED BY

ORDERED -PAIR ( )

FIRST PUT X-CO-ORDINATE THEN Y-CO-ORDINATE IN BRACKET

( X, Y )

Representation of points on plane

Co-ordinate of origin (0,0)

Distance formulaTo find out the distance

between two points in the plane

Let the two points are P(x1,y1) and Q(x2,y2)

P(x1,y1)

Q(x2,y2)

P(x1,y1)

Q(x2,y2)

x1 x2

y2

y1 R(x2,y1)

Then distance QR = y2 - y1 and PR = x2 - x1

• Since PRQ is a right triangle• therefore by using Pythagorus

theorem• PQ2 = PR2 + RQ2

• PQ2 = (x2 - x1) 2 + (y2 - y1) 2

• PQ = (x2 - x1) 2 + (y2 - y1) 2

PROBLEMS ON COLLINEARITY OF THREE

POINTS

POINTS A , B and C ARE said to be collinear if AB +BC = AC

PROBLEM 1Determine by distance formula , whether the points (2,5) , (-1,2)

and (4,7) are collinear.• Sol. We are given three points A (2,5)

• B (-1,2) and C (4,7)

AB= (-1-2) 2 +(2-5) 2

• AB= (-3) 2 +(-3) 2= 9+9= 18= 3 2

SIMILARLY

BC= (4+1) 2 +(7-2) 2

= (5) 2 +(5) 2

= 25+25= 50 = 25X2= 5 2

AC= (2-4) 2 +(5-7) 2

= (-2) 2 +(-2) 2 = 4+4= 8 =2 2

THUS 3 2+ 2 2 =5 2

AB +AC = BCPOINTS B , A C are collinear

Problem on equidistant

• Find the point on y-axis which is equidistant from (-5,-2) and (3,2)

• solution: let p(0,y) be a point on y-axis which is equidistant from

• A(-5,-2) and B (3,2)

PA = PB

• PA2 = PB2

• (-5-0)2 + (-2-y)2 =(3-0)2 +(2-y)2

• 25+4+ y2 +4y = 9+4 + y2 -4y• 8y = -16• y= -2 • so the required point is p(0,-2)

Note

• If we have to find the point which is on the x-axis and equidistant from the given two points then that point will be p(x,0) . Then find x by applying the similar procedure and you will get the required point.

Note • To show that the given points are

vertices of an equilateral triangle .find the length of all sides using distance formula and check all sides are equal.

• To show that the given points are vertices of a right angle triangle check whether sides LENGTHS are following Pythagorus theorem

For square1.all sides are equal and 2.diagonals are equal

For rhombus

1.all sides are equal but2. diagonals are not equal

For rectangle

1. Opposite sides are equal2. Diagonals are equal

For parallelogram1. Opposite sides are equal2. Diagonals bisect each other

It gives the co-ordinates of the point which divides the given line segment in the ratio m : n

Let AB is a line segment joining the points A(x1,y1) and B(x2,y2)

LET P(x,y) be a point which divides line segment AB

in the ratio m : n internally

therefore

(x1,y1)

(x2,y2)

(x1,y1)

(x2,y2)

(x,y1)(x-x1)

(x2,y)(x2 - x)

(y2 - y)

(y-y1)

x1 xx x2

Complete the figure as follows

Now AQP PRB …..by AA similarity

AP/PB = AQ/PR = PQ/BR

BY CPST

m/n = x-x1/x2-x =y-y1/y2-y

m/n = x-x1/x2-xsolving this equation for x we will get

x = mx2+nx1

m+n

Similarly solving the equation m/n =y-y1/y2-y

we will gety = my2+ny1

m+n

(x1,y1)

(x-x1)

(y-y1)

(y2 - y)

(x2 - x)

Thus if a point p(x,y) divides a line segment joining the points

A(x1,y1) and B(x2,y2) in the ratio m : n then the co-ordinates

of P are given by

• The co-ordinates of A and B are (1,2) and (2,3). Find the co-ordinates of R so that AR/RB = 4/3.

SOLUTION 1

Let co-ordinates of point R (X,Y)

X = 4[2]+3[1] 4+3

X= {8+3]/7 = 11/7

Here m=4 , n=3, x1=1 , y1=2 x2= 2 , y2 = 3

Y = 4[3]+3[2] 4+3

Y = [ 12+6]/7 = 18/7

Hence co-ordinates of R = (11/7 , 18/7)

• Find the ratio in which the point (11,15) divides the line segment joining the points (15,5) and (9,20).

Let k : 1 be the required ratio

Here x=11 , y=15 , x1= 15 , y1=5 ,x2=9 , y2=20

m= k , n= 1

Therefore using section formula

11 = 9k+15 k+1

and 15 = 20k+5 k+1

solve either of these two equation let’s take first equation

11k+11=9k+1511k-9k=15-11

2k=4

k=2So,the required ratio is 2:1

• Find the ratio in which the line segment joining the points (6,4) and (1,-7) is divided internally by axis of x.

Solution 3Let k:1 is the required ratio.

P(x,0) point on x-axis which divides AB in required ratio.

Here x=x , y=0 , x1= 6 , y1=4 ,x2=1 , y2= -7

Therefore using section formula

x = k+6 k+1

and0= -7k+4 k+1

solve either of these two equation let’s take second equation

0= -7k+4

7k=4

K= 4/7

So,the required ratio is 4:7.

AREA OF TRIANGLE

Let vertices of triangle are ( x1,y1) (x2,y2) and (x3,y3).

Area of Triangle =1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

A( x1,y1)

B(x2,y2) C(x3,y3).

•Find the area of triangle with vertices A(6,4) B(1,-7) C(2,3).

Here x1= 6 , y1=4 ,x2=1 , y2= -7, x3=2 , y3=3

Area of Triangle =1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))

Area of Triangle =1/2(6(-7-3)+1(3-4)+2(4+7)) =1/2(-60-1+22)

=1/2(-39) = -39/2

= -39/2

= 39/2

So,Area of Triangle is 19.5