COORDINATE GEOMETRY

•Distance between 2 points•Mid-point of 2 points

Distance between two points.

5 18

3

17

A(5,3)

B(18,17)

18 – 5 = 13 units

17 – 3 = 14 units

AB2 = 132 + 142

Using Pythagoras’ Theorem,

AB2 = (18 - 5)2 + (17 - 3)2

y

x

Distance between two points.In general,

x1 x2

y1

y2

A(x1,y1)

B(x2,y2)

Length = x2 – x1

Length = y2 – y1

AB2 = (x2-x1)2 + (y2-y1)2

Hence, the formula for Length of AB or Distance

between A and B is

y

x

212

212 )()( yyxxAB

Find the distance between the points (-1,3) and (2,-6)

• Simply by using the formula:

(-1,3) and (2,-6)

(x1,y1) and (x2,y2)

Since

= 9.49 units (3 sig. fig)

Given 3 points A,B and C, distance formula is used to check whether the points are collinear.If not we

may check for an isosceles, equilateral or right

angled triangle.Perform the check on the following sets of points :

i. (1,5), (2,3), (-2, -11)

ii. (1,-1),(-½, ½),(1,2)

iii. (a,a),(-a,-a), (-a ,a )

iv. (12,8),(-2,6),(6,0)

v. (2,5),(-1,2),(4,7)

3 3

Distance Formula can be used to check for special quadrilaterals !!

Given 4 points A,B,C,D If AB=CD, AD = BC,it is a PARALLELOGRAM.

(Opposite sides are equal) If AB = CD, AD = BC, AC = BD ,it is a

RECTANGLE.(Diagonals are also equal) If AB=BC=CD=DA, it is a RHOMBUS.(All sides

are equal) If AB=BC=CD=DA and AC=BD, it is a SQUARE.

Find the perimeter of the quadrilateral ABCD. Is ABCD a special quadrilateral?

APPLICATIONS OF DISTANCE FORMULA

Parallelogram

APPLICATIONS OF DISTANCE FORMULA

Rhombus

APPLICATIONS OF DISTANCE FORMULA

Rectangle

APPLICATIONS OF DISTANCE FORMULA

Square

SPECIAL QUADRILATERALS

1. Show that (1,1),(4,4),(4,8),(1,5) are the vertices of a parallelogram.

2. Show that A(2,-2),B(14,10),C(11,13) and D(-1,1) are the vertices of a rectangle.

3. Show that the points (1,2),(5,4),(3,8),(-1,6) are the vertices of a square.

4. Show that (1,-1) is the centre of the circle circumscribing the triangle whose angular points are (4,3),(-2,3) and (6,-1).

FINDING CO-ORDINATES

1. Find the point on x – axis which is equidistant from (2, -5) and (-2,9).

2. Find the point on y – axis which is equidistant from (2,-5) and ( -2, 9).

3. Find a relation between x and y so that the point (x, y) is equidistant from (2,-5) and ( -2, 9).

4. Find the value of k such that the distance between the points (2, -5) and (k, 7) is 13 units.

The mid-point of two points.

2

518xM

5 18

3

17

A(5,3)

B(18,17)

Look at it’s horizontal length

= 11.5

11.5

Look at it’s vertical length

2

317yM

= 10

10

(11.5, 10)

Mid-point of AB

y

x

The mid-point of two points.

221 xx

M x

221 yy

M y

x1 x2

y1 A(x1, y1)

B(, x2 ,y2)

Look at it’s horizontal length

Look at it’s vertical lengthMid-point of AB

y

x

y2

Formula for mid-point is

)2

,2

( 2121 yyxxM AB

Section Formula – Internal Division

A(x 1, y 1

)

B(x 2, y 2

)

XX’

Y’

O

Y

P(x, y)

m

n:

L N M

H

K

Clearly AHP ~ PKBAP AH PHBP PK BK

1 1

2 2

x x y ymn x x y y

2 1 2 1mx nx my nyP ,

m n m n

The co-ordinates of the point which divides the line segment joining (x1, y1) and (x2, y2) in the ratio m : n internally are

The ratio in which the point (x, y)divides the line segment joining (x1, y1) and (x2, y2) is

nm

nymy

nm

nxmx 1212 ,

yy

yy

xx

xx

n

m

2

1

2

1

• Find the co-ordinates of the point which divides the line segment joining the points (4, -3) and (8,5) in the ratio 3:1 internally.

• Find the co-ordinates of the point which divides the line segment joining the points (-1,7) and (4,-3) in the ratio 2:3 internally.

• In what ratio does the point (-4,6) divide the line segment joining the points

A(-6,10) and B (3,-8)?• Find the coordinates of the points of

trisection of the line segment joining (4,-1) and (-2,-3).

• Find the coordinates of the points which divide the line segment joining A(-2,2) and B(2,8) in four equal parts.

• In what ratio is the join of the points (-4,6) and (3, -8) divided by the

(i) x- axis. (ii) y-axis.

Also find the co-ordinates of the point of division.• Find the coordinates of the centroid of the

triangle whose vertices are (12,8),(-2,6) and (6,0).

• Find the coordinates of the vertices of a triangle whose midpoints are (4,3),(-2,3) and (6,-1).

Area of a triangle

• Area of a triangle with vertices (x1, y1),

(x2, y2), (x3, y3) is given by

)()()(

2

1213132321 yyxyyxyyx

Collinearity of points using area of triangles

• Three points (x1, y1), (x2, y2), (x3, y3) are collinear if and only if the area of the triangle with these points as vertices is 0.

0)()()( 213132321 yyxyyxyyx

• Find the area of the triangle formed by the following points:

(3,4),(2,-1),(4,-6)• Show that the following points are collinear

(-5,1),(5,5) and (10,7)• For what value(s) of x,the area of the triangle

formed by the points (5,-1),(x,4) and (6,3) is 5.5 square units.

• For what value(s) of x, will the following lie on a line : (x,-1),(5,7),(8,11)

• If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

• Find the ratio in which 2x + 3y – 30 =0, divides the join of A(3, 4) and B(7, 8) and also find the point of intersection.

)4,3(A )8,7(B

k 1