maths- straight lines

7/27/2019 Maths- Straight Lines

1/13

COORDINATE GEOMETRY

Introduction

Coordinate Geometry is the unification of algebra and geometry in which algebra

is used in the study of geometrical relations and geometrical figures arerepresented by means of equations. The most popular coordinate system is therectangular Cartesian system. Coordinates of a point are the real variablesassociated in an order to describe its location in space. Here we consider thespace to be two-dimensional. Through a point O, referred to as the origin, wetake two mutually perpendicular lines XOX and YOY and call them x and y axesrespectively. The position of a point is completely determined with reference tothese axes of means of an ordered pair of real numbers (x, y) called thecoordinates of P where |x| and |y| are the distances of the point P from the y-axis and the x-axis respectively. X is called the x-coordinate or the abscissa of Pand y is called the y-coordinate or the ordinate of P.

Representation of points in a plane

We are familiar with the representation of real numbers on a line, which we calla real line. In this representation we fix a point O (called origin) and represent areal number by a point A on this line such that its distance OA (see figure givenbelow) is equal to the value of real number. In the left side of O we representnegative real numbers and in the right side of O we represent positive realnumbers. Thus, not only the magnitude of OA but the direction of the line OA isalso considered for representation.

Hence OA = AO

Similarly ordered pairs are represented in a plane. To represent an ordered pair(a, b) we take two reference lines which are mutually perpendicular. The orderedpair (a, b) represents in such a plane, by a point P(a, b) such that (see figuregiven below) OA = a and OB = b.


2/13

This system is called Cartesian co-ordinate system. Since elements of an orderedpair are not inter changeable (i.e., (a, b) (b, a) unless a = b) so they arerepresented in particular order, the first element a is represented on horizontalline called abscissa and the second element b on a vertical line called ordinate.Like the real number notation the positive side of the x-axis is the right side of Oand positive side of O and positive side of y-axis is upper side of O.

So, the two lines divide the region in 4 parts. These are called quadrants. Thesequadrants are characterized as

I quadrant x > 0, y > 0

II quadrant x < 0, y > 0

III quadrant x < 0, y < 0

IV quadrant x > 0, y < 0

Here the point O represents x = 0 and y = 0, hence order pair becomes (0, 0).

There is a second type of representation called the polar co-ordinate system. Inthis system a reference is fixed to a line (Called the initial line), and a pointcalled the origin in the system. Any point P is represented by ordered pair (r, ).

Such that

OP = r; The distance of point from origin.


3/13

and POX = The angular displacement of line OP from fixed line i.e. the

initial line, (in the anticlockwise direction)

Clearly a = r cos and b = r sin (see figure given below)

Distance between two points

The distance between two points P(x 1 , y 1 ) and Q(x 2 , y 2 ) is (see the figuregiven below).

Length PQ = (x 2 x 1 ) 2 + (y 2 y1 ) 2

Proof:

Let P(x 1 , y 1 ) and Q(x 2 , y 2 ) be the two points and let the distance betweenthem be d. Draw PA, QR parallel to y-axis and PR parallel to x-axis.

Angle QRP = 90 o


4/13

d 2 = PR 2 + RQ 2

d 2 = (x 2 x 1) 2 + (y 2 y 1) 2

d = (x 2 x1 ) 2 + (y 2 y 1) 2 .

Section Formula

Let us say we want to know the co-ordinates of point which divides a linesegment between two points A(x 1 , y 1 ) and B(x 2 , y 2 ) in the ratio m : n.

The coordinates of such a point are given by

(nx 1 + mx 2 /m+n, ny 1 + my 2 /m+n) (for internal division)

Note:

This is called section formula.

Let P divide the line segment AB in the ratio m : n. If P is inside AB then itis called internal division; if it is outside AB then it is called externaldivision.

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n.

Proof:

Consider ? ABB

Since BB||PQ and AP:PB = m:n (see figure given below)

AQ/AB' = PQ/BB' = m/m+n (= AP/AB)


5/13

x x1 /x 2 x 1 = m/m+n

x = nx 1 + mx 2 /m+n and y = ny 1 + my 2 /m+n

If P is outside AB (less assume it is at P)

We have

x x 1 /x 2 x 1 = m/m+n

x = nx 1 + mx 2 /m+n and y = ny 1 + my 2 /m+n

Similarly if P is at P then

x = mx 2+m+n/n m, y = my 2+ny 1 /m+n

Note:

m:n can be written as m/n or :1. So any point on line joining A and Bwill be P(x 2 +x 1 /+1.y 2+y 1 /+1). It is useful to assume :1 because itinvolves only one variable.

illustration:

Find the ratio in which line segment A(2, 1) and B(5, 2) is divided by x-axis.

Solution:

Let x-axis intersect line at point P(x p , 0) such that AP/BP = /1 y P = 0 = y 2+1.ya/+1 = 2+(1)/+1 = 1/2 AP/BO = 1/2

Illustration:

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are

(x 1 tan A + x 2 tan B + x 3 tan C/tan A + tan B + tan C, y 1 tan A + y 2 tan B+ y 3 tan C/tan A + tan B + tan C),

Solution:

In triangle A(x 1 , y 1 ), B(x 2 , y 2 ) and C(x 3 , y 3 ), draw AD perpendicular to BC.Our effort now should be to find the co-ordinates of the point D.


6/13

To do that, we need to find BC/CD. (figure is given above)

tan B = AD/BD and tan C = AD/CDBD/DC = tan C/tan B

Now we apply section formulae.

xD = x 2 tan B + x 3 tan C/tan B + tan C (i)

y D = y 2 tan B + y 3 tan C/tan B + tan C (ii)

We know that orthocenter will lie on AD. We need to find this point andits co-ordinates.

We should select a point H 1 on AD and take the ratio AH 1 /H 1 D in such amanner so that x H1 and y H1 calculated form (i) should be symmetric in x 1 ,x 2 , x 3 , tan A, tan B and tan C. Think before you proceed

Let AH 1 /H 1 D = tan B + tan C/tan A

x H1 = x 1 tan A + x 2 tan B + x 3 tan C/tan A + tan B +tan C

and y H1 = y 1 tan A + y 2 tan B + y 3 tan C/tan A + tan B + tan C

Since the result is symmetric, this point H 1 will lie on other altitude aswell i.e. the altitudes are concurrent

xH = x H1 and y H = y H1

Illustration:


7/13

Prove analytically that in a right angled triangle the midpoint of thehypotenuse is equidistant from the three angular points.

Solution:

While proving a problem analytically take most convenient co-ordinates of known points.

In the present case triangle is assumed as AOB with coordinates as shownin figure given below, C is midpoint of AB.

So co-ordinates of C will be (a/2, b/2)

Now AB = a 2 + b 2 CA = CB = AB/2 (C is mid point of AB)

= a 2 + b 2

and, we know that the distance between two points C and O is given by

CO = (a/2 0) 2 + (b/2 0) 2 = a 2 + b 2 /2

Hence CA = CB = CO

Coordinates of the point P dividing the join of two points A(x 1 , y 1 ) and

B(x 2 , y 2) internally in the given ratio 1 : 2 i.e., AP/BP= 1 / 2 areP( 2x 1+ 1x 2 / 2+ 1 , 2y 1+ 1y 2 / 2+ 1 ).

Coordinates of the point P dividing the join of two points A(x 1 , y 1 ) andB(x 2 , y 2) externally in the ratio 1 : 2 i.e., Ao/BP= 1 / 2 areP( 2x 1+ 1x 2 / 2 1 , 2y 1+ 1y 2 / 2 1 ).

Centroid of Triangle


8/13

The centroid of a triangle is the point of concurrency of the medians. Thecentroid G of the triangle ABC, divides the median AD, in the ratio of 2 : 1.

Illustration:

Find the centroid of the triangle the coordinates of whose vertices are given byA(x 1 , y 1), B(x 2 , y 2) and C(x 3 , y 3) respectively.

Solution:

AG/AD = 2/1

Since D is the midpoint of BC, coordinates of D are (x 2+x 3 /2, y 2+y 3 /2)

Using the section formula, the coordinates of G are

(2(x 2+x 3 /2)+1.x 1 /2+1, 2(y 2+y 3 /2)+1.y 1 /2+1)

Coordinates of G are (x1+x2+x3/3, y1+y2+y3/3).

Incentre of Triangle

The incentre I of a triangle is the point of concurrency of the bisectors of theangles of the triangle.

Illustration:

Find the incentre of the triangle the coordinates of whose vertices are given byA(x 1 , y 1), B(x 2 , y 2), C(x 3 , y 3).


9/13

Solution:

By geometry, we know that BD/DC = AB/AC (since AD bisects A).

If the lengths of the sides AB, BC and AC are c, a and b respectively, thenBD/DC = AB/AC = c/b.

Coordinates of D are (bx 2+cx 3 /b+c, by 2+cy 3 /b+c)

IB bisects B. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.

Let the coordinates of I be (x, y).

Then x = ax 1+bx 2+cx 3 /a+b+c, y = ay 1+by 2 +cy 3 /a+b+c.

Circum Centre of Triangle

This the point of concurrency of the perpendicular bisectors of the sides of thetriangle. This is also the centre of the circle, passing through the vertices of thegiven triangle.

Orthocentre of Triangle

This is the point of concurrency of the altitudes of the triangle.

Excentre

Excentre of a triangle is the point of concurrency of bisectors of two exterior and

third interior angle. Hence there are three excentres I 1 , I 2 and I 3 opposite tothree vertices of a triangle.

If A(x 1 , y 1), B(x 2 , y 2) and C(x 3 , y 3) are the vertices of a triangle ABC,


10/13

coordinates of centre of ex-circle opposite to vertex A are given as

I 1(x, y) = ( ax 1+bx 2+cx 3 /a+b+c/ a+b+c, ay 1+by 2+cy 3 / a+b+c).Similarly co-ordinates of centre of I 2(x, y) and I 3(x, y) are

I 2(x, y) = (ax 1bx 2+cx 3 /a b+c, ay 1by 2+cy 3 /a b+c)

I 3(x, y) = (ax 1+bx 2cx 3 /a+b c, ay 1+by 2cy 3 /a+b c)

Area of a triangle

Let (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) respectively be the coordinates of thevertices A, B, C of a triangle ABC. Then the area of triangle ABC, is

|1/2[x 1 (y 2 y 3 ) + x 2 (y 3 + y 1 ) + x 3 (y 1 y 2 )]| = .


11/13

It follows that the three points (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) will be

collinear if = 0.

Area of a polygon of n sides

First of all we plot the points and see their actual order. Let A 1 (x 1 , y 1 ),A(x 2 , y 2), , An(x n , y n) be the vertices of the polygon in anticlockwiseorder.

Then area of the polygon =

.

Illustration:

Calculate area of a triangle shown in figures given below.

Solution:

Using the just derived formula

Area of a triangle

1/2 [3(2 4) + (1) (4 6) + 5 (6 2)]

= 1/2 [ 6 2 + 20] = 6


12/13

Similarly, area of the triangle shown in the figure given above.

Area of a ?ABC is

= 1/2 [3(4 2) + (5) (2 6) + (1) (6 4)]

=1/2 [6 20 + 2] = 6

Caution:

Thus we observe that the area of a triangle is positive vertices are takenin the anticlockwise direction and negative when the vertices are takenthe clockwise direction.

Note:

Area of a triangle can also be expressed as = 1/2


13/13

= 1/2 [x 1 y 2 y1 x 2 + x 2 y3 y 2 x 3 + x 3 y1 y 3 x1 ]

This form is important. It can be used to find area of a quadrilateral,pentagon, hexagon and polygons.

Important:

If three points P 1 , P 2 and P 3 are collinear then the determinant at mustvanish i.e. the area of triangle formed must be zero.

Note:

If the vertices are in clockwise order then take modulus.

Illustration:

Prove that the area of the triangle with vertices at (p 4, p + 5),(p + 3, p 2) and (p, p) remains constant as p varies.

Solution:

The area of the triangle is

which remains constant for all values of p.

maths- straight lines

Documents