Coordinate Geometry

 Mind Maps

Class X - Maths: Coordinate Geometry

content
• Coordinate Geometry
• Distance Formula
• Section Formula
• Area of a Triangle
• co-ordinate geometry
• Relation between Cartesian and Polar Co-Ordinates
• Distance between Two Points
• Examples on Distance between two Points
• Internal Division of line segment
• External Division of line segment

Which is called the Distance formula.
1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP2 =x2 +y2 .
2. We can also write, PQ2 =(x2 – x1)2 + (y2 – y1)2 .

Section Formula
((m_1x_2+ m_2 x_1)/(m_1+m_2),(m_1y_2+ m_2 y_1)/(m_1+m_2))
This is known as the section formula.

Area of a Triangle

The area of a triangle when its base and corresponding height (altitude) are given.
You have used the formula :
Area of a triangle =1/2 × base × altitude

Area of a trapezium =1/2 (sum of parallel sides)(distance between them)

If the area of a triangle is 0 square units, then its vertices will be collinear.

The area of a polygon, we divide it into triangular regions, which have no common area, and add the areas of these regions.

Co-ordinate Geometry is of two types:
(i) Two Dimensional or Plane Co-ordinate Geometry
(ii) Three Dimensional or Solid Co-ordinate Geometry.

What is Rectangular Cartesian Co-ordinates?
Let O be a fixed point on the plane

We can uniquely determine the position of any point on the plane of the page referred to co-ordinate axes drawn through O.

Relation between Cartesian and Polar Co-Ordinates

Internal Division of line segment
((mx_2 + nx_1)/(m + n), (my_2 + ny_1)/(m + n))
External Division of line segment
((mx_2 - nx_1)/(m - n),(my_2- ny_1)/(m - n))

Coordinate Geometry
• The position of a point on a plane, we require a pair of coordinate axes. 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).
• A set of a pair of perpendicular axes on a graph paper. Now plot the following points and join them as directed: Join the point A(4, 8) to B(3, 9) to C(3, 8) to D(1, 6) to E(1, 5) to F(3, 3) to G(6, 3) to H(8, 5) to I(8, 6) to J(6, 8) to K(6, 9) to L(5, 8) to A. Then join the points P(3.5, 7), Q (3, 6) and R(4, 6) to form a triangle. Also join the points X(5.5, 7), Y(5, 6) and Z(6, 6) to form a triangle.
• Now join S(4, 5), T(4.5, 4) and U(5, 5) to form a triangle. Lastly join S to the points (0, 5) and (0, 6) and join U to the points (9, 5) and (9, 6).
• René Déscartes, the great French mathematician of the seventeenth century, liked to lie in bed and think! One day, when resting in bed, he solved the problem of describing the position of a point in a plane.
• His method was a development of the older idea of latitude and longitude.
• In honour of Descartes, the system used for describing the position of a point in a plane is also known as the Cartesian system.

Distance Formula

• A town B is located 36 km east and 15km north of the town A. How would you find the distance
from town A to town B without B actually measuring it. Let us see. This situation 15km
can be represented graphically as shown in
Fig. 7.1. You may use the Pythagoras Theorem A
to calculate this distance.Fig 7.1
• Now, sup pose two points lie on the x-axis.
Can we find the distance between them? For
instance, consider two points A(4, 0) and B(6, 0)
in Fig. 7.2. The points A and B lie on the x-axis.
• From the figure you can see that OA = 4
units and OB = 6 units.
Therefore, the distance of B from A, i.e.,
AB = OB – OA = 6 – 4 = 2 units.
So, if two points lie on the x-axis, we can easily find the distance between them.
Now, suppose we take two points lying on
the y-axis. Can you find the distance between
them. If the points C(0, 3) and D(0, 8) lie on the
B y-axis,similarly we find that CD = 8 – 3 = 5 units
(see Fig. 7.2).
• Next, can you find the distance of A from C (in Fig. 7.2)? Since OA = 4 units and
OC = 3 units, the distance of A from C, i.e., AC2 = 32 + 42 = 25 units. AC = 5units.

• Similarly, you can find the distance of B from D = BD = 10 units.
In Fig. 7.3, the points P(4, 6) and Q(6, 8) lie in the first quadrant.
How do we use Pythagoras theorem to find the distance between them?
Let us draw PR and QS perpendicular to the x-axis from P and Q respectively.
Also, draw a perpendicular from P on QS to meet QS at T.
Then the coordinates of R and S are (4, 0) and (6, 0), respectively.
So, RS = 2 units. Also, QS = 8 units and TS = PR = 6 units.

• Therefore, QT = 2 units and PT = RS = 2 units. Now, using the Pythagoras theorem,
we have PQ2 = PT2 + QT2
= 22 + 22= 8 units,
• How will we find the distance between two points in two different quadrants?
• Consider the points P(6, 4) and Q(–5, –3)(see Fig. 7.4). Draw QS perpendicular to the
x-axis. Also draw a perpendicular PT from the point P on QS (extended) to meet y-axis at the point
• Then PT = 11 units and QT = 7 units.
• Using the Pythagoras Theorem to the right triangle PTQ, we get
• PQ2 = 112 +72 == 170 units
• Note that since distance is always non-negative, we take only the positive square root.
So, the distance between the points P(x1 , y1 ) and Q(x2 , y2 ) is PQ =sqrt[(x_2 – x_1)^2 + (y_2 – y_1)^2]
• Let us now find the distance between any two points P(x 1, y1) and Q(x 2, y2). Draw PR and QS
• perpendicular to the x-axis. A perpendicular from the point P on QS is drawn to meet it at the point
T (see Fig. 7.5).
• Then, OR = x1, OS = x2. So, RS = x2 – x1 = PT.
• Also, SQ = y2, ST = PR = y1. So, QT = y2 – y1.
• Now, applying the Pythagoras theorem in Δ PTQ, we get
PQ2= PT2+ QT2 = (x_2 – x_1)^2 + (y_2 – y_1)^2
• Therefore, PQ =sqrt[(x_2 – x_1)^2 + (y_2 – y_1)^2]
Note that since distance is always non-negative, we take only the positive square root.
So, the distance between the points P(x1 , y1 ) and Q(x2 , y2 ) isPQ =sqrt[(x_2 – x_1)^2 + (y_2 – y_1)^2 ]

• Which is called the Distance formula.
Remarks :

1. In particular, the distance of a point P(x, y) from the origin O(0, 0) is given by OP2 =x2 +y2 .
2. We can also write, PQ2 = (x_2 – x_1)^2 + (y_2 – y_1)^2 .
Example 1 :
Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so,
name the type of triangle formed.
Solution :

Let us apply the distance formula to find the distances PQ, QR and PR,
where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have

• PQ2 =(x_2 – x_1)^2 + (y_2 – y_1)^2
PQ2 = (3+2)2 +(2+3)2 = 7.07 (approx.)
QR2 = (–2 – 2)2 + (–3 – 3)2
• QR2 =(– 4) 2 +(– 6) 2 = 7.21 (approx.)
PR2 = (3 – 2)2 +(2 – 3)2
PR2 = 12+(-1)2 = 1.41 (approx.)
• Since the sum of any two of these distances is greater than the third distance, therefore,
• The points P, Q and R form a triangle.
Also, PQ2 + PR2 = QR2 , by the converse of Pythagoras theorem, we have ∠ P = 90°.
• Therefore, PQR is a right triangle.

Example 2 :
Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.
Solution :

Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way
of showing that ABCD is a square is to use the property that all its sides should be
equal and both its diagonals should also be equal. Now,
AB2 = (1 – 4)2 + (7- 2)2
9 +25 = 34
BC 2 = (4+ 1)2 +(2+ 1)2
= 25 +9
=34
CD2 = (–1+ 4)2 + (–1 – 4)2
= 9+ 25
= 34
DA2 = (1+ 4)2 + (7 – 4)2
=25+ 9
=34
AC2 = (1+ 1)2 + (7 +1)2
AC2 = 4+64
=68
BD2 = (4 +4)2 + (2+ 4)2
= 64+ 4= 68
Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral
ABCD are equal and its diagonals AC and BD are also equal. Therefore, ABCD is a square.

Alternative Solution:

• We find the four sides and one diagonal, say,AC as above.
• Here
AD2 + DC2 =34 + 34 = 68 = AC2 . Therefore, by the converse of Pythagoras theorem, ∠ D = 90°. A quadrilateral with all four sides equal and one angle 90° is a square. So, ABCD is a square.

Example 3 :
Fig. 7.6 shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively.Do you think they are seated in a line? Give reasons for your answer.
Solution :
Using the distance formula, we have
AB2 = (6- 3)2 + (4- 1) 2 =9 +9= 18
AB=3√2
BC 2 = (8 – 6)2 + (6 – 4)2 = 4+ 4= 8
BC=2√2
AC2 = (8 – 3)2 + (6 – 1)2 = 25+ 25= 50
AC=5√2

• Since, AB + BC = 3√2 + 2√2 = 5√2 = AC we can say that the points A, B and C
are collinear. Therefore, they are seated in a line.

AB + BC =3√2 + 2√2 = 5√2 = AC

Example 4 : Find a relation between x and y such that the point (x , y) is equidistant
from the points (7, 1) and (3, 5).
Solution :

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).
We are given that AP = BP. So, AP 2 = BP 2
i.e., (x – 7)2 + (y – 1)2 =(x – 3) 2 + (y – 5) 2
i.e., x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25
i.e., x – y =2
which is the required relation.

Remark:
Note that the graph of the equation x – y = 2 is a line. From your earlier studies,you know that a point which is equidistant from A and B lies on the perpendicular bisector of AB. Therefore, the graph of x – y = 2 is the perpendicular bisector of AB(see Fig. 7.7).

Example 5 :
Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).
Solution :

We know that a point on the y-axis is of the form (0, y). So, let the point
P(0, y) be equidistant from A and B.
Then (6 – 0) 2 + (5 – y)2 = (– 4 – 0) 2 + (3 – y)2
i.e., 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y
i.e., 4y = 36
i.e., y =9
So, the required point is (0, 9).
Let us check our solution : AP 2 = (6 – 0)2 + (5 – 9)2 =36 +16
AP 2 = 52
AP= sqrt(52)
BP2 = (– 4 – 0)2 + (3 – 9)2 =16 +36= 52
BP=sqrt(52).

Note :
Using the remark above, we see that (0, 9) is the intersection of the y-axis and the perpendicular bisector of AB.

Section Formula

Suppose a telephone company wants to position a relay tower at P between A and B is such a way that the distance of the tower from B is twice its distance from A. If P lies on AB, it will divide AB in the ratio 1 : 2(see Fig. 7.9). If we take A as the origin O and 1 km as one unit on both the axis, the coordinates of B will be (36, 15). In order to know the position of the tower, we must know the coordinates of P. How do we find these coordinates.

• Let the coordinates of P be (x, y). Draw perpendiculars from P and B to the x-axis, meeting it in D and E, respectively. Draw PC perpendicular to BE. Then, by the AA similarity criterion, studied in Chapter 6, Δ POD and Δ BPC are similar.
• Therefore ,(OD)/(PC) = (OP)/( PB) = 1/2
• and (PD)/(BC) = (OP)/(PB) = 1/2
• So, (X)/(36-X)= 1/2 and (y)/(15-y)= 1/2 .
• These equations give x = 12 and y = 5.
You can check that P(12, 5) meets the condition that OP : PB = 1 : 2.
• Consider any two points A(x1, y1) and B(x2, y2) and assume that P (x, y) divides AB internally in the ratio m1 : m2, i.e.,
(PA)/(PB)= m_1/m_2( Fig. 7.10)
.
• Draw AR, PS and BT perpendicular to the x-axis. Draw AQ and PC parallel to
the x-axis. Then, by the AA similarity criterion,
• Δ PAQ ~ Δ BPC
Therefore, (PA)/(BP) = (AQ)/(PC) = (PQ)/(BC) ---------------(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
m_1/m_2 = ((x-x_1)/(x_2-x))=((y-y_1)/(y_2-y))
m_1/m_2 = ((x-x_1)/(x_2-x)), we get x = (m_1x_2+ m_2 x_1)/(m_1+m_2)
m_1/m_2 =((y-y_1)/(y_2-y)), we get y = (m_1y_2+ m_2 y_1)/(m_1+m_2)
• So, 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
[(m_1 x_2+ m_2 x_1)/(m_1+m_2),(m_1y_2+ m_2 y_1)/(m_1+m_2)]
This is known as the section formula.

Example 6 :
Find the coordinates of the point which divides the line segment joining the points
(4, –3) and (8, 5) in the ratio 3 : 1 internally.
Solution :

Let P(x, y) be the required point. Using the section formula, we get
x ={3(8)+ 1(4)}/{3+1}=7
y ={3(5)+1(–3)}/{3+1}=3
Therefore, (7, 3) is the required point.

Example 7 :
In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?
Solution :
Let (– 4, 6) divide AB internally in the ratio m1 : m2. Using the section formula, we get
(– 4, 6) = [(3m_1-6m_2)/(m_1+m_2),(-8m_1+10m_2)/(m_1+m_2)]
Recall that if (x, y) = (a, b) then x = a and y = b.
So, – 4 = (3m_1-6m_2)/(m_1+m_2) and 6 = (-8m_1+10m_2)/(m_1+m_2)
-4 =(3m_1-6m_2)/(m_1+m_2) gives us
-4m_1 – 4m_2 = 3m_1 – 6m_2
2m_2 = 7m_1
i.e., m1 : m2 = 2 : 7
You should verify that the ratio satisfies the y- coordinate also.
Now,  (-8m_1+10m_2)/(m_1+m_2)
 = (-8(m_1/m_2)+10)/((m_1/m_2)+1)  (Dividing throughout by m2 )
= [-8(2/7)+10]/[(2/7)+1] [here (m_1/m_2)=2/7]
=[[-16+70]/7] /[9/7]
= (54)/9
=6

• Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and
B(3, – 8) in the ratio 2 : 7.
Alternatively : The ratio m1 : m2 can also be written as (m_1)/(m_2):1 ,or k : 1. Let (– 4, 6)
divide AB internally in the ratio k : 1. Using the section formula, we get
• (– 4, 6) = ((3k- 6) /(k+18),(-8k+10)/(k+1))---------------(2)
-4=(3k-6)/(k+1)
So, – 4 = (3k - 6)/(k+1)
i.e., – 4k – 4 = 3k – 6
i.e., 7k = 2
i.e., k : 1 = 2 : 7
• You can check for the y-coordinate also.
• So, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and
B(3, – 8) in the ratio 2 : 7.

Note :
You can also find this ratio by calculating the distances PA and PB and taking their ratios provided you know that A, P and B are collinear.

Example 8 :
Find the coordinates of the points of trisection (i.e., points dividing in
three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).

Solution :
Let P and Q be the points of
trisection of AB i.e., AP = PQ = QB
(see Fig. 7.11).
• Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by
applying the section formula, are
• [(1(- 7)+ 2(2))/(1+2), (1(4)+ 2(- 2))/(1+2)], i.e(-I,0)
• Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are
[(2(- 7)+ 1(2)) /(2+1) ,(2(4) +1(- 2)) / (2+1)], i.e., (– 4, 2)
• Therefore, the coordinates of the points of trisection of the line segment joining A and
B are (–1, 0) and (– 4, 2).

Note :
We could also have obtained Q by noting that it is the mid-point of PB. So, we could have obtained its coordinates using the mid-point formula.

Example 9 :
Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.
Solution:

Let the ratio be k : 1. Then by the section formula, the coordinates of the
point which divides AB in the ratio k : 1 are  [(-k+5)/(k+1) ,(-4k-6)/(k+1)]

• This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.
• Therefore, [(-k+5)/(k+1)]=0
So, k =5
• That is, the ratio is 5 : 1.Putting the value of k = 5, we get the point of intersection as
{ 0,(13)/3.}

Example 10 : If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a
parallelogram, taken in order, find the value of p.
Solution :

We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD
i.e.,

• [(6+ 9)/2, (1+ 4)/2] =[(8+p)/2,(2+3)/2]
• [(15)/2,5/2] =[(8+p)/2,(2+3)/2]
• (15)/2 =(8+p)/2
i.e., p =7

Area of a Triangle

• The area of a triangle when its base and corresponding height (altitude) are given. You have used the formula :
Area of a triangle =1/2 × base × altitude
• you have also studied Heron’s formula to find the area of a triangle.Now, if the coordinates of the vertices of a triangle are given, Well, you could find the lengths of the three sides using the distance formula and then use Heron’s formula. But this could be tedious, particularly if the lengths of the sides are irrational numbers. Let us see if there is an easier way out.
• Let ABC be any triangle whose vertices are A(x 1, y 1), B(x2, y2) and C(x 3, y3). Draw AP, BQ and CR perpendiculars from A, B and C,respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapezia(see Fig. 7.13).
• Now, from Fig. 7.13, it is clear that area of Δ ABC = area of trapezium ABQP + area of trapezium APRC– area of trapezium BQRC.
• Area of a trapezium = 1/2 (sum of parallel sides)(distance between them)
Therefore,Area of Δ ABC =1/2 (BQ + AP) QP +1/2 (AP + CR) PR – 1/2 (BQ + CR) QR
=1/2 (y_2+y_1)(x_1-x_2)+1/2(y_1+y_2)(x_3-x_1)-1/2(y_2+y_3)(x_3-x_2)
• =1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_3)]
Thus, the area of Δ ABC is the numerical value of the expression
1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_3)]

Example 11 :
Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5).
Solution :

The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and
C (–3, –5), by using the formula above, is given by
1/2[ (6+ 5)+ (- 4) (- 5+ 1)+ (- 3) (- 1- 6) ]
= 1 (11+ 16+ 21)= 24
So, the area of the triangle is 24 square units.

Example 12 :
Find the area of a triangle formed by the points A(5, 2), B(4, 7) and
C (7, – 4).
solution

–is by= 1/2[5(7+4)+4(-4-2)+7(2-7)]
=1/2(55-24-35)=-4/2=-2

• Since area is a measure, which cannot be negative, we will take the numerical value
of – 2, i.e., 2. Therefore, the area of the triangle = 2 square units.

Example 13 :
Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4).
Solution:
The area of the triangle formed by the given points is equal to

If the area of a triangle is 0 square units, then its vertices will be collinear.
=1/2[-1.5(- 2- 4)+ 6(4- 3)+ (- 3)(3+ 2)]
= 1/2 (9+ 6-15)= 0
If the area of a triangle is 0 square units, then its vertices will be collinear.

Example 14 :
Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.
Solution :

Since the given points are collinear, the area of the triangle formed by them must be 0,
i.e.,1/2[(k+ 3)+4 (- 3- 3)+ 6(3-k )]=0

• 1/2(-4k)=0
• Therefore, k = 0
• area of Δ ABC = 1/2[ 2(0+ 3) +4 (- 3- 3)+ 6 (3- 0)]= 0

Example 15 :
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.
Solution :

By joining B to D, you will get two triangles ABD and BCD.

• Now the area of Δ ABD = 1/2[- 5(- 5- 5)+(- 4)(5- 7)+ 4 (7+ 5)]
= 1/2[ 50+8+48) = 106/2 = 53 square units
• The area of Δ BCD = 1/2[-4(- 6- 5) – 1(5+ 5) + 4( -5+ 6)]
= 1/2(44-10+4) = 19 square units
So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.

Note :
To find the area of a polygon, we divide it into triangular regions, which have no common area, and add the areas of these regions.

Example 16 : If two vertices of an quadrilateral triangle be(0, 0), (3,√3),find the third vertex.
Solution:

O(0,0) and A(3,3) be the given points and let B(x,y) be the third vertex of equilateral triangleOAD.then,

• OA=0B=AB
• OA2 =OB2 =AB2
WE have OA=(3-0)2 +(sqrt3-0)2 =12
0B2 =X2 +Y2
• AB2=(X-3)2 +(Y-sqrt3)2
• AB2 =X2 +Y2 -6X-2sqrt3Y+12
OA2 =OB2 AND OB2 =AB2
X2 +y2 =12
X2 +y2 =x2 +y2 -6x-2√3Y+12
X2 +y2 =12 and 6x+2√3Y=12
[3x+sqrt3Y=6 therefore y=(6-3x)/sqrt3]
X^2+[(6-3x)/sqrt3]^2=12
[X^2+(6-3x)^2]/3=12
3x^2+(6-3x)^2 =36
12x^2-36x=0
X=0,3

• X=0; sqrt3Y=6
Y=(6/sqrt3)=2/sqrt3
• X=3;9+sqrt3y=6
y=((6-9)/sqrt3)=-sqrt3
Hence the coordinate of the third vertex B are(0,2sqrt3) or,(3,-sqrt3).

co-ordinate geometry
• The subject co-ordinate geometry is that particular branch of mathematics in which geometry is studied with the help of algebra.
• This branch of mathematics was first introduced by the great French Philosopher and Mathematician Rene’ Descartes and by his name the subject is also called Cartesian Co-ordinate Geometry.
• In co-ordinate geometry, the concept of algebra is introduced and as a result, the fundamental properties and theorems of geometry can be easily deduced. For this reason sometimes this branch is called Analytical Geometry.
• Co-ordinate Geometry is of two types:
(i) Two Dimensional or Plane Co-ordinate Geometry
(ii) Three Dimensional or Solid Co-ordinate Geometry.
• What is Rectangular Cartesian Co-ordinates?
• Let O be a fixed point on the plane of this page; draw mutually perpendicular straight line XOX’ and YOY’ through O.
• Clearly, these lines divide the plane of the page into four parts. Each of these parts is called a Quadrant; the parts XOY, YOX’, X’OX are respectively called the first,second, third and fourth quadrant.

• The fixed point O is called the origin and the straight Lines XOX’ and YOY’ are called the co-ordinate axes;
• separately the line XOX’ is called the x-axis and the line YOY’ is called the y-axis.
• We can uniquely determine the position of any point on the plane of the page referred to co-ordinate axes drawn through O.
• Let P be any point in the first quadrant. From P draw PM perpendicular on x-axis. If OM and MP measure 4 and 5 units respectively then the position of P on the plane is determined i.e.,
• to get the point P on the plane, we are to move from O through a distance of 4 unite along OX and then to proceed through a distance of 5 units in direction parallel to OY. Note that, we shall have points Q, R and S in the second, third and fourth quadrants
respectively and the distance of each of them along x-axis and y-axis are 4 and 5 units respectively.
• Therefore, it is possible to have four different point on the plane of the page at equal distances along the co-ordinate axes.
• To differentiate among the position of such points we introduce the following convention regarding the signs of distances along the co-ordinate axes:
• the distance measured from O along x-axis on the right side (i.e., in the direction OX or in direction parallel to OX is positive and the distance from O along x-axis on the left side (i.e., in the direction OX’ or in direction parallel to OX’ is negative;
• (i) the distance measured from O along y-axis in the upward direction (i.e., in the direction OY or in direction parallel to OY) is positive and the distance from y- axis in the downward direction (i.e., in the direction OY’ or in direction parallel to OY’) is negative.
• By the above convention of sign the distances along x-axis as well as along y- axis are positive for P, for the point Q,the distance along x-axis is negative and that along x-axis is negative and that along y- axis is positive, for R both these distances are negative and for S the distance along x-axis is positive and that along y is negative.
• From the above discussion it is evident that to determine uniquely the position of a point on a plane referred to mutually perpendicular co-ordinate axes drawn through an origin O we require two signed real numbers. These two signed real numbers together are called the rectangular Cartesian co-ordinates of the given point we write the two signed real number in braces putting a comma between them where the first number is the distance from origin along x-axis and the second number is the distance from origin along y-axis (or parallel to y-axis).
• Therefore, the Cartesian co-ordinate of a point on a plane may be defined as an ordered pair of signed real numbers. Thus, the co-ordinate of the points P, Q, R and S are (4, 5), (-4, 5), (-4, -5) and (4, -5) respectively.
• In general , the statement, the co-ordinate of a point A are (a, b) means that the point A is situated at distance a units from origin O along x-axis and at distance b units from origin along (or parallel) to y- axis.
• Depending on the signs of a and b the point A may be on the first or second or third of fourth quadrant. Here, a is called the abscissa or x co-ordinate of A and b is called the ordinate or y co-ordinate of A.
• clearly, abscissa and ordinate are both positive for any point lying in the first quadrant; abscissa and ordinate is positive for any point lying in the second quadrant;
• abscissa and ordinate are both negative for any point lying in the third quadrant while the abscissa is positive and ordinate is negative for any point lying in the fourth quadrant. Conversely, if x,y are real and positive then the point
• co-ordinate (x, y) lies in the first quadrant
co-ordinate (-x, y)lies in the second quadrant,
co-ordinate (-x, -y) lies in the third quadrant
co-ordinate (x, -y) lies in the fourth quadrant
• The co-ordinate axes through the origin O are said to be oblique if they are not inclined at right angles.
The co-ordinate of a point on a plane referred to oblique axes are called oblique co-ordinate.
The present treatise deals mainly with rectangular co-ordinates.

• Note:
That the ordinate of any point on x-axis is zero, abscissa of any point on y-axis is zero and both the abscissa and ordinate of the origin O are zero. Therefore, the co-ordinate of a point on x-axis are of the form A (x, 0), the co-ordinate of a point on y-axis are of the form B (0, y) and the co-ordinate of the origin O are always (0, 0).

In which quadrant do the following points lie?
(i) (4, -6)
Solution:

For the point (4, -6) we see that the abscissa = 4, is positive and ordinate = -6, is negative.
Therefore, the point (4, -6) lies in the fourth quadrant.

(ii) (2, 3)
Solution:

For the point (2, 3) we see that the abscissa and ordinate are both positive.
Hence, the point (2, 3) lies in the first quadrant.

(iii) (-2, 1 - √3)
Solution:

Since - sqrt(3) > 1, hence (1 - sqrt(3)) is negative. Hence, the abscissa and ordinate are both negative for the point (-2, 1 - sqrt(3)). Therefore, the point (-2, 1 - sqrt(3)) lies in the third quadrant.

(iv) (sqrt(3) - 2, 5)
Solution:

Since, sqrt(3) < 2, hence (sqrt(3) - 2) is negative. Thus abscissa is negative and ordinate is positive for the point (√3 - 2, 5).
Therefore, the point (sqrt(3) - 2, 5) lies in the second quadrant.

What is Polar Co-ordinates?

• Besides Cartesian co-ordinate system we have several other methods for locating position of a point on a plane.
• Of all these system we shall make here a brief discussion on Polar Co-ordinates only.
Polar Co-ordinates are widely used in higher mathematics as well as in other branches of science.
• In polar co-ordinate system the position of a point on the reference plane is uniquely determined referred to a fixed point on the plane and a half line drawn through the fixed point.
• The fixed point is called the Pole or Origin and the half line drawn through the pole is called the Initial Line.

Relation between Cartesian and Polar Co-Ordinates

• Let XOX’ and YOY’ be a set of rectangular Cartesian axes of polar Co-ordinates through the origin O.
now, consider a polar Co-ordinates system whose pole and initial line coincide respectively with the origin O and the positive x-axis of the Cartesian system.
• Let P be any point on the plane whose Cartesian and polar Co-ordinates are (x, y) and (r, θ) respectively.
Draw PM perpendicular to OX. Then we have,
• OM = x, PM = y, OP = r and /_ MOP = theta
Now, from the right-angled triangle MOP we get,
x/r = cos theta or, x= r cos theta …… (1)
and
y/r = sin theta or, y = r sin theta …… (2)
Using (1) and (2) we can find Cartesian Co-ordinates (x, y) of the point whose polar Co-ordinates (r, theta) are given. Again, from the right angled triangle OPM we get,
r^2 = x^2 + y^2
or, r = sqrt(x^2 + y^2) …… (3)
and tan theta = y/x or, theta = tan^(-1) (y/x) ……… (4)
Using (3) and (4) we can find the polar Co-ordinates (r, theta) of the points whose Cartesian Co-ordinates (x, y) are given.
• r2 = x2 + y2
y/r = sin theta

If the Cartesian Co-ordinates (x, y) of a point are given then to find the value of the vectorial angle theta by the transformation equation theta = tan^(-1) y/x we should note the quadrant in which the point (x, y) lies.

Distance between Two Points

Here we will discuss about distance between two points.
How to find the distance between two given points?
Or,
How to find the length of the line segment joining two given points?

• Let OX and OY be the rectangular Cartesian Co-ordinates axes on the plane of reference and the Co-ordinates of a point P on the plane be (x, y). to
find the distance of P from the origin O.
from P draw PM perpendicular on OX; then , OM = x and PM = y.
Now from the right angle triangle OPM we get,
Therefore OP =sqrt[(x^2+ y^2) (Since, OP is positive.)
• (B) To find the distance between two points whose rectangular Cartesian co-ordinates are given:
• PQ = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2]
• Let (x1, y1) and (x2, y2) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY.
• We are to find the distance between the points P and Q. Draw PM and QN perpendiculars from P and Q respectively on OX; then draw PR perpendicular from P on QN.
Clearly, OM = x1, PM = y1, ON = x2 and QN = y2.
Now, PR = MN = ON – OM = x2 – x1
and QR = QN - RN = QN - PM = y2 – y1
Therefore, from the right-angled triangle PQR we get,
PQ2 = PR2 + QR2 = (x2 - x1)2 + ( y2 - y1)2
Therefore, PQ = sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2] (Since, PQ is positive )∙

• Examples on Distance between two Points

1. Find the distance of the point (-5, 12) from the origin.
Solution:

We know that, the distance between two given points (x1, y1) and (x2, y2) is
sqrt[(x_2 - x_1)^2 + (y_2 - y_1)^2].
The required distance of the point (- 5, 12) from the origin = the distance between the points (- 5, 12) and (0, 0)
= sqrt[(- 5 - 0)^2+ (12 - 0)^2]
= sqrt(25 + 144)
= sqrt(169)
= 13 units.

2. Find the distance between the points (- 2, 5) and (2, 2).
Solution:

We know that, the distance between two given points (x1, y1) and (x2, y2) is
sqrt[(x_2 - x_1)^2+ (y_2 - y_1)^2].
The required distance between the given points (- 2, 5) and (2, 2)
= sqrt[(2 + 2)^2+ (2 - 5)^2]
= sqrt(16 + 9)
= sqrt(25)
= 5 units.

(i) Internal Division of line segment:

Let (x1, y1) and (x2, y2) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and

• the point R divides the line-segment PQ internally in a given ratio m : n (say), i.e.,
PR : RQ = m : n. We are to find the co-ordinates of R.

• Let, (x, y) be the required co-ordinate of R .
From P, Q and R, draw PL, QM and RNperpendiculars on OX.
Again, draw PT parallel to OX to cut RN at S and QM at T. Then,
• Now, by construction, the triangles PRS and PQT are similar; hence,
(PS)/(PT) = (RS)/(QT) = (PR)/(PQ)
• Taking, (PS)/(PT) = (PR)/(PQ_ we get,
• (x - x_1)/(x_2 - x_1) = m/(m + n)
or, x (m + n) – x_1(m + n) = mx_2 – mx_1
or, x ( m + n) = mx_2 - mx_1 + m x_1 + nx_1 = mx_2 + nx_1
Therefore, x = (mx_2 + n x_1)/(m + n)
Again, taking (RS)/(QT) = (PR)/(PQ) we get,
(y - y_1)/(y_2 - y_1) = m/(m + n)
or, ( m + n) y - ( m + n) y_1 = my_2 – my_1
• PS = LN = ON - OL = x – x_1
PT = LM = OM – OL = x_2 - x_1
RS = RN – SN = RN – PL = y - y_1
and QT = QM – TM = QM – PL = y_2 – y_1
Again, (PR)/(RQ) = m/n
or, (RQ)/(PR) = n/m
or, (RQ)/(PR) + 1 = (n/m) + 1
or, (RQ + PR)/(PR) = (m + n)/m
or, (PQ)/(PR) = (m + n)/m
• or, ( m + n)y = my_2– my_1 + my_1 + ny_1
( m + n)y= my_2 + ny_1
Therefore, y = (my_2 + ny_1)/(m + n)
Therefore, the required co-ordinates of the point R are
((mx_2 + nx_1)/(m + n), (my _2 + ny_1)/(m + n))

(ii) External Division of line segment:

• Let (x1, y1) and (x2, y2) be the Cartesian co-ordinates of the points P and Q respectively referred to rectangular co-ordinate axes OX and OY and the point R divides the line-segment PQ externally in a given ratio m : n (say) i.e., PR : RQ = m : n. We are to find the co-ordinates of R.
• Let, (x, y) be the required co-ordinates of R. Draw PL, QM and RN perpendiculars on OX.
Again, draw PT parallel to OX to cut RN at S and QM and RN at S and T respectively, Then,
PS = LM = OM - OL = x2 – x1;
PT = LN = ON – OL = x – x1;
QT = QM – SM = QM – PL = y2 – y1
and RT = RN – TN = RN – PL = y — y1
Again, (PR)/(RQ) = m/n
or, (QR)/(PR) = n/m
or, 1 - [(QR)/(PR)] = 1 - [n/m]
or, (PR - RQ)/(PR) = (m - n)/m
or, (PQ)/(PR) = (m - n)/m
Now, by construction, the triangles
• PQS and PRT are similar; hence,
(PS)/(PT) = (QS)/(RT) = (PQ)/(PR)
Taking, (PS)/(PT) = (PQ)/(PR) we get,
(x_2 - x_1)/(x - x_1) = (m - n)/m
or, (m – n)x - x_1(m – n) = m (x_2 - x_1)
or, (m - n)x = mx_2 – mx_1 + mx_1 - nx_1 = mx_2- nx_1.
Therefore, x = (mx_2 - nx_1)/(m - n)
Again, taking (QS)/(RT) = (PQ)/(PR) we get,
(y_2 - y_1)/(y - y_1) = (m - n)/m
or, (m – n)y - (m – n)y_1 = m(y_2 - y_1)
or, (m - n)y = my_2 – my_1 + my_1 - ny_1 = my_2 - ny_1
Therefore, y = (my_2 - ny_1)/(m - n)
Therefore, the co-ordinates of the point R are
((mx_2 - nx_1)/(m - n), (my_2 - ny_1)/(m - n))
• Rene Descartes, the great French mathematician of the seventeenth century, liked to lie in bed and think! One
day, when resting in bed, he solved the problem of describing the position of a point in a plane.
• His method was a development of the older idea of latitude and longitude.
• In honour of Descartes, the system used for describing the position of a point in a plane is also known as the Cartesian system.
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