René Descartes introduced a coordinate system that represented geometric figures using algebra. This allowed geometry and algebra to be linked by applying the methods of one field to problems in the other. The document then discusses several ways that geometry is used in daily life and provides formulas to calculate distances and areas using coordinate systems. It also defines key geometric concepts like slope, parallel and perpendicular lines, and the angle between two lines in terms of their slopes. Overall, the document outlines Descartes' contributions to representing geometry algebraically and defines important geometric relationships using coordinates.

3. René Descartes ,
Descartes introduced a
method of representing
geometric figures
within a coordinate
system. His work
forged a link between
geometry and algebra
by showing how to
apply the methods of
one discipline to the
other.
Rene Descartes (1596-
1650) also known as
father of modern
geometry

4. Introduction
Geometry, branch of mathematics that
deals with shapes and sizes.
Basic geometry allows us to determine
properties such as the areas and
perimeters of two-dimensional shapes
and the surface areas and volumes of
three-dimensional shapes.

5. Geometry used in daily
Life
People use formulas derived from geometry
in everyday life for tasks such as figuring
how much paint they will need to cover the
walls of a house or calculating the amount of
water a fish tank holds.

6. By the mean of coordinate system we are now
able to find the dimensions between the two
points :
 We can find the distance between the two
points whose coordinates are given.
 we can find the coordinate of the point which
divides a line segment joining two point in a
given ratio.
 And to find the area of the triangle formed by
three given points.

7. 1. distance between two point can be found by the
formula called Distance Formula :
distance between the point P(X1 , y1) and
Q (x2 , y2) is
PQ=  (X2 –X1)² + (Y2 –Y1)²

8. this can be found out by Section Formula :
the coordinate of a point dividing the line segment
joining the point (X1 , Y1) and (X2, Y2) internally
in the ratio m : n are :
x-axis – m1x2+m2x1 y-axis – m1y2+m2y1
m1 + m2 m1 + m2
2. we can find the coordinate of the point which
divides a line segment joining two point in a given
ratio.

9. Area of triangle whose vertices are (x1, y1) ,
(x2, y2) and (x3, y2) is :
½ {x1(y2 –y3)+ x2(y3 –y1) +x3(y1 –y2)}
3. the area of the triangle formed by three given points

10. Straight line
 A straight line is a curve
such that every point on the
line segment joining any two
points on it lies on it.

11.  Despite its simplicity, the line is a
vital concept of geometry and enters
into our daily experience in
numerous interesting and useful
ways. Main focus is on representing
the line algebraically, for which
slope is most essential.

12. Slope of a line
A line in a coordinate plane forms two angles with
the x-axis, which are supplementary. The angle
(say)  made by line l with positive direction of
x-axis and measured anticlockwise is called the
inclination of the line. Obviously 0    180

180 - 
y
x
l
o

13. definition 1:-
If  is the inclination of line l, then tan  is
called the slope or gradient of the line l.
The slope of a line whose inclination is 90 is not
defined
The slope of a line is denoted by m
thus, m = tan ,   90
It may be observed that the slope of x-axis is zero
and the slope of y-axis is not defined.

14. illustration 1:
Find the slope of line whose inclination to the (+) ve
direction of x-axis in anticlockwise sense is
(1). 60
(2). 150
Sol:- (1) .slope = tan 60 =3
(2). slope = tan 150 = -cot 60 = -1
3

15. Slope of a line – coordinates are given
we know that a line is completely determined
when we are given two points on it. Hence, we
proceed to find the slope of a line in terms of
the coordinates of two points on the line.
The inclination of the line l may be acute or
obtuse . Lets us take each cases

16. Let P(x1, y1) and Q(x2, y2) be two point on non
vertical line l whose inclination is .
Draw perpendicular QR to x-axis and PM
perpendicular to RQ as shown in fig,2.
P(x1, y1)
Q(x2, y2)
R
y
O X
M
l


Fig. 2.

17. CASE 1 :-
when angle  is acute
in fig,2. MPQ = 
therefore, slope of line l = m = tan 
but in MPQ, we have tan
= MQ = y2 – y1
MP x2 – x1
so, m = y2 – y1
x2 – x1
)(

18. CASE 2 :-
when  is obtuse
in the fig,3. we have
MPQ = 180 - 
therefore,  = 180 - MPQ
Y
P (x1 , y1)
R
X
O
180 - 
M
Q (x2 , y2)
l

Fig.3

19. now, slope of line l
m = tan
= tan( 180 - MPQ) = -tan MPQ
= - MQ = - y2 – y1 = y2 – y1
MP x1 – x2 x2 – x1

20. consequently, we see that in both the case
the slope m of the line trough the point
(x1 , y1) and (x2 , y2) is given by
m = y2 – y1
x2 – x1

21. illustration 2 :
find the slope of a line which passes through point
(3, 2) and (-1, 5).
Sol :- we known that the slope of a line passing
through two point (x1, y1) and (x2, y2) is given by
m = y2 – y1
x2- x1
here the line is passing through the point (3 ,2) and
(-1 , 5). So, Its slope is given by
m = 5 – 2 = - 3
-1 – 3 4

22. Condition for parallelism and
perpendicularity of lines in terms of slopes
1. condition for parallelism of two
lines
if two lines m1 and m2 are perpendicular, then
the angle  between them is 90
In the coordinate plane , suppose that non –
vertical line l1 and l2 have slope m1 and m2
respectively. Let their inclination be  and
, respectively.

23. if the line l1 is parallel to line l2, from the
fig4,. Then their inclination are equal ,i.e.,
 = ,
hence, tan  = tan
Therefore, m1 = m2, i.e., their slopes are equal
l1
l2
 
O
Y
X
Fig.,4

24. conversely, if the slope of two line l1 and l2 is
same, i.e.,
m1 = m2
then, tan  = tan .
By the property of tangent function (between
0 and 180 ),  = 
therefore, the line are parallel,
Hence, two non vertical lines l1 and l2 are
parallel if and only if there slopes are equal.

25. 2. condition for perpendicularity of two
lines
if two lines of slope m1 and m2 are perpendicular,
then the angle  between them is of 90
Y
X
l1
l2
 
O
Fig.,5

26. if the line l1 and l2 are perpendicular from the fig
.5, then  =  + 90
Therefore, tan  = tan( + 90)
=- cot = - 1
tan 
i.e., m2 = - 1 or m1 m2 = - 1
m1

27. Conversely, if m1 m2 = -1, i.e., tan tan = -1
Then tan = - cot = tan ( + 90) or tan (  - 90)
therefore,  and  differ by 90.
Thus, line l1 and l2 are perpendicular to each other.
hence, two non–vertical lines are perpendicular to
each other if and only if their slope are negative
reciprocal of each other,
m2 = -1 or, m1 m2 = -1
m1

28. illustration 3 :
find the slope of line:
1. passing through (3, -2 )and (-1, 4)
2. passing through (3, -2) and (7, -2)
Sol :- 1. slope of line through (3, -2) and (-1, 4)
m = 4 –(-2) = 6 = 3
-1 -3 -4 2
2. The slope of line through (3, -2) and (7, -2)
m = -2 – (- 2) = 0 = 0
7 – 3 4


29. angle between two lines
When we think about more then one line in a
plane then we find that these lines are either
parallel or intersecting. Here we will discus
the angle between two line in terms of there
slopes.

30. let l1 and l2 be two vertical lines with slope m1 and
m2, respectively . If  and  are the inclination of
lines l1 and l2, respectively. Then
m1 = tan  and m2 = tan 
Y
X
l1
l2

 
Fig.,6


31. We know that two lines intersect each other, they
make two pairs of vertically opposite angle such
that sum of any two adjacent angle is 180. let
and  be  the adjacent angle between the line l1
and l2 (fig., 4) then
 =  -  and  ,   90

32. Therefore, tan  = tan( - ) = tan - tan = m2 – m1
1 + tan tan 1+ m1 m2 1+ m1 m2
And,  = tan (180 - ) = - tan = - m2 – m1
1+m1 m2
Now, there arises two cases:
1. when , m2 – m1 ( is positive)
1+ m1 m2
2. When , m2 – m1 ( is negative)
1+ m1 m2

33. Case 1 :
if, m2 – m1 ( is positive )
1+ m1 m2
Here tan  will be positive and
 will be negative
which means  will be acute and
 will be obtuse

34. Case 2:
if, m2 - m1 (is negative)
1+ m1 m2
here tan will be negative and
 will be positive
which means that  will be obtuse and
 will be acute

35. thus ,acute angle  between line l1 and l2 which slope
m1 and m2 respectively, is given by
tan  = m2 – m1
1+ m1 m2
the obtuse angle  can be found out by
 = ( 180 -  )

36. illustration 4 :
If A (-2, 1) , B (2, 3) and C (-2, -4) are three points, find
the angle between BA and BC
Sol :- let m1 and m2 be the slope of BA and BC
respectively. then,
m1 = 3 – 1 = 2 = 1 m2 = -4 -3 = 7
2- (- 2) 4 2 -2 -2 4
7 - 1
Tan  = m2 – m1 = 4 2 = 10/8 = ± 2/3
1+ m1 m2 1+ 7  1 15/8
4 2
 = tan ( 2/3 )-1

38. By :- Sahil Puri
Class :- XI ‘B’
Roll no :- 21
School :- Kendriya Vidyalaya NO.2
Ambala Cantt