Coordinate Geometry
Coordinate Geometry is
the unification of algebra and geometry in which algebra is used in the study
of geometrical relations and geometrical figures are represented by means of
equations. The most popular coordinate system is the rectangular cartesian system.
Coordinates of a point are the real variables associated in an order to
describe its location in space. Here we consider the space to be
twodimensional. Through a point O, referred to as the origin, we take two
mutually perpendicular lines X O X¢ and Y O Y¢
and call them x and y axes respectively. The position of a point is completely
determined with reference to these axes by means of an ordered pair of real
numbers (x, y) called the coordinates of P where x and y are the distances
of the point P from the yaxis and the x  axis respectively. x is called the
xcoordinate or the abscissa of P and y is called the ycoordinate or the
ordinate of P.
Distance between two
points:
Let A and B be two
given points, whose coordinates are given by A(x_{1}, y_{1})
and B(x_{2}, y_{2}) respectively. Then.
AB = .
Section formula:
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 l_{1} : l_{2} are P.
Coordinates of the point P dividing
the join of two points A(x_{1}, y_{1}) and B(x_{2}, y_{2})
externally in the ratio of l_{1} : l_{2} are .
In both the cases, l_{1}/l_{2} is positive.
Notes:
(i) If the ratio, in which a given
line segment is divided, is to be determined, then sometimes, for convenience
(instead of taking the ratio l_{1} : l_{2}), we take the ratio k : 1. If the value of k turns out
to be positive, it is an internal division otherwise it is an external
division.
(ii) The coordinates of the
midpoint of the linesegment joining (x_{1}, y_{1}) and
(x_{2}, y_{2}) are .
Definitions:
·
Centroid: The
point of concurrency of the medians of a triangle is called the centroid of the
triangle. The centroid of a triangle divides each median in the
ratio 2 :1. The coordinates are given by G = .
·
Orthocentre: The
point of concurrency of the altitudes of a triangle is called the orthocentre
of the triangle. The coordinates of the orthocentre of the triangle
A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3},y_{3})
are .
.
·
Incentre : The
point of concurrency of the internal bisectors of the angles of a triangle
is called the incentre of the triangle. The coordinates of the incentre are
given by .
I = where a, b
and c are length of the sides BC, CA and AB respectively
·
Circumcentre :
The point of concurrency of the perpendicular bisectors of the sides of a
triangle is called circumcentre of the triangle. The coordinates of the
circumcentre of the triangle with vertices A(x_{1}, y_{1}), B(x_{2},
y_{2}), C(x_{3}, y_{3}) are .
.
Note:
Centriod G, orthocentre H and
Circumcentre P of a DABC are
collinear and G Divides HP in the
ratio 2 :1. .
i.e.
HG:GP =2:1 .
Also, AH
= 2PD


Example 1
If O be the origin
and the coordinates of any two points Q_{1} and Q_{2} be (x_{1},
y_{1}) and (x_{2}, y_{2}) respectively, then prove that
OQ_{1}. OQ_{2} . cos∠Q_{1}OQ_{2} = x_{1}x_{2}
+ y_{1}y_{2}.
Solution:
Here OQ_{1}^{2}
= x_{1}^{2}+y_{1}^{2} and OQ_{2}^{2}
= x_{2}^{2}+y_{2}^{2} Using
cosine formula in DOQ_{1}Q_{2}, we
have
Q_{1}Q_{2}^{2} = OQ_{1}^{2}+OQ_{2}^{2}2OQ_{1}.OQ_{2}
. cos∠Q_{1}OQ_{2}.
&⇒ (x_{2}x_{1})^{2}+(y_{2}y_{1})^{2}
= x_{1}^{2}+y_{1}^{2} + x_{2}^{2}+y_{2}^{2
}2OQ_{1}.OQ_{2} . cos∠Q_{1}OQ_{2}.
&⇒
OQ_{1}.OQ_{2} . cos∠Q_{1}OQ_{2} = x_{2}x_{1}
+ y_{2}y_{1}.


Example 2
Prove that the
centroid of the triangle whose vertices are given by
A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3},
y_{3}) respectively is
Solution:
Since D is the midpoint of BC,
coordinates of D are
Using
the section formula, the


coordinates of G are
&⇒
Coordinates of G are .
Example 3
Prove that the incentre of the
triangle whose vertices are given by A(x_{1}, y_{1}), B(x_{2},
y_{2}), C(x_{3}, y_{3}) is where a, b,
and c are the sides opposite to the angles A, B and C respectively.
Solution:
By
geometry, we know that (since
AD bisects ∠A).
If
the lengths of the sides AB, BC and AC are c, a and b respectively, then &⇒
Coordinates of D are .


Since
IB bisects ∠B.
Hence
Let the coordinates of
I be .
Then
(using section
formula).
Area of a triangle:
Let (x_{1}, y_{1}),
(x_{2}, y_{2}) and (x_{3}, y_{3}) respectively
be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of
triangle ABC, is.
…. (1)
= ….(2)
While using formula
(1) or (2), order of the points (x_{1}, y_{1}), (x_{2},
y_{2}) and (x_{3}, y_{3}) has not been taken into
account. If we plot the points A(x_{1}, y_{1}),
B(x_{2}, y_{2}) and C (x_{3}, y_{3}) , then
the area of the triangle as obtained by using formula (1) or (2) will be
positive or negative as the points A, B, C are in anticlockwise or
clockwise directions,.











So, while finding the area of
triangle ABC, we use the formula:
Area of DABC
=
= Modulus of
Note:
(i) If three points (x_{1}, y_{1}),
(x_{2}, y_{2}) and (x_{3}, y_{3}) are
collinear, then =
0.
(ii) Equation of straight line
passing through (x_{1}, y_{1}) and (x_{2}, y_{2})
is given by
= 0.
(iii) In case of polygon with (x_{1},
y_{1}), (x_{2}, y_{2}), . . . . (x_{n}, y_{n})
the area is given by
Example 4
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 and explain the result . .
Solution:
The area of triangle is
(R_{1}→
R_{1} – R_{3}, R_{2} → R_{2}
–R_{3})
=.Which remains
constant for all values of p.
Explanation:
As p varies, the same triangle shifts its position in space. So area
remains the same.