Class 9 – CoOrdinate Geometry
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CoOrdinate Geometry 

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Introduction: Analytical geometry was invented by the French philosopher Rene Descartes (15961650) and the year of invention is generally set as 1637, when he published his La Geometrica, which provided a new tool for unifying the two branches of mathematics, algebra and geometry. Prior to this, mathematicians confined themselves to Euclidean geometry and did not know how to apply algebra advantageously to the study of geometrical relationships. Coordinate geometry: The study of that branch of Mathematics which deals with the interrelationship between geometrical and algebraic concepts is called Coordinate geometry (or) Cartesian geometry in honour of the French mathematician Rene Descartes. Coordinate geometry is that branch of geometry in which two numbers, called coordinates. These are used to indicate the position of a point in a plane and which makes use of algebraic methods in the study of geometric figures. 
Coordinate axes: In graphs, two number lines at right angles to each other are used as reference lines. These lines are called axes. The direction to the right of the vertical axis is denoted by a positive sign, and to the left by a negative sign, while direction upward from the horizontal axis is positive and downward is negative. The horizontal line is usually termed as the xaxis and the vertical as the yaxis. Cartesian Coordinates of a point: Let and be the coordinate’s axes, and let P be any point in the plane. Draw perpendiculars PM and PN on x and yaxis respectively. The length of the directed line segment OM in the units of scale chosen is called the xcoordinate (or) abscissa of point P. Similarly, the length of the directed line segment ON on the same scale is called the ycoordinate (or) ordinate of point P. 

Let OM = x and ON = y. Then the position of the point P is the plane with respect to the coordinate axes is represented by the ordered pair .The ordered pair is called the coordinates of point P.
i) is called x – axis.
is positive x  axis and is negative x – axis.
ii) is called y – axis
is positive y – axis and is negative y – axis
Abscissa: The perpendicular distance of any pointfrom the yaxis is called the abscissa of the point.
Ordinate: The perpendicular distance of any pointfrom the xaxis is called the ordinate of the point.
Coordinates: The abscissa and ordinate of a pointare together called its coordinates.
Example: The coordinates of the point A in the figure above are (4, 5), where 4 is abscissa and 5 is ordinate and it is written as A (4, 5). Similarly, the other points are
B (3, 2), C (4, 7), D (3, 4).
Quadrants: The two axes namely x and y axes divide the plane into four equal parts called quadrants, numbered I, II, III and IV as shown in the figure.
Origin: The intersection of the two axes is called the origin. It is usually denoted by O. The coordinates of the originare (0, 0).
I Quadrant (Q1): x >0, y > 0 II Quadrant (Q2): x <0, y > 0
III Quadrant (Q3): x <0, y < 0 IV Quadrant (Q4): x >0, y < 0
The coordinates of the origin are taken as (0, 0). The coordinates of any point on xaxis are of the form (x, 0) and the coordinates of any point on yaxis are of the form (0, y). Thus, if the abscissa of a point is zero, it would lie some where on the yaxis and if its ordinate is zero it would lie on xaxis.
It follows from the above discussion that by simply looking at the coordinates of a point we can tell in which quadrant it would lie.
Example:.
Distance formula: The distance d between the points and is given by the formula
Proof: Let and be the two points. From P, Q draw PL,QM perpendiculars on the xaxis and PR perpendicular on MQ.
Then,
units
Distance from origin: Let O (0, 0) are two points then
Collinearity: If A, B, C are three points, then
i) are collinear.
ii) are collinear.
iii) are collinear.
iv) are collinear.
Example: Show that the points A(1,1), B(2,7)and C(3,3) are collinear .
AB =
BC =
AC =
Clearly, BC = AB + AC
Hence C, A, B are collinear.
Important points:
a) Let A, B, C are three noncollinear points:
i) ABC forms a scalene triangle if any two sides are not equal .
ii) ABC forms an isosceles triangle if any two sides are equal.
iii) ABC forms an equilateral triangle i fall the three sides are equal .
iv) ABC forms a right angle triangle if square of longest side is equal to sum of squares of other two sides.
v) ABC forms an acute angled triangle if (or) (or) .
vi) ABC forms an obtuse angled triangle if (or) (or).
vii) ABC forms an isosceles right angled triangle if any two sides are equal and square of the unequal side = sum of the squares of equal sides.
Example: Find the distance between the following pairs of points:
i) A(14, 3) andB(10, 6) ii) M(1, 2) and N(0, 6)
Solution:
i) Distance between two points units
ii) Here
Midpoint formula: Let P, Q be the points with coordinates respectively and (x, y) is there quired coordinates of M, which is the midpoint of PQ. Draw PD and QF and ME perpendiculars to OX. Through M, draw NML parallel to OX and meet DP producedat N and QF at L.
Then, from congruent triangles PMN and QML, we get
NM = ML DE= EF
OE– OD = OF – OE
Again, from the same congruent triangles, we get
PN = LQ DN DP = FQ – FL
EM EM
Hence, the coordinates of the midpoint of the join of P and Q is
Example: Find the midpoint of the following:
i) A (2, 3) ,B (3, 5) ii) P (5, 3) , Q (3, 4)
Solution:
i) A (2, 3) B (3, 5)
Midpoint of
Midpoint of.
ii) P (5,3) Q (3, 4)
Midpoint of
Midpoint of
Important points:
Let A, B, C, D are four noncollinear points taken in order:
i) ABCD forms a Parallelogram, if AB = CD;AD = BC and AC ≠ BD. ( Opposite sides are equal and diagonal sare not equal).
ii) ABCD forms a Rectangle, if AB = CD; AD =BC and AC = BD. ( Opposite sides are equal and diagonal sare equal).
iii) ABCD forms a Rhombus, if AB = BC = CD =AD and AC ≠ BD.
( All sides are equal and diagonals are not equal).
iv) To prove ABCD forms a Square, AB = BC =CD = AD and AC = BD.
( All sides are equal and diagonals are equal).
1. Show that the pointsP (–1, –1), Q (2, 3) and R (–2, 6) are the vertices of a right – angled triangle.
Solution:
2. Show that the points A (1, 2), B (4, 5) and C (1, 0) lie on a straight line.
Solution: Here,
i.e.BA + AC = BC
Hence, B, A, C lie on a straight line. In other words B, A, C are collinear.
3. Prove that the points (2a, 4a),(2a, 6a) and are the vertices of an equilateral triangle whose side is 2a.
Solution:Let the points be A(2a,4a), B(2a, 6a) and C
4. Show that four points (0,1), (6, 7), (2, 3) and (8, 3) are the vertices of a rectangle.
Solution: Let A (0,1), B (6, 7),C (2, 3) and D (8, 3) be the given points.
units
BC units
units
AC = units
BD =units
By observation AB = CD, BC = AD and AC = BD (Opposite sides and diagonals are equal).
ABCDis a rectangle.
Find the midpoints of i) A (5, 0) , B (9, 0) ii) P(0, 3) , Q (0, 5)
Solution:
i) A (5, 0) B(9, 0)
Midpoint of
Midpoint of
ii) P(0, 3) Q(0, 5)
Midpoint of
Midpoint of