Class 10 – 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, mathematician’s 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 axes: In graph, two number lines at right angles to each other are used as reference lines. These lines are called axes. The horizontal line is called as the xaxis and the vertical line as the yaxis and both together are called coordinate axes.

Cartesian Coordinates of a point:Let A be any point in the plane. Draw perpendiculars AP and AQ on x and yaxis respectively. The length of the directed line segment OP in the units of scale chosen is called the xcoordinate (or) abscissa of point A. Similarly, the length of the directed line segment OQ on the same scale is called the ycoordinate (or) ordinate of point A.
Let OP = x and OQ = y. Then the position of the point A is the plane with respect to the coordinate axes is represented by the ordered pair . The ordered pair is called the coordinates of point A.

iii) In , x is called coordinate (or) abscissa. is called y coordinate (or) ordinate.
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 origin are (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 somewhere 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
If A, B, C are three points, such that
i) AB + BC = AC then A, B, C are collinear.
ii) If then ABC is an Isosceles triangle.
iii) If AB = BC = AC, then ABC is an equilateral ∆le.
iv) If , then ABC is right angled ∆le.
v) If AB = BC and then ABC is Isosceles right angled ∆le.
Example:Find the distance between the following pairs of points:
i) A(4, 3) and B(0, 6) ii) M(1, 5) and N(9, 6)
Solution:
i) Distance between two points units
ii) Here
i) The distance between and origin is units.
ii) If are two points then the midpoint of AB is .
Example:Find the midpoint of the following.
i) A (5, 2) B (3, 8) ii) P (2, 5) Q (0, 4)
Solution:
i) A (5, 2) B (3, 8)
Midpoint of
Midpoint of.
ii) P (2, 5) Q (0, 4)
Midpoint of
Midpoint of
Let A, B, C, D are four noncollinear points such that:
i) If AB = CD; AD = BC and AC ≠ BD, then ABCD is a Parallelogram.
ii) If AB = CD; AD = BC and AC = BD, then ABCD is a Rectangle.
iii) If AB = BC = CD = AD and AC ≠ BD, then ABCD is a Rhombus.
iv) If AB = BC = CD = AD and AC = BD, then ABCD is a Square.
Example:Verify if the points A (2, 5), B (5, –2) and C (7, –3) are the vertices of a right – angled triangle.
Solution:
Example: Show that the points P (2, 1), Q (5, 4) and R (0, –1) lie on a straight line.
Solution: Here,
i.e., QP + PR = QR
Hence, Q, P, R lie on a straight line. In other words Q, P, R are collinear.
Example: Prove that the points (2a, 4a), (2a, 6a) and are the vertices of an equilateral triangle.
Solution: Let the points be A(2a, 4a), B(2a, 6a) and C
Example:Show that the points are the vertices of a square.
Solution: Let 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 digonals should also be equal. Now.
So sides are
Fourth side DA is also equal and diagonals are equal.
Hence the given points belong to a square.
i) Midpoint of and origin is .
ii) Midpoint of and is .
iii) Midpoint of and is .
iv) Midpoint of and is .