Class 9 – Co-Ordinate Geometry

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Co-Ordinate Geometry
  • Distance formula
  • Section formula
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Co-Ordinate Geometry
  • Slope and intersection of two lines
  • Equation of a line in various forms
  • Area of a triangle
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Introduction: Analytical geometry was invented by the French philosopher Rene Descartes (1596-1650) 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.

Co-ordinate geometry: The study of that branch of Mathematics which deals with the interrelationship between geometrical and algebraic concepts is called Co-ordinate geometry (or) Cartesian geometry in honour of the French mathematician Rene Descartes.

http://images.clipartpanda.com/bloc-clipart-note-md.pngCo-ordinate 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.

Co-ordinate 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 x-axis and the vertical as the y-axis.

Cartesian Co-ordinates of a point: Let and be the co-ordinate’s axes, and let P be any point in the plane. Draw perpendiculars PM and PN on x and y-axis respectively. The length of the directed line segment OM in the units of scale chosen is called the x-coordinate (or) abscissa of point P. Similarly, the length of the directed line segment ON on the same scale is called the y-coordinate (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 co-ordinate axes is represented by the ordered pair .The ordered pair is called the coordinates of point P.

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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 y-axis is called the abscissa of the point.

Ordinate: The perpendicular distance of any pointfrom the x-axis is called the ordinate of the point.

Co-ordinates: The abscissa and ordinate of a pointare together called its co-ordinates.

Example: The co-ordinates 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 co-ordinates 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 x-axis are of the form (x, 0) and the coordinates of any point on y-axis are of the form (0, y). Thus, if the abscissa of a point is zero, it would lie some where on the y-axis and if its ordinate is zero it would lie on x-axis.

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 x-axis 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 non-collinear 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 co-ordinates respectively and (x, y) is there quired coordinates of M, which is the mid-point 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 mid-point of the join of P and Q is

Example: Find the mid-point 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)

Mid-point of

Mid-point of.

ii) P (-5,-3) Q (3, -4)

Mid-point of

Mid-point of

Important points:

Let A, B, C, D are four non-collinear 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.

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Find the mid-points of i) A (5, 0) , B (9, 0) ii) P(0, -3) , Q (0, 5)

Solution:

i) A (5, 0)     B(9, 0)

Mid-point of

Mid-point of

ii) P(0, -3)     Q(0, 5)

Mid-point of

Mid-point of