# Class 10 – Co-ordinate Geometry

Take practice tests in Co-ordinate Geometry

## Online Tests

Topic Sub Topic Online Practice Test
Co-ordinate Geometry
• Distance formula
• Section formula
Take Test See More Questions
Co-ordinate Geometry
• Area of triangle and quadrilateral
• Equation of various lines
• Slope, parallel and perpendicular lines
Take Test See More Questions

## Study Material

 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, mathematician’s 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.   Co-ordinate 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 x-axis and the vertical line as the y-axis and both together are called coordinate axes. Cartesian Co-ordinates of a point:Let A be any point in the plane. Draw perpendiculars AP and AQ on x and y-axis respectively. The length of the directed line segment OP in the units of scale chosen is called the x-coordinate (or) abscissa of point A. Similarly, the length of the directed line segment OQ on the same scale is called the y-coordinate (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  co-ordinate (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 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 somewhere 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

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 mid-point of AB is .

Example:Find the mid-point 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)

Mid-point of

Mid-point of.

ii) P (-2, -5)   Q (0, -4)

Mid-point of

Mid-point of

Let A, B, C, D are four non-collinear 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)               Mid-point of and origin is .

ii)             Mid-point of  and  is .

iii)            Mid-point of  and  is .

iv)            Mid-point of  and  is .