10 Math Coordinate Geometry

Finish Line & Beyond COORDINATE GEOMETRY x-coordinate, or abscissa: The distance of a point from the y-axis is called its xcoordinate or abscissa. y-coordinate, or ordinate: The distance of a point from the x-axis is called its ycoordinate, or ordinate. The coordinates of a point on the x-axis are of the form (x, 0), and of a point on the y-axis are of the form (0, y). Distance Formula The distance between the points P(x1, y1) and Q(x2, y2) is PQ =

(x2 - x1)² + (y2 - y1)²

In fact this formula is derived from Pythagoras theorem, which is h² =

p² + b²

Y

C

X A

B

In the given figure suppose there are three towns. The distance between town A and Town B is 20 Kms. The distance between town B and town C is 15 Kms. Given these data the distance between town A and town C can be calculated using Pythagoras formula. Now the same information is given in terms of x and y coordinates as follows: x1=0 and x2=20 y1=0 and y2=15 Hence Distance AC = =

(x2 - x1)² + (y2 - y1)² (20 - 0)² + (15 - 0)²

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond 400 + 225 = 625 =

= 25 Example: Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed. Solution: Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have PQ =

(3 + 2)² + (2 + 3)² = 50 =7.07 (approx.)

QR =

(-2 - 2)² + (-3 - 3)² = 52 = 7.21 (approx.)

PR =

(3 - 2)² + (2 - 3)² =

2 =1.41 (approx.)

Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. Now Since

52 = 50 + 2

So this is a Right Angled Triangle. Example: Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively. Do you think they are seated in a line? Give reasons for your answer. Solution: Using the distance formula, we have AB =

(6 - 3)² + (4 - 1)² =3 2

BC =

(8 - 6)² + (6 - 4)² =2 2

AC =

(8 - 3)² + (6 - 1)² =5 2

Since, AB + BC = 3 2 +2 2 =5 2 =AC we can say that the points A, B and C are collinear. Therefore, they are seated in a line. Section Formula The coordinates of the point P(x, y) which divides the line segment joining the points A(x1, y1) and B(x2, y2), internally, in the ratio m1 : m2 are

 m1x2 + m2x1 m1y2 + m2y1  ,   m1 + m2   m1 + m2 Special Case: The mid-point of a line segment divides the line segment in the ratio 1 : 1. Therefore, the coordinates of the mid-point P of the join of the points A(x1, y1) and B(x2, y2) is

 1.x 2 + 1.x1 1. y 2 + 1. y1   x1 + x 2 y1 + y 2  , ,   =   1+ 1  2 2   1+ 1  Example: Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. Solution:

www.excellup.com ©2009 send your queries to [email protected]

Finish Line & Beyond Using the formula we get

 24 + 4 15 − 3  ,   4   4

Or, P(x, y) = (7, 3) Area of a Triangle Area=

1 [ x1( y 2 − y3) + x2( y3 − y1) + x3( y1 − y 2)] 2

Example: Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). Solution:

1 [1(6 + 5) + − 4(− 5 + 1) + − 3(− 1 − 6)] 2 1 = (11+16+21) =24 2

www.excellup.com ©2009 send your queries to [email protected]