Many shapes in a plane are well represented by equations. Circles, parabolas and hyperbolas are some of the shapes which can be represented by equations.
But to deal with them we need to know the basics of geometry. The posts related to basics of coordinate geometry is given in the related posts "Preliminaries in plane geometry". The posts will be given in many parts. each part will cover two to three topics. In this post we will discuss about Distance formula and Section formula.
Distance formula
The Cartesian coordinates are used to represent points in a plane. Every point in a place has a one to one relationship to a coordinate point. The distance (d) between two points A(x1, y1) and B(x2, y2) is given by the formula
d = √[(x2 − x1)²+(y2 − y1)²]
The above formula can be found with the help of Pythagoras theorem. Draw a right angled triangle as shown in the graph. When we draw a right angled triangle we draw it such that the two edges including the right angle are parallel to the two axis. This helps us to find the coordinate of the vertex. Name the right angled vertex as C. The coordinate of the point C is (x2, y1). a line parallel to the coordinate axis has its other coordinate same. Suppose the line is parallel to x-axis then all the points on the line has its y-coordinate equal. The distance between the points A and B is d. The distance between the points A and C is AC =(x2 − x1) and distance between the points B and C is BC =(y2 − y1). By Pythagoras Theorem we have (AB)² = (AC)² + (BC)². Substituting the value of AC and BC we get
d² = (x2 − x1)²+(y2 − y1)² Taking square root and as distance cannot be negative, we get
d = √[(x2 − x1)²+(y2 − y1)²]
Section Formula
A point C(x,y) is present between A(x1,y1) and B(x2,y2). C divides AB internally in the ratio m:n. Then the coordinate of C is given by
C≡( [mx2+nx1)/(m+n)] , [(my2+ny1)/(m+n)] )
C≡( [mx2− nx1)/(m − n)] , [(my2− ny1)/(m − n)] )
But to deal with them we need to know the basics of geometry. The posts related to basics of coordinate geometry is given in the related posts "Preliminaries in plane geometry". The posts will be given in many parts. each part will cover two to three topics. In this post we will discuss about Distance formula and Section formula.
Distance formula
The Cartesian coordinates are used to represent points in a plane. Every point in a place has a one to one relationship to a coordinate point. The distance (d) between two points A(x1, y1) and B(x2, y2) is given by the formula
The above formula can be found with the help of Pythagoras theorem. Draw a right angled triangle as shown in the graph. When we draw a right angled triangle we draw it such that the two edges including the right angle are parallel to the two axis. This helps us to find the coordinate of the vertex. Name the right angled vertex as C. The coordinate of the point C is (x2, y1). a line parallel to the coordinate axis has its other coordinate same. Suppose the line is parallel to x-axis then all the points on the line has its y-coordinate equal. The distance between the points A and B is d. The distance between the points A and C is AC =(x2 − x1) and distance between the points B and C is BC =(y2 − y1). By Pythagoras Theorem we have (AB)² = (AC)² + (BC)². Substituting the value of AC and BC we get
Example:
Distance between A(3,5) and B(1,2) is
AB = d
d = √[(1 − 3)²+(2 − 5)²]
= √(4+9)
= √13
Distance between A(3,5) and B(1,2) is
AB = d
= √(4+9)
= √13
Section Formula
A point C(x,y) is present between A(x1,y1) and B(x2,y2). C divides AB internally in the ratio m:n. Then the coordinate of C is given by
Example:
C divides AB internally in the ratio 1:2. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=1, n=2, x1= 3, x2 = 1, y1 = 5, y2 = 2.
x = (1×1 + 2×3)/(2+1)
= 7/3
y= (1×2 + 2×5)/(2+1)
= 12/3
= 4. Hence the coordinate of C is (7/3,4).
If C divides AB externally in the ratio m:n. Then the coordinate of C is given byC divides AB internally in the ratio 1:2. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=1, n=2, x1= 3, x2 = 1, y1 = 5, y2 = 2.
= 7/3
y= (1×2 + 2×5)/(2+1)
= 12/3
= 4.
Example:
C divides AB externally in the ratio 2:1. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=2, n=1, x1= 3, x2 = 1, y1 = 5, y2 = 2.
x = (2×1 − 1×3)/(2 − 1)
= −1
y = (2×2 − 1×5)/(2 − 1)
= −1
Hence the coordinate of C is (−1,−1).
C divides AB externally in the ratio 2:1. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=2, n=1, x1= 3, x2 = 1, y1 = 5, y2 = 2.
= −1
y = (2×2 − 1×5)/(2 − 1)
= −1
No comments:
Post a Comment