Coordinate geometry.

Basics Revisited

Points on a Cartesian Plane

Points on a plane are located by a pair of numbers called the coordinates.  The distance of a point from the y-axis is called its x-coordinate, or abscissa. The distance of a point from the x-axis is called its y-coordinate, or ordinate

Representation of $(x,y)$ on the cartesian plane

Distance Formula

Distance between Two Points on the Same Coordinate Axes

The distance between two points which are on the same axis (x-axis or y-axis), is given by the difference between their ordinates if they are on the y-axis, else by the difference between their abscissa if they are on the x-axis.

Distance AB = 6 - (-2) = 8 units

Distance CD = 4 - (-8) = 12 units

Distance between Two Points Using Pythagoras Theorem

Transparent BG

Finding distance between 2 points using
Pythagoras Theorem

Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be any two points on the cartesian plane.
Draw lines parallel to the axes through P and Q to meet at T. $\Delta PTQ$  is right angled at T. From Pythagoras Theorem,
$P{Q}^2=P{T}^2+Q{T}^2=(x_2-x_1{)}^2+(y_2-y_1{)}^{2}PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Distance Formula

Distance between any two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by
$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
Where $d$ is the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$.

 

Section Formula

Section Formula

If the point P(x,y) divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ internally in the ratio m:n, then, the coordinates of P are given by the section formula as
$P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Finding ratio given the points

To find the ratio in which a given point P(x,y) divides the line segment joining $A(x_1,y_1)$ and $B(x_2,y_2),$

• Assume that the ratio is k :1
• Substitute the ratio in the section formula for any of the coordinates to get the value of k.

$x=\frac{k{x}_2+x_1}{k+1}$
Since, $x_1,x_2$ and $x$ are known, k can be calculated. The same can be calculated from the y- coordinates also.

Mid Point

The midpoint of any line segment divides it in the ratio 1:1.
The coordinates of the midpoint(P) of line segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ is given by
$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Points of Trisection

To find the points of trisection P and Q which divides the line segment joining
$A(x_1,y_1)$ and $B(x_2,y_2)$ into three equal parts:

i) AP : PB = 1 : 2
So, $P=(\frac{x_2+2x_1}{3},\frac{y_2+2y_1}{3})$

ii) AQ : QB = 2 : 1
So, $Q=(\frac{2x_2+x_1}{3},\frac{2y_2+y_1}{3})$

Centroid of a triangle

If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of a $\Delta ABC,$ then the coordinates of its centroid(P) is given by
$P(x,y)=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

 

Area from Coordinates

Area of a triangle given its vertices

If $A(x_1,y_1),B(x_2,y_2)$ and $C(x_3,y_3)$ are the vertices of a $\Delta$ ABC, then its area is given by

$A=\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
Where A is the area of the $\Delta$ ABC.

Collinearity Condition

If three points A, B and C are collinear and B lies between A and C, then,

• AB + BC = AC.  AB, BC, and AC can be calculated using the distance formula.
• The ratio in which B divides AC, calculated using section formula for both the x and y coordinates separately will be equal.
• Area of triangle formed by the three points is zero.