Basic Concepts of Coordinate Geometry

DISTANCE BETWEEN TWO POINTS

Let P and Q are two points whose coordinates are (x₁, y₁) and (x₂, y₂), then

PQ = $\sqrt{PQ{{}^{\prime }}^2+QQ{{}^{\prime }}^2}$ = $\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$

Note: Distance is always positive. Therefore we often write PQ instead of |PQ|.

Key tools

In order to prove that a given figure is a

1. Square: Prove that the four sides are equal and the diagonals are equal.

2. Rhombus (but not a square): Prove that the four sides are equal but the diagonals are not equal.

3. Rectangle: Prove that the opposite sides are equal and the diagonals are also equal.

4. Parallelogram (but not a rectangle): Prove that the opposite sides are equal but diagonals are not equal.

Note: That in each of these cases diagonals bisects each other.

5. Equilateral Triangle: Prove all three sides are equal.

6. Isosceles Triangle: Prove two opposite side are equal. (Opposite angle also equal)

SECTION FORMULA

The coordinates of a point which divides the line segment joining two given points A(x₁, y₁) and B(x₂, y₂) internally in the given ratio m : n are

$x=\frac{m{x}_2+n{x}_1}{m+n},y=\frac{m{y}_2+n{y}_1}{m+n}$

The coordinates of the point P(x, y) which divides A (x₁, y₁) and B (x₂, y₂) externally in the ratio m:n are

$x=\frac{m{x}_2-n{x}_1}{m-n},y=\frac{m{y}_2-n{y}_1}{m-n}$

Illustration 1: Find the ratio in which x-axis and y-axis divides the line segment joining (–2, – 3) and (–1, 2).

Key concept: If the ratio, in which a given line segment is divided, is to be determined, then for convenience instead of taking m:n, we take the ratio k:1. If the value of k turns out to be positive, it is an internal division and if k is negative it is an external division.

Solution: Let x-axis divides the line segment in k : 1. Then the coordinate of point on x-axis which divides PQ in K : 1 is

$\left(\frac{-2-k}{k+1},\frac{-3+2k}{k+1}\right)$. But if any point is lying on the x-axis, then its y-coordinate will be zero

$\Rightarrow \frac{-3+2k}{k+1}=0\Rightarrow k=\frac{3}{2}$

Hence x – axis divides PQ in 3:2 internally. Similarly coordinates of any point on y-axis which divides PQ in k :1 is

$\left(\frac{-2-k}{k+1},\frac{-3+2k}{k+1}\right)$. But if any point is lying on the y-axis, then it's x-coordinate will always zero

$\Rightarrow \frac{-2-k}{k+1}=0\Rightarrow k=-\frac{2}{1}$

Hence y-axis divides PQ in 2:1 externally

SOME BASIC DEFINITIONS

Centroid (G): The centroid of a triangle divides each median in the ratio 2:1(2 from the vertex and 1 from the opposite side). The coordinates of centroid are given by

G º $\left(\frac{x_1+x_2+x_3}{3},\frac{{\text{y}}_{\text{1}}+y_2+y_3}{\text{3}}\right)$.

Incentre (I): The coordinates of the in-centre of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) are given by I º $\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$ where a, b and c are length of the sides BC, CA and AB respectively.

Orthocentre (H): The co-ordinates of the orthocentre of the triangle A(x₁, y₁), B(x₂, y₂), C(x₃,y₃) are $\left(\frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C},\frac{{\text{y}}_{\text{1}}\tan A+y_2\tan B+y_3\tan C}{\text{tanA+tanB+tanC}}\right)$.

Circumcentre (C):

$\bullet$The coordinates of the circum-centre of the triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) is given by

O = $\left(\frac{x_1\sin 2A+x_2\sin 2B+x_3\sin 2C}{\sin 2A+\sin 2B+\sin 2C},\frac{{\text{y}}_{\text{1}}\sin 2A+y_2\sin 2B+y_3\sin 2C}{\text{sin2A+sin2B+sin2C}}\right)$.

$\bullet$Note: The circum centre of a right angled triangle is the mid point of its

hypotenuse.

$\bullet$In an equilateral triangle centroid, in-centre, orthocentre and circum centre all coincides.

$\bullet$Centroid divides the line joining orthocentre and circum centre in 2:1 internally.

Illustration 2: If the vertices of a triangle be (2$\sqrt{3}$, 6), (3, –$\sqrt{3}$) and (0, 0). Then find its orthocentre and circumcentre.

Key concept: In a right angled triangle orthocentre is the right angled vertex, and circumcentre is the mid point of the hypotenuse.

Solution: Clearly, the given triangle is right angled at vertex(0, 0), so its orthocentre is (0, 0) and circumcentre will be the mid point of hypotenuse which is $\left(\frac{2\sqrt{3}+3}{2},\frac{6-\sqrt{3}}{2}\right)$.

AREA OF A TRIANGLE

Let (x₁, y₁), (x_2, y₂) and (x₃, y₃) respectively be the coordinates of the vertices A, B, C of a triangle ABC. Then the area of triangle ABC, is \text{ }\dfrac{\text{1}}{\text{2}}{{\left| \begin{align}  {{x}_{1}} {{y}_{1}} 1 \\  {{x}_{2}} {{y}_{2}} 1 \\  {{x}_{3}} {{y}_{3}} 1 \\ \end{align} \right|}^{{}}}

LOCUS

The equation to a locus is the relation which exists between the coordinates of any point on the path, and which holds for no other point except those lying on the path.

Working rule to find the locus of a point:

Step 1: Let the coordinates of the moving point be P(h, k).

Step 2: Write down the given geometrical condition and express these conditions in terms of h and k.

Step 3: Eliminate the variable to get the relation in h and k i.e. this relation must contain only h, k and known quantities.

Step 4: Express the given relation in h and k in the simplest form and then put x for h and y for k. The relation thus obtained, will be the required equation of the locus of P(h,k).

Illustration 3: Find the locus of a point P which divides the line joining (1, 0) and (2 cosq, 2 sinq) internally in the ratio 2 : 3 for all q.

Solution: Let the coordinate of point P which divides the line joining (1, 0) and (2cosq, 2sinq) in the ratio 2 : 3 be (h, k) then

$h=\frac{4\cos \theta +3}{5},k=\frac{4\sin \theta }{5}$

cosq = $\frac{5h-3}{4}$, sinq = $\frac{5k}{4}$

${\cos }^2\theta +{\sin }^2\theta =1={\left(\frac{5h-3}{4}\right)}^2+{\left(\frac{5k}{4}\right)}^2$

= (5h-3)² + (5k)² = 16

Hence locus is (5x – 3)² + (5y)² = 16