**CBSE Class 10 Maths Coordinate Geometry Notes:-**Download PDF Here

The complete notes on coordinate geometry class 10 are provided here. Go through the below article and learn the points on the coordinate plane, distance formulas, section formulas and so on with the detailed explanation.

## Basics of Coordinate Geometry

#### For More Information On Basics of Coordinate Geometry, Watch The Below Video.

### Points on a Cartesian Plane

A pair of numbers locate points on a plane called the** coordinates**.Â The distance of a point from the y-axis is known asÂ **abscissa **or x-coordinate. The distance of a point from the x-axis is calledÂ **ordinates **or y-coordinate.

## 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

Let P(x1,Â y1) and Q(x2,Â y2) be any two points on the cartesian plane.

Draw lines parallel to the axes through P and Q to meet at T.

Î”PTQÂ is right-angled at T.

ByÂ **Pythagoras Theorem**,

PQ2Â =Â PT2Â +Â QT2

= (x2Â – x1)^{2Â }+ (y2Â – y1)^{2}

PQ = âˆš[x2Â – x1)^{2Â }+ (y2Â – y1)^{2}]

### Distance Formula

Distance between any two points (x1,Â y1) and (x2,Â y2) is given by

dÂ = âˆš[x2Â – x1)^{2}+(y2Â – y1)^{2}]

Where d is the distance between the pointsÂ (x1,y1) and (x2,y2).

## Section Formula

#### For More Information On Section Formula, Watch The Below Videos.

### Section Formula

If the point P(x, y) **divides** the line segment joining A(x1,Â y1) and B(x2,Â y2)** internally**Â in the **ratio m:n**, then, the coordinates of P are given by the **section formula **as:

### Finding ratio given the points

To find the ratio in which a given point P(x, y) divides the line segment joining A(x1,Â y1) and B(x2,Â y2),

- 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.

WhenÂ x1,Â x2 and x are known, k can be calculated. The same can be calculated from the y- coordinate also.

### MidPoint

The **midpoint **of any line segment divides it in the ratio** 1 : 1**.

The coordinates of the midpoint(P) of line segment joining A(x1,Â y1) and B(x2,Â y2) 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(x1,Â y1) and B(x2,Â y2) into three equal parts:

i) **AP : PB = 1 : 2**

ii)Â **AQ : QB = 2 : 1**

\(Q=(\frac{2x_{2}+x_{1}}{3},\frac{2y_{2}+y_{1}}{3} )\)

### Centroid of a triangle

If A(x1,Â y1),B(x2,Â y2) and C(x3,Â y3) are the vertices of a Î”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(x1,Â y1),B(x2,Â y2) and C(x3,Â y3) are the vertices of a Î” ABC, then its area is given by

A = 1/2[x1(y2Â âˆ’Â y3)Â +Â x2(y3Â âˆ’Â y1)Â +Â x3(y1Â âˆ’Â y2)]

Where A is the area of the Î” ABC.

### Collinearity Condition

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

- AB + BC = AC.Â 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 a triangle formed by three collinear points is zero.

It helped me a lot. thank you

Yes it helped me a lot too ðŸ˜Š