CBSE Class 10 Chapter 7 Coordinate Geometry Notes

Introduction:

Coordinate System:

• System used to locate points on a plane.
• Consists of a pair of perpendicular axes: x-axis and y-axis that intersect at the origin (0,0).
• It is used to pinpoint location on a plane.
• Comprises x-axis and y-axis.
• Points on the plane are represented by ordered pairs (x, y), where x is the distance along the horizontal x-axis, and y is the distance along the vertical y-axis.

Coordinates:

• Pair of numbers (x, y) represents the position of a point in the plane.
• Expressed as (x,y).
• 
  

Abscissa (x-coordinate):

• The distance of a point from the y-axis is called Abscissa (x-coordinate).
• It is denoted as x.
• A point on the x-axis has coordinate (x,0).

Ordinate (y-coordinate):

• The distance of a point from the x-axis is called Ordinate (y-coordinate).
• It is denoted as y.
• A point on the y-axis has coordinate (0,y).

Quadrants

Quadrant I:

• Located in the upper right portion of the coordinate plane.
• Both x and y coordinates are positive (+,+).
• Points in this quadrant have positive x and y values.

Quadrant II:

• Located in the upper left portion of the coordinate plane.
• x coordinate is negative and y coordinate is positive.
• Points in this quadrant have negative x and positive y values.

Quadrant III:

• Located in the lower left portion of the coordinate plane.
• Both x and y coordinates are negative.
• Points in this quadrant have negative x and y values.

Quadrant IV:

• Located in the lower right portion of the coordinate plane.
• x coordinate is positive and y coordinate is negative.
• Points in this quadrant have positive x and negative y values.

Distance Formula

Distance between Two Points on the Same Coordinate Axes

If we have two points on the same coordinate axis (either the x-axis or the y-axis), finding the distance between them is straightforward because only one coordinate changes.

Points on the x-axis:

If we have two points (x₁,0) and (x₂,0) on the x-axis, the distance between them is the difference of their x-coordinates:

$$d=\mid x_{2}-x_{1}\mid$$

This is because the y-coordinates are both 0, and we are essentially measuring the horizontal distance along the x-axis.

Points on the y-axis:

If we have two points (0, y₁) and (0, y₂) on the y-axis, the distance between them is the difference of their y-coordinates:

$$d=\mid y_{2}-y_{1}\mid$$

This is because the x-coordinates are both 0, and we are essentially measuring the horizontal distance along the y-axis.

The distance is always non-negative.

Example: 

In above given figure:

Distance between point R and S is the difference of their x-coordinates, so

$$d=\mid x_{2}-x_{1}\mid$$

$$RS=7-(-2)$$

$$RS= 9unit$$

Distance between point P and Q is the difference of their y-coordinates, so

$$d=\mid y_{2}-y_{1}\mid$$

$$PQ=5-(-1)$$

$$RS= 6unit$$

Distance between any two point by Pythagoras Theorem:

Let's takes two points on a Cartesian plane: P(x₁,y₁) and Q(x₂,y₂).

The line segment connecting these two points forms the hypotenuse of a right triangle.

The distance between the points (P(x₁,y₁) and Q(x₂,y₂)) is the length of the hypotenuse, which can be found using the Pythagorean Theorem.

If Hypotenuse of right triangle is c and sides are a and b, then

$$c^{2}=a^{2}+b^{2}$$

As per figure, In right angle triangle-

$$a=x_{2}-x_{1}$$

$$b=y_{2}-y_{1}$$

Putting value in Pythagoras Theorem

$$c^{2}=(x_{2}-x_{1})+(y_{2}-y_{1})$$

$$c=\sqrt{(x_{2}-x_{1})+(y_{2}-y_{1})}$$

or

$$PQ=\sqrt{(x_{2}-x_{1})+(y_{2}-y_{1})}$$

Here c is distance between point P(x₁,y₁) and Q(x₂,y₂). if d is distance which is equal to c then

$$d=\sqrt{(x_{2}-x_{1})+(y_{2}-y_{1})}$$

The distance of point P(x,y) from origin O(0,0) can be given as:

$$OP=\sqrt{x^{2}+y^{2}}$$

Section Formula:

Derivation:

To develop Section Formula, let we take two points: A(x₁,y₁) and B(x₂,y₂) and P(x,y) divide AB internally in ratio m₁ : m₂ As shown in figure. by construction we can say

$$\frac{PA}{PB}=\frac{m_{1}}{m_{2}}$$

Now, Draw AR, PS and BT Perpendicular on x-axis and AQ and PC parallel to x-axix.

By AA similarity criterion we can say that

$$\triangle PAQ\sim\triangle BPC$$

Therefore, we can say that

$$\frac{PA}{BP}=\frac{AQ}{PC}=\frac{PQ}{BC}.........(i)$$

Now, we can write as

$$AQ=RS=OS-OR=x-x_{1}$$

$$PC=ST=OT-OS=x_{2}-x$$

$$PQ=PS-QS=PS-AR=y-y_{1}$$

$$BC=BT-CT=BT-PS=y_{2}-y$$

Put these value in equation (i)

$$\frac{m_{1}}{m_{2}}=\frac{x-x_{1}}{x_{2}-x}=\frac{y-y_{1}}{y_{2}-y}$$

Take

$$\frac{m_{1}}{m_{2}}=\frac{x-x_{1}}{x_{2}-x}$$

after Cross multiply we get

$$m_{1}x_{2}-m_{1}x=m_{2}x-m_{2}x_{1}$$

combine like term

$$m_{1}x_{2}+m_{2}x_{1}=m_{1}x+m_{2}x$$

$$m_{1}x_{2}+m_{2}x_{1}=x(m_{1}+m_{2})$$

$$x=\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}$$

Similarly take

$$\frac{m_{1}}{m_{2}}=\frac{y-y_{1}}{y_{2}-y}$$

after Cross multiply we get

$$m_{1}y_{2}-m_{1}y=m_{2}y-m_{2}y_{1}$$

combine like term

$$m_{1}y_{2}+m_{2}y_{1}=m_{1}y+m_{2}y$$

$$m_{1}y_{2}+m_{2}y_{1}=y(m_{1}+m_{2})$$

$$y=\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}$$

Thus, the coordinates of the point P(x, y) which divides the line segment and joining the points A(x₁,y₁) and B(x₂,y₂) in the ratio m₁ : m₂ are

$$\left(\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right)$$

This is known as Section Formula.

How to find Ratio for given points

As per figure, given point P(x, y) divides the line segment A(x₁,y₁) and B(x₂,y₂)

Let assume that the ratio is k : 1

then the coordinates of the point P will be

$$\left(\frac{kx_{2}+x_{1}}{k+1},\frac{ky_{2}+y_{1}}{k+1}\right)$$

we can calculate k if we know x₁, x₂ and x. and same can be calculated for y-coordinates.

How to find coordinate of the mid point:

If mid point of line segment P(x,y) divide line segment A(x₁,y₁) and B(x₂,y₂) in ratio 1 : 1

Then the coordinate of mid point P is

$$\left(\frac{1.x_{1}+1.x_{2}}{1+1},\frac{1.y_{1}+1.y_{2}}{1+1}\right)=\left(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}\right)$$

Points of Trisection

If mid point of line segment P(x,y) divide line segment A(x₁,y₁) and B(x₂,y₂) in ratio 1 : 2

Then the coordinate of P is

$$\left(\frac{2.x_{1}+1.x_{2}}{2+1},\frac{2.y_{1}+1.y_{2}}{2+1}\right)=\left(\frac{2x_{1}+x_{2}}{3},\frac{2y_{1}+y_{2}}{3}\right)$$

If mid point of line segment P(x,y) divide line segment A(x₁,y₁) and B(x₂,y₂) in ratio 2 : 1

Then the coordinate of P is

$$\left(\frac{1.x_{1}+2.x_{2}}{1+2},\frac{1.y_{1}+2.y_{2}}{1+2}\right)=\left(\frac{x_{1}+2x_{2}}{3},\frac{y_{1}+2y_{2}}{3}\right)$$

Centroid of a Triangle

If A(x1, y1), B(x2, y2), and C(x3, y3) are the vertices of a ΔABC, then the coordinates of its centroid (P) are given by