COORDINATE GEOMETRY-Notes

A Cartesian plane is formed by two perpendicular lines:

• The horizontal line called the x-axis
• The vertical line called the y-axis

These two axes intersect at a point called the origin (0, 0).
Each point on the plane is written as an ordered pair (x, y), where:

• x-coordinate (abscissa):
  Horizontal distance from the y-axis
• y-coordinate (ordinate):
  Vertical distance from the x-axis

The axes divide the plane into four quadrants, each having a specific sign pattern:

1. Quadrant I:
   \(\Rightarrow\) (+,+)
2. Quadrant II:
   \(\Rightarrow\) (-,+)
3. Quadrant III:
   \(\Rightarrow\) (-,-)
4. Quadrant IV:
   \(\Rightarrow\) (+,-)

Understanding signs helps in placing points correctly on the graph.

One of the core concepts of this chapter is the distance between two points in a plane.
For two points \((x_1,y_1),\ (x_2,y_2)\), the distance between them is: \[\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

Why it Works

This formula comes from the Pythagorean theorem.
Imagine the line joining the two points as the hypotenuse of a right-angled triangle— the horizontal difference is one side and the vertical difference is the other.

Uses

• Verifying the type of a triangle (isosceles, equilateral, right-angled)
• Checking whether points form a straight line
• Finding lengths of sides of polygons

For a point \((x,y)\), the distance from the origin \((0,0)\) is:

\[\sqrt{x^2+y^2}\] This is simply a special case of the distance formula.

Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

Solution:

Let us apply distance Formla to the points (say) \(A\left( 3,2\right) ,B\left( -2,-3\right) ,C\left( 2,3\right)\)

$$\begin{aligned} AB&=\sqrt{\left( 3-\left( -2\right) \right) ^{2}+\left( 2-\left( -3\right) ^{2}\right) }\\ &=\sqrt{\left( \left( 3+2\right) ^{2}+2+3\right) ^{2}}\\ &=\sqrt{5^{2}+5^{2}}\\ &=\sqrt{50}\\\\ BC&=\sqrt{\left[ \left( 2-2\right) ^{2}+\left( -3-3\right) ^{2}\right] }\\ &=\sqrt{\left[ \left( 4\right) ^{2}+\left( -6\right) ^{2}\right] }\\ &=\sqrt{16+36}\\ &=\sqrt{52}\\\\ CA&=\sqrt{\left( 3-2\right) ^{2}+\left( z-3\right) ^{2}}\\ &=\sqrt{1^{2}+\left( -1\right) ^{2}}\\ &=\sqrt{2}\end{aligned}$$

Sum of any two distances is greater than the third distance therefore the points A, B and C forms a triangle

Also, $$\begin{aligned}AB^{2}+ CA^{2}&=BC^{2}\\ \left( \sqrt{50}\right) ^{2}+\left( \sqrt{2}\right) ^{2}&=\left( \sqrt{52}\right) ^{2}\\ 52&=52\end{aligned}$$

Hence, ABC is a right angled triangle

Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Solution:

Lets apply distance formula to the points (say) \(A\left( 1,7\right),\ B\left( 4,2\right),\ C\left( -1,-1\right),\ D\left( -4,-4\right) \)

$$\begin{aligned}AB^{2}&=\sqrt{\left( 1-4\right) ^{2}+\left( 7-2\right) ^{2}}\\ &=\sqrt{\left( -3\right) ^{2}+\left( 5\right) ^{2}}\\ &=\sqrt{9+25}\\ &=\sqrt{34}\\\\ BC^{2}&=\sqrt{4-\left( -D\right) ^{2}+\left( 2-\left( -1\right) \right) ^{2}}\\ &=\sqrt{(4+1)^{2}+\left( +1\right) ^{2}}\\ &=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34}\\\\ CD^{2}&=\sqrt{\left[ \left( 1-4\right) ^{2}+\left( 1+4\right) ^{2}\right] }^{2}\\ &=\sqrt{\left( -3\right) ^{2}+5^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34}\\\\ DA^{2}&=\sqrt{\left( 1-\left[ \left( -4\right) ^{2}+\left( -4\right) ^{2}-?\right] \right) }\\ &=\sqrt{\left( 5\right) ^{2}+\left( -3\right) ^{2}}\\ &=\sqrt{25+9}\\ &=\sqrt{34}\\\\ &\text{Diagonals}\\\\ AC^{2}&=\sqrt{\left( 1+1\right) ^{2}+\left( 7+1\right) ^{2}}\\ &=\sqrt{2^{2}+8^{2}}\\ &=\sqrt{4+64}\\ &=\sqrt{68}\\\\ BD^{2}&=\sqrt{\left( 4+4\right) ^{2}+\left( 2-4\right) ^{2}}\\ &=\sqrt{8^{2}+2^{2}}\\ &=\sqrt{64+4}\\ &=\sqrt{68}\end{aligned}$$ Since AB= BC = CD = DA, all sides of quadrilaterals are equal and diagonals also equal, therefore, ABCD is a square

Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

Solution:

(x, y) is equidistant from points (7, 1) and (3,5), therefore,

$$\begin{aligned}\sqrt{\left( x-7\right) ^{2}+\left( y-1\right) ^{2}}&=\sqrt{\left( x-3\right) ^{2}+\left( y-5\right) ^{2}}\\ \left( x-7\right) ^{2}+\left( y-1\right) ^{2}&=\left( x-3\right) ^{2}+\left( y-5\right) ^{2}\\ \left( x-7\right) ^{2}-\left( x-3\right) ^{2}&=\left( y-5\right) ^{2}-\left( y-1\right) ^{2}\\ x^{2}-14x+49-\left( x^{2}-6x+9\right) &=y^{2}-10y+25-\left( y^{2}+1-2y\right) \\ x^{2}-14x+49-x^{2}+6x-9&=y^{2}-10y+25-y^{2}-1+2y\\ -8x+40&=-8y+24\\ -8x+8y&=20-40\\ 8\times \left( x-y\right) &=16\\ x-y=2\end{aligned}$$ \[\begin{aligned}&\boxed{x-y=2}\\\text{is re}&\text{quired relation}\end{aligned}\]

If a point \(P(x,y)\) divides the line segment joining \(A(x_1,y_1)\) and \(B(x_2,y_2)\) in the ratio \(m:n\) internally, then:

\[\begin{aligned}\boxed{x=\dfrac{mx_2+nx_1}{m+n}}\\\\ \boxed{y=\dfrac{my_2+ny_1}{m+n}}\end{aligned}\]

Applications

• Dividing a line in a given ratio
• Locating a point between two known positions
• Constructing geometric shapes using analytic methods

A direct application of the section formula is the midpoint formula.
For endpoints \((x_1, y_1)\) and \((x_2, y_2)\)

\[\boxed{\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}}\]

This gives the point exactly halfway between the two coordinates.

Why Important

• Useful for calculating the center of a line segment
• Helps in symmetry-based problems
• Often asked in board examinations

Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.

Solution:

Points (4,-3) and (8,5)

Let coordinates \((x,y)\) divides line segment in ratio of 3: 1 Let say \(m =3\) and \(n = 1\)

$$\begin{aligned}x&=\dfrac{mx_{2}+nx_{1}}{m_{1}+m_{2}}\\ &=\dfrac{3\times \left( 8\right) +1\times \left( 4\right) }{3+1}\\ &=\dfrac{24+4}{4}\\ &=\dfrac{28}{4}\\ &=7\\\\ y&=\dfrac{my_{2}+ny_{1}}{m+n}\\ &=\dfrac{1\times \left( -3\right) +3\times \left( 5\right) }{3+1}\\ &=\dfrac{-3+15}{4}\\ &=\dfrac{12}{4}\\ &=3\\\\ \left( x,y\right) &=\left( 7,3\right) \end{aligned}$$

In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Solution:

Let point (4,6) divides line segment joining the point \(A(- 6,10)\) and \(B (3,-8)\) in ratio of \(m: n\)

$$\begin{align}-4&=\dfrac{m\left( 3\right) +n\left( -6\right) }{m+n}\\ -4&=\dfrac{3m-6n}{m+n}\\ -4m-4n&=3m-6n\\ -7m&=4n-6n\\ 7m&=2n\\ 7m&=2n\\ \dfrac{m}{n}&=\dfrac{2}{7}\\ m:n&=2:7\tag{1}\\\\ 6&=\dfrac{-8m+1m}{m+n}\\ 6m+6n&=-8m+10n\\ 14m&=4n\\ 7m&=2n\\ \dfrac{m}{n}&=\dfrac{2}{7}\\ m:n &=2:7\tag{2} \end{align}$$

from equation-1 and equation-2, \(m: n = 2: 7\), hence \(2: 7\) is correct ratio

Coordinate geometry isn’t just a school topic—it is used widely in:

• Mapping and GPS technology
• Computer graphics and animation
• Engineering designs
• Navigation and robotics
• Astronomy for locating celestial bodies

The ability to convert shapes into coordinates makes it valuable across many fields.

• Mixing up signs:
  Always check the quadrant carefully.
• Reversing coordinates:
  Remember (x, y), not (y, x).
• quaring negative numbers incorrectly:
  After squaring, negatives become positive.
• Using wrong ratios in section formula:
  Maintain correct order: first point is linked with the coefficient of the other point.