Coordinate Geometry is considered to be one of the interesting concepts of Mathematics. Coordinate Geometry describes the link between geometry and algebra through graphs involving curves and lines. It provides geometric aspects in Algebra and enables to solve geometric problems. It is a part of geometry where the position of points on the plane is described using an ordered pair of numbers.

## What is a Co-ordinate and a Co-ordinate Plane?

You must be familiar with plotting graphs on a plane from the tables of numbers for both linear and nonlinear equations. The number line which is also known as a Cartesian plane is divided into four quadrants by two axes perpendicular to each other, labeled as the x-axis (horizontal) and the y-axis(vertical line).

The four quadrants along with their respective values are represented in the graph below-

Quadrant 1 : (+x, +y)

Quadrant 2 : (-x, +y)

Quadrant 3 : (-x, -y)

Quadrant 4 : (+x, -y)

The point at which the axes intersect is known as the origin. The location of any point on a plane is expressed by a pair of values (x, y) and these pairs are known as the coordinates.

The figure below shows the Cartesian plane with coordinates (4,2). If the coordinates are identified, the distance between the two points and the interval’s midpoint that is connecting the points can be computed.

### Distance Between two Points

Let the two points be A and B, having coordinates to be \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) respectively.

Thus, the distance between two points is given as-

\(\large d = \sqrt{(x_{2} -x_{1})^{2} + (y_{2} – y_{1})^{2}}\)### Midpoint of a line connecting two points:

Consider the same points A and B, having coordinates to be \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) respectively. Let M(x,y) be the midpoint of lying on the line connecting these two points A and B. The coordinates of the point M is given as-

\(\large M (x,y)= \left ( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \right )\)**Equation of a Line- **Equation of a line can be represented in many ways, few of which is given below-

**(i) General form- **The general form of a line is given as Ax + By + C = 0.

**(ii) Slope intercept form- **Let x,y be the coordinate of a point through which a line passes, m be the slope of a line, and c be the y-intercept, then the equation of a line is given as-

**(iii) Intercept form of a line- **Consider a and b be the x-intercept and y-intercept respectively, of a line, then the equation of a line is represented as-

**Slope of a line- **Consider the general form of a line Ax + By + C = 0, the slope can be found by converting this form to the slope intercept form.

or \(\large \Rightarrow y = -\frac{A}{B}x – \frac{C}{B}\)

Comparing the above equation with y = mx + c,

\(\large \boldsymbol{m = -\frac{A}{B}}\)Thus we can directly find the slope of a line from the general equation of a line.

**Angle between two lines-**

Consider two lines A and B, having their slopes to be \(\large m_{1} \;\; \) & \(\;\; m_{2}\) respectively.

Let \(\large \theta\) be the angle between these two lines, then the angle between them can be represented as-

\(\large \tan \theta = \frac{m_{1} – m_{2}}{1 + m_{1} m_{2}}\)**Special Cases- **

**Case 1- **When the two **lines are parallel** to each other,

\(\large m_{1} = m_{2}\) = m

Substituting the value in the equation above,

\(\large \tan \theta = \frac{m – m }{1 + m^{2}} = 0\) \(\large \Rightarrow \theta = 0\)**Case 2- **When the two **lines are perpendicular **to each other,

**\(\large m_{1} . m_{2} = -1\)**

Substituting the value in the original equation,

\(\large \tan \theta = \frac{m_{1} – m_{2}}{1 + (-1)} = \frac{m_{1} – m_{2}}{0}\) which is undefined.

\(\large \Rightarrow \theta = 90^{\circ}\)**Section Formula- **

Consider a line A and B having coordinates \(\large (x_{1}, y_{1})\;\;\) & \( \;\; (x_{2}, y_{2})\) respectively. Let P be a point that which divides the line in the ratio m:n, then the coordinates of the coordinates of the point P is given as-

**(i) When the ratio m:n is internally- **

**(ii) When the ratio m:n is externally- **

Students can follow the link provided to learn the proof for the section formula.

**Examples- Find the distance between the points M (4,5) and N (-3,8).**

Solution- Applying the distance formula we have,

\(d = \sqrt{(-3 – 4)^{2} + (8 – 5)^{2}}\) \(\Rightarrow d = \sqrt{(- 7)^{2} + (3)^{2}} = \sqrt{49 + 9}\) \(\Rightarrow d = \sqrt{58}\)**Example- Find the equation of a line parallel to 3x+4y = 5 and passing through points (1,1).**

**Solution- **For a line parallel to the given line, the slope will be of the same magnitude.

Thus the equation of a line will be represented as 3x+4y=k

Substituting the given points in this new equation, we have

\(k = 3 \times 1 + 4 \times 1 = 3 + 4 = 7\)Therefore the equation is 3x+4y=7

**Area of a triangle:**

The area of the triangle whose vertices are \((x_1,y_1 ),(x_2,y_2)\) and \((x_3,y_3)\) is

\(\frac{1}{2}|x_1 (y_2~ -~ y_3)~ + ~x_2(y_3~ – ~y_1)~ +~ x_3(y_1~ – ~y_2)|\)

If the area of a triangle whose vertices are \((x_1,y_1),(x_2,y_2)\) and \((x_3,y_3)\) is zero, then the three points are collinear.

For detailed, chapter wise solutions for NCERT questions, students can compare their answers with NCERT sample answers given here by Byju’s –NCERT Solutions

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