# Coordinate Geometry Class 10 Notes: Chapter 7

Coordinate geometry class 10 notes i.e. for Chapter 7 are provided here to help the students of class 10 learn this topic in a more efficient way. This concise notes can also help during revision as students can quickly check the important points and recall the concepts. The points that are covered in these notes are-

• Distance Formula
• Section Formula
• Mid-Point Theorem
• Area of a triangle
• Practice Questions

In simple words, coordinate geometry is used to represent a point on a plane. The distance of any given point from y-axis is called as its ‘x-coordinate’ or ‘abscissa,’ whereas the distance from the x-axis is called as its ‘y-coordinate’ or ‘ordinate.’

### Distance Formula

Consider a line having two point $A (x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, then the distance of these points is given by-

$AB = \sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} }$

The above formula is said to be distance formula.

### Section Formula

Section formula is used to divide any line into two parts which are in the ratio m:n.

Let us consider a line AB whose coordinates are given as $A (x_{1}, y_{1})$ and $B(x_{2}, y_{2})$,

then the coordinate of the point which divides a line in the given ratio of m:n is given as:
$\left ( \frac{mx_{2} + nx_{1}}{m + n} , \frac{my_{2} + ny_{1}}{m + n} \right )$

Alternatively, to ease the method of section formula consider the ratio m:n = k,
thus the new ratio becomes ‘k:1′
The section formula is then given as-
$\left ( \frac{kx_{2} + x_{1}}{k + 1} , \frac{ky_{2} + y_{1}}{k + 1} \right )$

### Mid-Point Theorem

As the name suggests, if a line segment is divided in the ratio 1:1, then the point of division is called as the midpoint of the line segment. The coordinate of the mid-point is given as-
$\left ( \frac{x_{2} + x_{1}}{2} , \frac{y_{2} + y_{1}}{2} \right )$

### Area of a Triangle

Consider the triangle formed by the points $(x_{1}, y_{1}), (x_{2}, y_{2}) \;\; and \;\; (x_{3}, y_{3})$, then the area of a triangle is given as-
$A = \frac{1}{2}\left [x_{1}(y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{3}) \right ]$

### Practice Questions

1. In what ratio does the line 2x + y – 4 = 0 divides the line segment joining the
points A(2, – 2) and B(3, 7).
2. Calculate the area of the triangle whose vertices are at (2, 3), (–1, 0), (2, – 4).
3. What would be the value of X if the points A(2, 3), B(4, X) and C(6, –3) are
collinear.

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#### Practise This Question

In ΔABC, DE || BC. Then which of the following is stated by Basic Proportionality Theorem?