 # AP SSC Class 10 Maths Chapter 7 Coordinate Geometry

AP SSC Class 10 Maths Chapter 7 Coordinate Geometry is worth a mention. Coordinate Geometry is an important branch of mathematics that provides a connection between geometry and algebra in the form of graphs of lines and curves. It helps us locate points on a graph and its applications are spread across fields like dimensional geometry, calculus, etc.

Students preparing for the board exams can refer to this AP SSC 10th Class Maths Chapter 7 Coordinate Geometry notes and solutions to know how to answer the questions and to understand the concepts well.

## Important Formulas in Coordinate Geometry

• The distance between two points $$A(x_{1},y_{1})$$ and $$B(x_{2},y_{2})$$ is calculated using the formula $$\sqrt{(x_{2}-x_1)^{2}+(y_{2}-y_1)^2}$$
• The distance between a pointP(x,y) and the origin is given by $$\sqrt{x^2+y^2}$$
• Distance between two points $$x_{1},y_{1}$$ and $$x_{2},y_{2}$$ on line parallel to Y-axis is $$\left | y_{2}-y_{1} \right |$$.
• Distance between two points $$x_{1},y_{1}$$ and $$x_{2},y_{2}$$ on line parallel to X-axis is $$\left | x_{2}-x_{1} \right |$$.
• The coordinates of the point P(x, y) which divides the line segment joining the points A $$x_{1},y_{1}$$ and B$$x_{2},y_{2}$$) internally in the ratio $$m_{1}:m_{2}$$ are $$\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}},$$
• The mid-point of the line segment joining the points P$$x_{1},y_{1}$$ and $$x_{2},y_{2}$$ is given by $$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$$
• The Heron’s formula or the area of the triangle is given as $$A=\sqrt{S(S-a)(S-b)(S-c))}$$ where $$S=\frac{a+b+c}{2}$$.
• A slope of a line is determined as follows $$m=\frac{y_2-y_1}{x_2-x_1}$$

In the next section, let us look at a few solved chapter questions to better understand coordinate geometry.

### Class 10 Maths Chapter 7 Coordinate Geometry Solved Questions

1. Find the distance between the points (0, 0) and (36, 15).

Solution:

The distance between two points (0,0) and (36,15) can be calculated using the formula

D = $$\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^2}$$

Substituting the values in the equation, we get

D = $$\sqrt{(36-0)^2+(15-0)^2}$$

D = $$\sqrt{1296+225}$$

D = $$\sqrt{1521}$$

D = 39

The distance between two points is 39.

1. Find the midpoint of the line segment joining points (5, 0) and (-3, 6).

Solution: The mid-point of the line segment joining the points P$$x_{1},y_{1}$$ and $$x_{2},y_{2}$$ is given by $$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$$

Substituting the values in the formula, we get

$$M(x,y)=(\frac{5+(-3)}{2},\frac{0+6}{2})$$ $$M(x,y)= (\frac{2}{2},\frac{6}{2})$$ $$M(x,y)= (1,3)$$

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