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 Yaxis is \(\left  y_{2}y_{1} \right \).
 Distance between two points \(x_{1},y_{1}\) and \(x_{2},y_{2}\) on line parallel to Xaxis 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 midpoint 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 \(A=\sqrt{S(Sa)(Sb)(Sc))}\) where \(S=\frac{a+b+c}{2}\).
 A slope of a line is determined as follows \(m=\frac{y_2y_1}{x_2x_1}\)
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

 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 = \(\sqrt{(x_2x_1)^{2}+(y_2y_1)^2}\)
Substituting the values in the equation, we get
D = \(\sqrt{(360)^2+(150)^2}\)
D = \(\sqrt{1296+225}\)
D = \(\sqrt{1521}\)
D = 39
The distance between two points is 39.

 Find the midpoint of the line segment joining points (5, 0) and (3, 6).
Solution: The midpoint 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
\(M(x,y)=(\frac{5+(3)}{2},\frac{0+6}{2})\) \(M(x,y)= (\frac{2}{2},\frac{6}{2})\) \(M(x,y)= (1,3)\)Stay tuned to BYJU’S to get the latest notification on SSC exam along with AP SSC model papers, exam pattern, marking scheme and more.