NCERT Solutions for Class 10 Mathematics Chapter 7 Coordinate Geometry
Coordinate Geometry in Class 10 is an interesting topic, but students need to practise a lot to understand it thoroughly. NCERT textbook for Class 10 Mathematics has to practise questions at the end of every chapter so that students can gauge their understanding of the chapter, clarify their doubts, and prepare better for exams.
Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 7 to help students solve textbook problems easily. The solutions are prepared by subject experts, which ensures they are reliable and accurate.
Access NCERT Solutions for Class 10 Mathematics Chapter 7 – Coordinate Geometry
The NCERT Solutions for Class 10 Mathematics Chapter 7 have step-by-step answers explained with diagrams, wherever required. The answers are written in simple language, making it easier for students to understand the solution/theorem explained therein..
Students can refer to NCERT solutions for revision, last-minute preparation, and while solving the textbook exercises.
NCERT Solutions for Class 10 Mathematics
Here are the chapters included in NCERT Class 10 Mathematics:
- Chapter 1 – Real Numbers
- Chapter 2 – Polynomials
- Chapter 3 – Pair of Linear Equations in Two Variables
- Chapter 4 – Quadratic Equations
- Chapter 5 – Arithmetic Progressions
- Chapter 6 – Triangles
- Chapter 7 – Coordinate Geometry
- Chapter 8 – Introduction to Trigonometry
- Chapter 9 – Some Applications of Trigonometry
- Chapter 10 – Circles
- Chapter 11 – Constructions
- Chapter 12 – Areas Related to Circles
- Chapter 13 – Surface Areas and Volumes
- Chapter 14 – Statistics
- Chapter 15 – Probability
Students can access the NCERT Solutions for Class 10 Mathematics for all the chapters on Extramarks.
Coordinate Geometry
Coordinate Geometry is a branch of Mathematics that uses an ordered pair of integers to locate a given point. It is also known as Cartesian Geometry. It aids in determining the distance between two points whose coordinates are known. You can also determine the coordinates of the point that divides the line segment that connects two points in the given ratio. Students will also learn how to calculate the area of a triangle using the coordinates of its vertices.
Chapter 7 Coordinate Geometry has four sections:
- Introduction to coordinate geometry
- Formula to calculate the distance between two points
- Finding the area of a triangle in the form of coordinates of their vertices
- Finding the coordinate of the point that divides a line in a perpendicular ratio, also called the section formula
What is Coordinate Geometry?
Coordinate geometry is a branch of Mathematics that uses an ordered pair of integers to assist us in precisely locating a given position. Coordinate geometry is a problem-solving technique that combines geometry and mathematics.
Terms Related to Coordinate Geometry
While studying Coordinate Geometry, students should be aware of a few essential terms which are covered in depth in this chapter.
Distance Formula
The distance formula is used to calculate the distance between two places. When the two coordinates of the points are given, we can get the distance between them using the formula.
Section Formula
The section formula assists us in determining the coordinates of a point dividing a given line segment into two parts such that their lengths are in a given ratio.
Area of Triangle
The formula allows us to calculate the area of any triangle using the coordinates of its vertices. This formula will be used to calculate the area of quadrilaterals as well.
Table of All Formulae of Coordinate Geometry
| General Form of a Line |
Ax + By + C = 0
Where A , B, C are real numbers and x and y are variables |
| Slope Intercept Form of a Line |
y = mx + c
Where x and y are variables, c is constant and m is the slope |
| Point-Slope Form |
y− y1= m(x − x1)
x1 ,y1,x2,y2 are the X Y coordinates and m is the slope |
| The slope of a Line Using Coordinates |
m = Δy/Δx = (y2 − y1)/(x2 − x1) |
| The slope of a Line Using General Equation |
m = −(A/B) |
| Intercept-Intercept Form |
x/a + y/b = 1 |
| Distance between two points O(x1,y1) and B(x2, y2) |
OB =√[(x2−x1)²+(y2−y1)²] |
| For Parallel Lines |
m1 = m2 |
| For Perpendicular Lines |
m1m2 = -1 |
| Midpoint Formula/Section Formula |
M (x, y) = ½(x1+x2),½(y1+y2) |
| Angle between Two Lines |
θ = tan-1 (m1–m2)/1+m1m2 |
| Area of a Triangle |
½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)| |
| Perpendicular Distance from a Point to a Line |
d = |ax1+by1+c|/√(a2+b2) |
Historical Facts
- The Great French Mathematician of the seventeenth century, Rene Descartes, preferred to meditate while lying in bed. He solved the challenge of defining the position of a point in a plane one day while resting in bed. His method was based on the old latitude and longitude notion.
- Analytic geometry is another name for Cartesian geometry.
NCERT Solutions provide the following benefits
- It provides adequate resources that are complete in every way so that the students can answer any question easily.
- During exams, students are under stress and face deadlines to complete their tasks. The solution comes in handy during last-minute preparation and revision.
- It helps to build their mathematical and problem-solving abilities and enhances their confidence in answering questions in exams.
What is special about Extramarks NCERT Solutions?
The NCERT Solutions by Extramarks are written by subject matter experts, thus students can be assured that they are referring to reliable and authentic learning material. As the answers are written in a simple yet comprehensive manner, students find no difficulty in understanding the concept applied.
Why is Coordinate Geometry Important?
Coordinate geometry aids in finding a point’s exact placement in a plane. It provides a link between algebra and geometry through lines, graphs, curves, and equations.
Deeper into Exercises
The exercises in Chapter 7 are designed to assess the conceptual understanding of students.
Exercise 7.1
The first exercise is based on finding the distance between two points on a plane. It has a total of ten questions, each one of them is of different type.
Exercise 7.2
The section formula is used in Exercise 7.2 to find the coordinates of a point that divides a given line by a certain ratio. A well-defined theorem is used to generate the section formula. There are ten questions in this exercise also.
Exercise 7.3
The exercise has a total of five questions based on calculating the area of triangle and quadrilateral, finding collinearity of three points, and finding results when three points are collinear. The distance formula and Heron’s formula are mostly used in these exercise questions.
Exercise 7.4
Exercise 7.4 (optional) is based on the several principles discussed, such as the distance formula, section formula, and triangle area. There are eight questions, which will allow students to assess their understanding level. .
Summary
- Two perpendicular lines are required to locate the position of an object or a point in a plane. The line on the left is horizontal, and the line on the right is vertical.
- The plane is known as the cartesian plane or coordinate plane, while the lines are known as coordinate axes.
- The X-axis is the horizontal line, while the Y-axis is the vertical line.
- A given point’s abscissa and ordinate are its distances from the Y-axis and X-axis, respectively.
- Any point on the x-axis has coordinates of the type (x, 0).
- Any point on the y-axis has coordinates of the type (0, y).
- The distance between points P (x1, y1) and Q (x2, y2) is derived by
PQ = √(x2−x1)2+(y2−y1)2
- Distance of point P( x, y) from the origin(0, 0) is given by OP = (x2 + y2)
- The coordinates of the centroid of triangle formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) are ( x1 + x2 + x3/3 , y1 + y2 + y3/ 3)
- The area of the triangles formed by the points A(x1 , y1) , B( x2, y2) and C(x3, y3) is the Area of a Triangle = ½ |x1(y2−y3)+x2(y3–y1)+x3(y1–y2)|
- If points A(x1 , y1) , B( x2, y2) and C(x3, y3) are collinear, then x1(y2−y3)+x2(y3–y1)+x3(y1–y2) = 0
Related Questions
Q1. The graph of 2x+y=3 passes through the origin. Is this statement true or false?
Ans. The statement is false.
Explanation:
The given equation is 2x+y=3.
If the point satisfies the equation, the graph passes through that point.
As we know the point of origin is (0,0)
Now, substitute the value of x as 0 and the value of y as 0 in the equation.
⇒2⋅0+0=3
On further simplification, you will get
⇒0=3
0 cannot be equal to 3. So, we can say that the point does not satisfy this line. The statement i.e. the graph of 2x+y=3 passes through the origin is false.