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Coordinate geometry is the NCERT chapter which deals with distance formula. The NCERT Class 10 Maths chapter 7 notes covers an outline of the chapter coordinate geometry. The main topics covered in coordinate geometry Class 10 notes are section formulae, abscissa and ordinates. Class 10 Maths chapter 7 notes includes FAQ or frequently asked questions about the chapter. These topics can also be downloaded from the Coordinate Geometry Class 10 notes pdf download option.
Also, students can refer,
In the Cartesian coordinate system, two number lines that are perpendicular to each other,, the x-axis (horizontal) and y-axis (vertical) represents the two variables. These two perpendicular lines are called the coordinate axis.
The intersection point of these two lines is known as the center or the origin of the coordinate plane. Its coordinates are (0,0).
Point on this coordinate plane is represented by an ordered pair of numbers. Let (a, b) be an ordered pair then a is the x-coordinate and b is the y-coordinate.
The distance of any point from the y-axis is called x-coordinate or abscissa and the distance of any point from the x-axis is called y-coordinate or ordinate.
The Cartesian plane has four quadrants I, II, III and IV.
Equation of a Straight Line
An equation of line is used to plot a graph of that line on a cartesian plane.
Slope intercept form
y = mx +b
Where, m is the slope of the line, b = y-intercept.
m=change in ychange in x
Distance formula
The distance between two points A(x_{1},y_{1}) and B(x_{2},y_{2}) is calculated by the formula:
To find the distance of any point from the origin, one point is P(x,y) and the other point is the origin itself, which is O(0,0). According to the above distance formula, it will be:
Let F(x, y) is any point on the line segment AB, that divides AB in the ratio of m: n, then the coordinates of the point F(x, y) will be:
x=(mx_{2}+nx_{1})_{/(}m+n) and
y=(my_{2}+ny_{1})/m+n
If F(x, y) is the mid-point of the line segment AB, that divides AB in the ratio of 1:1, then the coordinates of the point F(x, y) will be:
x=(x_{2}+x_{1})/2 and y=(y_{2}+y_{1})/2
Area of triangle ABC = Area of trapezium ABQP + Area of trapezium APRC – Area of trapezium BQRC.
Area of a trapezium=12(sum of parallel sides) (distance between them)
Therefore,
Area of ABC
=0.5[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]
When the area of the triangle is zero then the given three points must be collinear.
The point where all the three medians of the triangle meet with each other, is known as the centroid of a triangle
The coordinates of the centroid are :
X=(X_{1}+X_{2}+X_{3})/3, y=(y_{1}+y_{2}+y_{3})/3
No. | Figures made of four points | Prove |
1. | Square | Four sides are equal and the diagonals are also equal. |
2. | Rhombus | Four sides are equal. |
3. | Rhombus but not square | Four sides are equal and the diagonals are not equal. |
4. | Rectangle | Opposite sides are equal and the diagonals are equal. |
5. | Parallelogram | Opposite sides are equal. |
6. | Parallelogram but not a rectangle | Opposite sides are equal but the diagonals are not equal. |
No. | Figures made of three points | Prove |
1. | Scalene triangle | None of its sides are equal. |
2. | Isosceles triangle | Any two sides are equal. |
3. | Equilateral triangle | All three sides are equal. |
4. | Right triangle | The sum of the squares of any two sides is equal to the square of the third side. |
Coordinate Geometry Class 10 notes will help to understand the formulas, statements, rules in detail.
This NCERT Class 10 Maths chapter 7 notes also contains previous year’s questions and NCERT TextBook pdf. It contains FAQ’s or frequently asked questions along with the topic-wise explanation
In offline mode, Class 10 Maths chapter 7 notes pdf download can be used to prepare.
Using the formula given above,
A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)] A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)] A = (1/2) [-3 + 12]
Area = 9/2 square units. Therefore, the area of a triangle ABC is 9/2 square units.
Using the formula given above,
A = (1/2) [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)] A = (1/2) [1(2 – 5) + 4(5 – 2) + 3(2 – 2)] A = (1/2) [-3 + 12]
Area = 9/2 square units. Therefore, the area of a triangle ABC is 9/2 square units.
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