Coordinate Geometry teaches us that even the smallest point can help complete the big picture. It is the study of geometry using coordinate points. Coordinate geometry is used to determine the distance between two points. Coordinate geometry is also called analytic geometry or Cartesian geometry. It is a part of geometry. It describes the links between two points in graphs and algebra. Coordinate geometry is used in many areas, like navigation and mapping, including map projection, GPS, air traffic control, computer graphics, interior design, building construction, art and design, etc. It deals with the geometrical figures that have points on the x-axis and y-axis to determine the exact position in a 2-dimensional plane. These NCERT class 10 Maths chapter 7 notes turn difficult diagrams into easy answers. For full syllabus coverage and solved exercises as well as a downloadable PDF, please visit this link: NCERT.
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These notes include basic definitions of coordinate geometry, its subtopics like definition of points, formulae required to calculate the distance between two points, section formula, mid point theorem, what a coordinate is and a coordinate plane, equation of a line in the Cartesian plane, etc. CBSE Class 10 chapter Coordinate Geometry includes all the topics covered in these notes. Students must practice questions and check their solutions using the NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry. These NCERT class 10th maths notes cover the complete syllabus of class 10, all the required definitions, formulas, and examples. These NCERT notes are designed according to the latest syllabus of NCERT
Students who wish to access the Coordinate Geometry Class 10 Maths notes can click on the link below to download the entire notes in PDF.
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Coordinate Geometry: Coordinate geometry is a branch of geometry that is used to determine the position of points using coordinates.
Coordinates: Coordinates are the set of values that are used to define the exact position of a point in the coordinate plane.
Coordinate Plane: It is a 2-dimensional plane that has two perpendicular axes called the x-axis and y-axis that intersect at a point. The point where both axes intersect is called the origin.
The 2-dimensional coordinate plane has four quadrants that are shown below:
1. Quadrant 1: (+x, +y)
2. Quadrant 2: (-x, +y)
3. Quadrant 3: (-x, -y)
4. Quadrant 4: (+x, -y)
A pair of numbers is generally given in the terms of (x, y), and these numbers are called coordinates, and these coordinates define the position in the coordinate plane. In this plane, the distance from the point on the y-axis is called the abscissa or x-coordinate, and the distance of a point from the x-axis is called the y-coordinate or ordinate.
Example: Consider the coordinate (4, 2) as shown in Figure 4 is the abscissa, and 2 is the ordinate. 4 represents the distance from the y-axis, and 2 represents the distance of a point from the x-axis.
The distance between the given points is calculated differently based on the coordinates and the position of the points.
If the two points are on the same axis, x-axis or y-axis, then the difference will be determined by subtracting the value of the x-axis from the x-axis and the y-axis from the y-axis.
Example: Consider the figure shown below:
Distance between PQ = 3 - (-2) = 5
Distance between RS = 3 - (-1) = 4
The distance between two points can be calculated using the Pythagorean theorem by extending the lines and creating a right-angle triangle.
Example: Consider points A(0, 3), B(4, 3), C(-4, -4), and D(-4, 0), determine the distance between C and D.
Given:
A(0, 3), B(4, 3), C(-4, -4), and D(-4, 0)
Extend the BA up to E and join C, D and E to make a right-angle triangle as shown below.
The distance between CD by using Pythagoras' Theorem is:
CD2 = BE2 + EC2
⇒ CD2 = (4 - (-4))2 + (0 - (-4))2
⇒ CD2 = 82 + 42
⇒ CD2 = 64 + 16
⇒ CD = $\sqrt{80}$ units
If two points (x1, y1) and (x2, y2) then the distance can be calculated as:
$d= \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
If a point P divides the line segment in the ratio of m : n, then the point P can be determined by the section formula. Suppose point P(x, y) divides the line segment A(x1, y1) and B(x2, y2) in m : n ratio, then to determine the P(x, y) the section formula is as follows:
$P(x, y)$ = $ (\frac{mx_2 + nx_1}{m + n}$, $ \frac{my_2 + ny_1}{m + n})$
Example: Find the coordinates of the point which divides the line segment joining the points (3, 1) and (7, 4) in the ratio 2 : 1 internally.
Given:
m : n = 2 : 1
x = $ (\frac{2(7) + 1(1)}{2 + 1})$, y = $ (\frac{2(4) + 1(-2)}{2 + 1})$
⇒ x = $ (\frac{14 + 1}{3})$, y = $ (\frac{8 - 2}{3})$
⇒ x = 5, y = 2
If P(x, y) is the point that divides the line segments A(x1, y1) and B(x2, y2) internally, then the ratio of the line can be determined as follows:
Assume that the ratio is k : 1
Substitute the ratio in the section formula for any of the coordinates to get the value of k.
x = $ (\frac{kx_2+ x_1}{k + 1})$
or y = $ (\frac{ky_2+ y_1}{k + 1})$
Example: Find the ratio when point (– 3, 5) divides the line segment joining the points A(– 5, 9) and B(2, – 7)?
Given:
A(– 5, 9) and B(2, – 7)
P (– 3, 5)
$\frac mn$ : 1 = k : 1
P(– 3, 5) divide the line into k : 1 segment,
x = $ (\frac{kx_2+ x_1}{k + 1})$, y = $ (\frac{ky_2+ y_1}{k + 1})$
After substituting the values we get,
⇒ -3 = $ (\frac{2k+ (-5)}{k + 1})$
⇒ -3 = $ (\frac{2k - 5}{k + 1})$
⇒ -3k - 3 = 2k - 5
⇒ 5k = 2
⇒ k = $ (\frac{2}{5})$
Therefore, the required ratio is 2 : 5.
Midpoint is a point in a line that divides the line into two equal parts or in 1 : 1 ratio. If the P is the point that divides the line A(x1, y1) and B(x2, y2).
Therefore,
$P(x, y)$ = $ (\frac{x_1 + x_2}{2}$, $ \frac{y_1 + y_2}{2})$
Example: What is the midpoint of line segment AB whose coordinates are A(-4, 4) and B(2, 6), respectively?
Given:
A(-4, 4) and B(2, 6)
$P(x, y)$ = $ (\frac{-4 + 2}{2}$, $ \frac{4 + 6}{2})$
$⇒P(x, y)$ = $ (\frac{2}{2}$, $ \frac{10}{2})$
$⇒P(x, y)$ = $(1, 5)$
Therefore, the midpoint is (1, 5).
If P(x1, y1), Q(x2, y2), and R(x3, y3) are the vertices of a ΔPQR, then the coordinates of its centroid(A) are defined by,
$A(x, y)$ = $ (\frac{x_1 + x_2 + x_3}{3}$, $ \frac{y_1 + y_2 + y_3}{3})$
Example: Find the coordinates of the centroid of a triangle whose vertices are given as (-2, -4), (5, 1) and (9, -3)
Given:
(-2, -4), (5, 2) and (9, -3)
$A(x, y)$ = $ (\frac{x_1 + x_2 + x_3}{3}$, $ \frac{y_1 + y_2 + y_3}{3})$
After substituting the values we get,
$A(x, y)$ = $ (\frac{-2 + 5 + 9}{3}$, $ \frac{-4 + 1 + (-3)}{3})$
$⇒A(x, y)$ = $ (\frac{12}{3}$, $ \frac{-6)}{3})$
$⇒A(x, y)$ = $(4, -2)$
Therefore, the centroid of the triangle is (4, -2)
If P(x1, y1), Q(x2, y2), and R(x3, y3) are the vertices of a ΔPQR, then the area of ΔPQR can be defined as,
$Area$ = $ \frac{1}{2}$ $[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
Example: Find the area of the triangle ABC whose vertices are A(2, 3), B(5, 3) and C(4, 6).
Given:
A(2, 3), B(5, 3) and C(4, 6)
$Area$ = $ \frac{1}{2}$ $[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$
After substituting the values, we get:
$Area$ = $ \frac{1}{2}$ $[2(3 - 6) + 5(6 - 3) + 4(3 - 3)]$
$⇒Area$ = $ \frac{1}{2}$ $[2(-3) + 5(3) + 4(0)]$
$⇒Area$ = $ \frac{1}{2}$ $[-6 + 15 + 0]$
$⇒Area$ = $ \frac{1}{2}\times[9]$
$⇒Area$ = $ \frac{9}{2}$
Therefore, the area of the triangle is $ \frac{9}{2}$ square units.
Question 1:
Find the coordinates of the points where the graph $57x – 19y = 399$ cuts the coordinate axes.
Solution:
$57x-19y=399$
⇒ $\frac{57x}{399}-\frac{19y}{399} = 1$
⇒ $\frac{x}{7}-\frac{y}{21} = 1$ -------------(i)
Comparing it with the equation of a line:
$\frac{x}{a}+\frac{y}{b} = 1$
x-intercept = $a$ = 7
y-intercept = $b$ = –21
The line, $57x-19y=399$, cuts the x-axis at (7, 0) and the y-axis at (0, –21)
Hence, the correct answer is 'x-axis at (7, 0) and y-axis at (0, –21)'.
Question 2:
The graph of the equation $x=a(a \neq 0)$ is a_____.
Solution:
Given: $x=a(a \neq 0)$
A diagram that shows the change of one variable to one or more other variables, such as a collection of one or more points, lines, line segments, curves, or regions.
The value of $x$ = The value of $a$ = 1, 2, 3, and so on to infinity.
The value of $y$ = constant (assume $y=4$)
As a result, the graph these points create will be a straight line parallel to the y-axis.
Hence, the correct answer is a line parallel to the y-axis.
Question 3:
The area of triangle formed by the straight line $3x + 2y = 6$ and the co-ordinate axes is:
Solution:
The area of a triangle formed by a line and the coordinate axes $=\frac{1}{2} \times \text{base} \times \text{height}$
The line $3x + 2y = 6$
At $y = 0⇒x = \frac{6}{3} = 2$
At $x = 0⇒y = \frac{6}{2} = 3$
So, the base and the height of the triangle are $2$ units and $3$ units, respectively.
$\text{Area of the triangle formed} = \frac{1}{2} \times 2 \times 3 = 3$
Hence, the correct answer is 3 square units.
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Frequently Asked Questions (FAQs)
Students can determine the area of a triangle using the formula
$Area$ = $ \frac{1}{2}$ $[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]$
The important formulas are,
Distance formula $d$ = √[x2 – x1)2+(y2 – y1)2],
Section Formula $P(x, y)$ = $ (\frac{mx2 + nx1}{m + n}$, $ \frac{my2 + ny1}{m + n})$,
Ratio Formula x = $ (\frac{kx2+ x1}{k + 1})$ or y = $ (\frac{ky2+ y1}{k + 1})$,
Mid Point Formula $P(x, y)$ = $ (\frac{x1 + x2}{2}$, $ \frac{y1 + y2}{2})$
and Centroid of a triangle $A(x, y)$ = $ (\frac{x1 + x2 + x3}{3}$, $ \frac{y1 + y2 + y3}{3})$
Students can determine the distance between two points by using the distance formula
$d$ = √[x2 – x1)2+(y2 – y1)2].
The section formula is
$P(x, y)$ = $ (\frac{mx2 + nx1}{m + n}$, $ \frac{my2 + ny1}{m + n})$
Coordinate geometry is a branch of geometry that is used to determine the position or distance of points using coordinates.
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