## Coordinate Geometry Graph – Definition, Facts, Examples | How do you Plot the Coordinates of a Graph?

Are you confused about plotting the coordinate geometry graph maker? If yes then don’t worry we will help you plot the points on the coordinate geometry graph paper. The coordinate geometry graph the points and states whether they are collinear are or not. Coordinate geometry is one of the branches of geometry. The graph is plotted using the coordinates (x, y).

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## Coordinate Geometry Graph – Definition

Coordinate geometry graphs deal with plotting and evaluating points, lines on the coordinate plane. The graph has two axes horizontal line is known as the x-axis and the vertical line is known as the y-axis. The coordinate graph is divided into four parts quadrant 1, quadrant 2, quadrant 3, and quadrant 4. The point of intersection of two lines is known as the origin with the ordinates (0, 0).

### Coordinate Geometry Graph Questions and Answers

Example 1.
Plot each of the following points on a graph?
a. (6, 2) b. (7, 0) c. (-5, -2) d. (8, -1)
Solution:
Given points are (5, 2), (8, 0), (-5, -2), (9, -1)
Coordinates of the point (6, 2) both the x coordinate and y coordinate are positive so the point lies in the first quadrant. On the x-axis, take 6 units to the right of the y-axis and then on the y-axis, take 2 units above the x-axis.
Therefore, we get the point (6, 2).
Coordinates of the point (7, 0) both x coordinate and y coordinate are positive so the point lies in the first quadrant.
On the x-axis take 7 units and take 0 units on the y-axis to get the point (7, 0).
For the point (-5, -2), both x coordinate and y coordinate are negative so the point lies in the third quadrant. Take -5 units on the x-axis and take -2 units on the y-axis
and therefore we get the points (-5,-2).
The point (8, -1) lies in the fourth quadrant because the x-coordinate is positive and the y-coordinate is negative.
To plot this point, take 8 units on the x-axis and take -1 units on the y-axis.

Example 2.
Draw the graph of the linear equation y = x + 1?
Solution:
Given linear equation is y = x + 1
The given equation is in the form of y = mx + c
slope m = 1, and constant c = 1
By using the trial and error method, find the value of y for each value of x.
If x = 0, y = 0 + 1, then y = 1
If x = 1, then y = 1 + 1 = 2
If x = 2, then y = 2 + 1 = 3
X    0      1      2
y    1      2      3
Plot the graph using the above table
Mark the points (0, 1), (1, 2), (2, 3) on the graph.

Example 3.
Find the equation of a line parallel to 3x+6y = 5 and pass-through points (2,2).
Solution:
For a line parallel to the given line, the slope will be of the same magnitude.
Thus the equation of a line will be represented as 3x+6y=k
Substituting the given points in this new equation, we have
k = 3 × 2 + 6 × 2 = 6 + 12 = 18
Therefore the equation is 3x + 6y = 18

Example 4.
Find the ratio in which the y−axis divides the line segment joining the points (5,-6) and (-1,-4). Also, find the point of intersection.
Solution:
Let the point be A(3,-6), B(-1,-4) and P(0,y)
Point P is on the y−axis,hence its x coordinate is 0.
So, it is in the form of P(0,y)
Now, we have to find a ratio.
Let the ratio be k:1
Hence m
m1=k, m2 =1, x1 = 3, y1= -6, x2 = -1,y2 = -4, x=0, y=0
Using sections formula x= m1x2 + m2x1/ m1 + m2
​0= -k+3/k+1
Therefore k = 3
Again y= m1y2+m2y1/m1 + m2
​= -4k-6/k+1
= -20-6/6
for k=3
= -18/4
Hence the coordinates of the point is P(0, -18/4).

Example 5.
Plot the points (3,-4) and (-7,6)?
Solution:
Given that the points are (3,-4) and (-7,6)
Coordinates of the point (3, -4). Here the x coordinate is positive and the y coordinate is negative so the point lies in the fourth quadrant.
On the x-axis, take 3 units to the right of the y-axis and then on the y-axis, take 4 units above the x-axis.
Therefore, we get the point (3, -4).
Coordinates of the point (-7, 6). Here the x coordinate is negative and the y coordinate is positive so the point lies in the second quadrant.
On the x-axis, take 7 units to the right of the y-axis and then on the y-axis, take 6 units above the x-axis.
Therefore, we get the point (-7, 6).

### FAQs on Coordinate Geometry Graph

1. What is the linear equation of the coordinate geometry?

The general form of a line in the coordinate geometry is Ax + By + C = 0
Where
A is the coefficient of x
B is the coefficient of y
C is the constant value.
Intercept form of a line is y = mx + c.
Where (x, y) is a point on the line m is the slope.

2. What is meant by Coordinate and Coordinate Planes?

A coordinate plane could be a 2-dimensional plane created by the intersection of two axes named horizontal axis (x-axis) and also the vertical axis (y-axis). These lines are perpendicular to every difference and meet at the purpose known as origin or zero. The axes divide the coordinate plane into four equal sections, and every section is understood because of the quadrant. The quantity line that has quadrants is additionally referred to as the sheet.

3. What are the Four different quadrants and their respective signs?

Quadrant 1: In this quadrant both x and y are positive. The point represented in this quadrant is (+x, +y).
Quadrant 2: In this quadrant X-axis is negative and the y-axis is positive. So, the point is shown as (-x, +y).
Quadrant 3: In this quadrant both the x and y-axis are negative. So the point in this quadrant is represented as (-x, -y).
Quadrant 4: In this quadrant, x is positive, and y is negative. The points in these Coordinates are (+x, -y).

## Worksheet on Plotting Points in the Coordinate Plane | Plotting Points on a Graph Worksheet PDF

Plotting points on a coordinate plane worksheet pdf in coordinate geometry is available here. In the worksheet on plotting points in the coordinate plane, the students can learn how to plot the points on graph paper. Practice plotting points on a coordinate plane worksheet answers help you to identify axes, quadrants, ordered pairs on the x-y plane. This plotting points on a graph worksheet pdf will help you to score better grades in the exams.

## Plotting Points in the Coordinate Plane Worksheet PDF with Answers

Example 1.
Plot the points (−2,3) in the coordinate plane.

Solution:

Given that the point is (-2,3)
Here the x coordinate is -2 and the y coordinate is 3.
Here the second coordinate is positive and the first coordinate is negative
The x-coordinate is -2 moves two units to the left from the origin. The y-coordinate 3 moves three units up in the positive y-direction.

Example 2.
Plot the points (4,4) in the coordinate plane.

Solution:

Given that the point is (4,4)
Here the x coordinate is 4 and the y coordinate is 4 both the coordinates are positive.
The x-coordinate is 4 moves four units to the right. The y-coordinate is also 4 moves four units up in the positive y-direction.

Example 3.
Plot the points (0,-6) in the coordinate plane.

Solution:

Given that the points are (0,-6)
Here the x-coordinate is 0 and the y-coordinate is -6.
Here the x-coordinate is positive and the y-coordinate is negative at the point (0,-6) beginning at the origin.
The x-coordinate is 0 not to move in either direction along the x-axis.
The y-coordinate is -6 moves six units down in the negative y-direction.

Example 4.
Plot the point (-6,5) in the coordinate plane.

Solution:

In the given points -6 is the x coordinate and 5 is the y coordinate.
Here the x coordinate is negative and the y coordinate is positive so the point lies in the second quadrant.
We move 6 units to the left from the origin and then 5 units vertically up to plot the point (−6,5).

Example 5.
Plot the points (−4,6) in the coordinate plane.

Solution:

Given that the point is (-4,6)
Here the x coordinate is -4 and the y coordinate is 6.
Here the second coordinate is positive and the first coordinate is negative
The x-coordinate is -4 moves four units to the left from the origin. The y-coordinate 6 moves six units up in the positive y-direction.

Example 6.
Plot the point (-9,-5) in the coordinate plane.

Solution:

In the given points -9 is the x coordinate and -5 is the y coordinate.
Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 9 units to the left from the origin and then 5 units vertically up to plot the point (−9,-5)

Example 7.
Plot the point (1,7) in the coordinate plane.

Solution:

In the given points 1 is the x coordinate and 7 is the y coordinate.
Here the x coordinate is positive and the y coordinate is also positive so the point lies in the first quadrant. We move 1 unit to the left from the origin and then 7 units vertically up to plot the point (1,7)

Example 8.
Plot the point (9,-11) in the coordinate plane.

Solution:

In the given points 9 is the x coordinate and -11 is the y coordinate.
Here the x coordinate is positive and the y coordinate is negative so the point lies in the fourth quadrant. We move 9 units to the left from the origin and then 11 units vertically up to plot the point (9,-11)

Example 9.
Plot the point (-1,-2) in the coordinate plane.

Solution:

In the given points -1 is the x coordinate and -2 is the y coordinate.
Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 1 unit to the left from the origin and then 2 units vertically up to plot the point (−1,-2)

Example 10.
Plot the point (-8,7) in the coordinate plane.

Solution:

In the given points -8 is the x coordinate and 7 is the y coordinate.
Here the x coordinate is negative and the y coordinate is also negative so the point lies in the third quadrant. We move 8 units to the left from the origin and then 7 units vertically up to plot the point (−8,7)

## Reflection in Lines Parallel to Axes – Definition, Facts, Examples | How do you Reflect Over the Axes?

The complete information on reflection in lines parallel to axes is available on this page. Thus the students who wish to learn in detail about the reflection in lines parallel to axes can refer to our page and practice the problems. Let us discuss how to solve the problems on Reflection in Lines Parallel to Axes (x-axis and y-axis) here. The solutions seen at the end of this page are prepared by the math experts that helps you to score the highest marks in the exams.

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## Reflection in Lines Parallel to Axes Examples

Example 1.
The reflection of the point (4, -1) about the line 5x + y + 6 = 0 is
Solution:
Given that
The reflection point be A (h, k)
Now, the mid point of the line joining (h, k) and (4, -1) will lie on the line
5x + y + 6 = 0
Therefore
5(h+4)/2 + (k−1)/2 + 6 = 0
5h + 20 + k – 1 + 12 = 0
5h + k + 31 = 0……(i)
Now, the slope of the line joining points (h, k) and (4, -1) are perpendicular to the line is
5x + y + 6 = 0
Slope of the line=−5
Slope of the joining by points (h, k) and (4, -1)
k+1/h−4
Therefore
k+1/h−4(−5) = -1
5k − h + 69 = 0…..(ii)
Solving (i) and (ii), we get
5h + k + 6 – 5k + h – 69 = 0
6h – 4k – 63
h = 6 and k = -4

Example 2.
The point (-3, 0) on reflection in a line is mapped to (3, 0) and the point (4, -6) on reflection in the same line is mapped to (-4, -6).
(i) State the name of the mirror line and write its equation.
(ii) State the coordinates of the image of (-7, -5) in the mirror line.
Solution:
(i) We know the reflection of a point (x, y) in the y-axis is (-x, y).
Hence, the point (-3, 0) when reflected in the y-axis is mapped to (3, 0).
Therefore, the mirror line is the y-axis and its equation is x = 0.
(ii) Coordinates of the image of (-7, -5) in the mirror line in the y-axis are (7, -5).

Example 3.
The point (-4, 0) on reflection in a line is mapped as (4, 0) and the point (-7, -6) on reflection in the same line is mapped as (2, -6).
(a) Name the line of reflection.
(b) Write the coordinates of the image of (5, -9).
Solution:
(a) We know that reflection in line x = 0 is the reflection in the y-axis.
Given that the point is
Point (-4, 0) on reflection in a line is mapped as (4, 0).
Point (-7, -6) on reflection in the same line is mapped as (7, -6).
Hence, the line of reflection is x = 0.
(b) we know that
My (x, y) = (-x, y)
Coordinates of the image of (5, -9) in the line x = 0 are (-5, -9).

Example 4.
Point P(4, -3) is reflected as P’ in the y-axis. Point B on reflection in the x-axis is mapped as Q’ (-2, 7). Write the coordinates of P’ and Q.
Solution:
Given that the points are P(4,-3) and Q(-2,7)
Reflection in y-axis is given by
My (x, y) = (-x, y)
P’ = Reflection of P(4, -3) in y-axis
P’= (-4, -3)
Reflection in x-axis is given by
Mx (x, y) = (x, -y)
Q’ = Reflection of Q in x-axis = (-2, 7)
Therefore Q = (-2, -7)

Example 5.
The point (-3, 0) on reflection in a line is mapped as (3, 0) and the point (-8, -6) on reflection in the same line is mapped as (4, -6).
(a) Name the line of reflection.
(b) Write the coordinates of the image of (5, -6).
Solution:
(a) We know that reflection in line x = 0 is the reflection in the y-axis.
Given that the point is
Point (-3, 0) on reflection in a line is mapped as (3, 0).
Point (-8, -6) on reflection in the same line is mapped as (8, -6).
Hence, the line of reflection is x = 0.
(b) we know that
My (x, y) = (-x, y)
Coordinates of the image of (5, -6) in the line x = 0 are (-5, -6).

### FAQs on Reflection in Lines Parallel to Axes

1. What does parallel to the axes mean?

If a line is parallel to the x-axis or y-axis either the x-coordinate or y-coordinate is constant or fixed throughout the line and it should pass through either (0, a) or (a, 0).

2. How do you reflect over the axis?

A reflection of a point over the y -axis is. The rule for a reflection over the y -axis is (x,y) = (-x,y). And the reflection over the x- axis is (x,y) = (x,-y).

3. What Does Parallel to the Axes Mean?

Parallel to axes means that the lines are parallel to either the x-axis or y-axis.
If a line parallel to the x-axis is called a horizontal line whose equation is in the form of y = k,
where ‘k’ is the distance of the line from the x-axis.
Similarly, a line parallel to the y-axis is called a vertical line whose equation is in the form of x = k,
where ‘k’ is the distance of the line from the y-axis.

## Reflection of a Point in a Line Parallel to the y-axis – Rules, Formula, Examples | How do you find the Reflection of a Point Parallel to y-axis?

In order to score well in the exams, one must understand the concepts in maths. That is possible with us here we provide quick and easy learning tricks in reflection here. Reflection of a Point in a Line Parallel to the y-axis is a scoring topic from the chapter reflection. Scroll down this page to know what is meant by Reflection of a Point in a Line Parallel to the y-axis with suitable examples. Step by step explanation is given for the problems by the math experts.

Also Refer:

## What is the Reflection of a Point in a Line Parallel to the y-axis?

Let P be the point on the x-axis with the coordinates (x, y). Let the image of P be P’ in the horizontal line drawn on the y-axis. The image of the point (x, y) in the line parallel to the y-axis. When it comes to the refection of a point in a line parallel to the y-axis the sign of the y-axis will be changed and the sign of the x-axis remains the same.

### Reflection of a Point in a Line Parallel to the y-axis Examples

Example 1.
The point P ( -4, -7) on reflection in the y-axis is mapped on P’. The point P’ on reflection in the origin is mapped on P”. Find the coordinates of P’ and P”. Write down a single transformation that maps P onto P”.
Solution:
Given that the point P (-4, -7)
And, P’ is the image of point P in the y-axis
The coordinates of P’ will be (4, -7)
Again,
P” is the image of P’ under reflection in origin.
Thus, the coordinates of P” will be (-4, 7).
The single transformation that maps P onto P” is the x-axis.

Example 2.
Point A(5, 7) is first reflected in the origin to point A’. Point A’ is then reflected in the y-axis to point A”.find the coordinates of A”.
Solution:
Given that the point is A(5,7)
Coordinates of A’ after reflection from origin =(-5,-7)
And after reflection from y-axis coordinates becomes =A”=(5,-7)

Example 3.
A point P (a, b) is reflected in the x-axis to P’ (4, 3). Write down the values of a and b. P” is the image of P, reflected in the y-axis. Write down the coordinates of P”. Find the coordinates of P”, when P is reflected in the line, parallel to the y-axis, such that x = 4.
Solution:
The coordinates of the point P be(4,-3)
Hence the value of a and b are
4 and -3

after reflection coordinates of P”=(-4,3)
Coordinates of p after reflection from x=4 are (8,-3)

Example 4.
The reflection of point (-7, -9) through the y-axis is.
Solution:
When reflected through the y axis, the coordinates x value reverses its sign
This means the positive becomes negative, and the negative becomes positive.
Now we have, the coordinates (-7, -9) its reflection through the y-axis gives us (-7, 9).

Example 5.
The reflection of point (-3, -5) through the y-axis is.
Solution:
When reflected through the y axis, the coordinates x value reverses its sign
This means the positive becomes negative, and the negative becomes positive.
Now we have, the coordinates (-3, -5) its reflection through the y-axis gives us (-3, 5).

### FAQs on Reflection of a Point in a Line Parallel to the y-axis

1. When a line is parallel to the Y-axis What is the slope?

The slope of the line parallel to the y-axis is undefined.
The line parallel to the y axis is at an angle of 90º to the x-axis
Then the slope of this line is tan90º, which is undefined.

2. How do you find the slope of the y-axis?

The slope-intercept form of a line is: y=mx+b
where m is the slope and
b is the y-intercept.

3. When a line is parallel to the y-axis?

A line is parallel to the x-axis and y-axis either the x-coordinate or y-coordinate is fixed or constant throughout the line and the line passes from either (0, a) or (a, 0).

## Reflection of a Point in a Line Parallel to the x-axis – Definition, Rules, Formula, Examples | How to find the Reflection of a Point across the x-axis?

In this article, we will explain in detail the reflection of a point in a line parallel to the x-axis by taking an example. We will show how to plot the points on the graph to know the reflection of a point in the x-axis. The y-coordinate of each point on the line parallel to the x-axis is constant i.e., remain the same.

## What is Reflection of a Point in a Line along the x-axis?

Let P be the point on the x-axis with the coordinates (x, y). Let the image of P be P’ in the horizontal line drawn on the x-axis. The image of the point (x, y) in the line parallel to the x-axis. When it comes to the refection of a point in a line parallel to the x-axis the sign of the x-axis will be changed and the sign of the y-axis remains the same.

### Reflection of a Point in a Line Parallel to the x-axis Examples

Let us discuss the concept of Reflection of a Point in a Line Parallel to the x-axis with some examples.

Example 1.
Point P (a, b) is reflected in the x-axis to P’ (-5, 3). Write down the values of a and b.
Solution:
Given points are (-5, 3)
We know Mx (x, y) = (x, -y)
P'(-5, 3) = reflection of P (a, b) in x-axis.
Thus, the coordinates of P are (5, 3).
Hence, a = 5 and b = 3

Example 2.
Find the Reflection of the point (-5,7).
Solution:
Given that the point is (-5,7)
When a point is reflected in the x-axis, the sign of its ordinate changes.
Reflection of the point
(-5,7) in the x-axis is (5,7).

Example 3.
The image of a point P reflected in the x-axis is (-4,4). Find the coordinates of P.
Solution:
Given that the point is P(-4,4)
When a point is reflected in the x-axis, the sign of its ordinate changes. Hence, the coordinates of P are (4,4)

Example 4.
A point P is reflected in the x-axis. Coordinates of its image are (-3,9). Find the coordinates of the image of P under reflection in the x-axis.
Solution:
Given that the point is P(-3,9)
The Coordinates of the image of P are under reflection in the x-axis (3,9).

Example 5.
Find the Reflection of the point (-3,8).
Solution:
Given that the point is (-3,8)
When a point is reflected in the x-axis, the sign of its ordinate changes.
Reflection of the point
(-3,8) in the x-axis is (3,8).

### FAQs on Reflection of a Point in a Line Parallel to the x-axis

1. What does it mean if a point is parallel to the x-axis?

The point Parallel to the axis means the lines that are parallel to either the x-axis or y-axis.
A line parallel to the x-axis is called the horizontal line.
The equation is in the form of y = k, where ‘k’ is the distance of the line from the x-axis.

2. What is the rule for reflecting a point across the x-axis?

The rule for a reflection of the x -axis is (x,y) = (x,−y).

3. What is the image of a point after a reflection in the x-axis?

The point M is reflected in the x-axis the image M’ is formed in the fourth quadrant and their coordinates are (h, -k). Hence the point is reflected in the x-axis the x-coordinate remains the same, but the y-coordinate becomes negative. Thus, the image of point M (h, k) is M’ (h, -k).

## Reflection of a Point in a Line – Definition, Formula, Examples | How do you find the Image of a Point Reflected over a Line?

In geometry, reflection is the type of transformation in the coordinate plane. A reflection is a mirror image of the given shape. In the previous articles, we have seen the reflection of a point in a plane, line, x-axis, y-axis, and origin. Let us discuss here the reflection of a point about a line formula from this page.  Go through the examples given below and try to solve the problems in an easy manner.

Also, Refer:

## Reflection of a Point in a Line – Definition

Reflection of a point is an interesting topic in the coordinate system. The reflection of the point in the x-axis means the x-coordinate remains the same and the y-coordinate sign will be changed. The reflection of the point in the y-axis means the y-coordinate remains the same and the x-coordinates sign will be changed.

### Reflection of a Point in a Line Examples

Example 1.
What is the reflection of the point (-1,3) on the line x = -3?
Solution:
The line x = -3 is a straight line parallel to the y-axis and at a distance of 3 units in the negative direction of the x-axis
Point (-1,3) is 3 units away from the given line x = -3.
Therefore the distance of the image from the line is also 3 units.
Thus the distance of image from y axis = 3+4+1 = 7.
Units in the negative direction of the x-axis.
Thus, the reflection of point (-1,3) in the line x = -3 is (-6,3).

Example 2.
What is the reflection of the point (1,2) in the line y = 4
Solution:
The line y = 4 bisects the line joining A and A’ the midpoint of these two points must lie on the line y = 4.
If we take point A’ as (a,b) the midpoint of A and A’ are
(1+1)/2, (b+3)/2 must satisfy the equation y = 4.
= (b+2)/2 = 4.
= b + 2 = 8
b = 6.
The required point is (1,6).

Example 3.
What is the reflection of the point (5,3) if the line is y = -6?
Solution:
The perpendicular distance between line y = -6.
And the point A(5,3) must be the same as the distance between line y = -6 and required reflection point P’
y = -6 is a line parallel to the x-axis.
Since the reflection is on the x-axis the x coordinate of P’ remains the same as 5.
y coordinate must be |-6| + 3 = 9 units away from the line y = -6.
The required y coordinate is -15.
Required point A’ = (5,-15).

Example 4.
What is the reflection of the point (-1,2) on the line x = -2.
Solution:
The line x = -2 is a straight line parallel to the y-axis and at a distance of 2 units in the negative direction of x-axis
Point (-1,2) is 2 units away from the given line x = -2.
Therefore the distance of the image from the line is also 2 units.
Thus the distance of image from y axis = 2+2+1 = 5.
Units in the negative direction of the x-axis.
Thus, the reflection of point (-1,2) in the line x = -2 is (-4,3).

Example 5.
What is the reflection of the point (1,3) in the line y = 5
Solution:
The line y = 5 bisects the line joining P and P’ the midpoint of these two points must lie on the line y = 5.
If we take point P’ as (a,b) the midpoint of P and P’ are
(1+1)/2, (b+3)/2 must satisfy the equation y = 4.
= (b+2)/2 = 5.
= b + 2 = 10
b = 8
Required point is (1,8).

### FAQs on Reflection of a Point in a Line

1. What is a reflection line?

A reflection is a transformation, it means flipping of a figure. Figures may be reflected in a point, line, or a plane. When reflecting a figure in a line or in a point, then the image is congruent to the preimage. The fixed line is called the line of reflection.

2. How do you write a line of reflection?

The line of reflection is given in the form of y = mx + b. And draw the line of reflection using the point in the starting figure that is the same perpendicular distance from the line of reflection as its corresponding point in the image.

3. What does reflection across y = 2 mean?

The x-value of the mirror image will be the same. At the y-values. the y-values must be the same number of units below the line y = 2 as above the line y = 2.

## Area of a Circular Ring – Definition, Formula, Examples | How do you find the Area of a Circular Ring?

In this article, you will learn about how to find the Area of a Circular Ring. A circular ring is a plane figure which is bounded by the circumference of two concentric circles of two different radii. A circle is made of multiple points arranged equidistant from a single central point and that point is called the center of the circle. The outer circle and inner circles define that the ring is concentric, that shares a common center point. The best way to think of it is a circular disk with a circular hole in it.

On this page, we will discuss the definition of the area of a circular ring, formulas, solved example problems, and so on.

### Area of a Circular Ring – Definition

The area of a circular ring will be subtracted from the area of a large circle to the area of a small circle. A ring-shaped object is bounded by the circumference of two concentric circles of two different radii. The dimensions of the two radii are R, r, which are the radii of the outer ring and the inner ring respectively.

### Formula:

To find the area of a circular ring or annulus, to multiply the product of the sum and the difference of the two radii. It will be bounded by two concentric circles of radii R and r (R>r).
Therefore, the area of a circular ring is, area of the bigger circle – the area of the smaller circle.
= π(R + r) (R – r) = πR²- πr²
where R and r are the outer circle radius and the inner circle radius. So, it will be as R=√ r2+Aπ  and r = √ R2+Aπ.

### Area of a Circular Ring Examples

Problem 1:

Find the area of a flat circular ring formed by two concentric circles, circles with the same center whose radii are 8cm and 4cm?

Solution:
Given in the question, the values are
The radius (r1) of the bigger circle is 8cm.
The smaller circle radius (r2) is 4cm.
Now, we have to find the area of a circular ring.
As we see the required area is between the two circles as shown in the figure.
Using the formula, we can find the value.
The area of the shaded portion is = Area of the bigger circle – Area of the smaller circle.
A = πR²- πr²
Substitute the given values within the above formula, we get
A = π(8×8) – π(4×4) = 64π – 16π = 48π
48π = 48 x 22/7 = 6.85 x 22 = 150.7 cm²
Therefore, the area of the circular ring formed by two concentric circles is 150.7 cm².

Problem 2:

A path is 21cm wide surrounds a circular lawn with a diameter of 240cm. Find the area of the path?

Solution:
As given in the question,
A circular dawn diameter is 240cm.
So, the radius of the inner circle(r) is 120 cm.
The wide of a path is 21cm.
The radius of the outer circle(R) is 120+21 = 141cm.
Now, we have to find the area of a path.
We know the formula, Area of a path is π(R+ r) (R – r).
After the substitution of the value, we get,
Area of the path = 22/7(141+120)(141-120).
= 22/7(161)(21) = 22 x 23x 21=10616Sq.cm
Thus, the Area of the path is 10616sq.cm.

Problem 3:

The inner diameter and the outer diameter of the circular path are 628 m and 600m respectively. Find the breadth of a circular path and the area of the circular path. Consider the π value as 22/7.

Solution:
As given in the question,
The outer radius of a circular path (R) is, 628/2 = 314m.
The inner radius of a circular path (r) is, 600/2 = 300m.
Now, we have to find the breadth of a circular path and the area of a circular path.
So, the breadth of a circular path is R-r = 314m – 300m = 14m.
Next, the Area of a circular path is, π(R + r)(R – r)
Area = 22/7 (314+300)(314-300)
A= 22/7(614)(14)m² = 27016m².
Therefore, the area of the circular path is 27016m².

Problem 4:

Find the area of a circular ring formed by two concentric circles whose radii are 6.2cm and 5.8cm respectively. Take the π value as 3.14.

Solution:
As we know the radii of the outer circle and inner circles will be R and r respectively.
The concentric circle (R) value is 6.2 cm and the concentric circle (r) value is 5.8 cm.
Now, we will find the area of a circular ring value.
We know the formula, Area of a circular ring is π(R+ r) (R – r).
Substitute the given values in the above formula, we get
A=π(R+ r) (R – r) = 22/7 (6.2 +5.8)(6.2 – 5.8)
sq. cm
sq. cm
Hence, the area of a circular ring value is 15.072 sq. cm

Problem 5:
A circular ring is 8cm wide. Find the difference between the outer circle radius and inner circle radius?

Solution:
As given in the question, the circular ring wide is 8cm.
We have to find the difference value of the outer circle and inner circle radius.

The difference between the outer circle radius and inner circle radius is R-r.
So, the value of R-r is 8 cm.
Thus, the difference value of an inner circle radius and an outer circle radius is 8cm.

## Problems on Plotting Points in the x-y Plane | Plotting Points on a xy Plane Questions and Answers

Problems on Plotting Points in the x-y Plane in Coordinate Geometry with answers are available here. Let us learn how to plot the points in the x-y Plane with examples from this page. So, have a look at this article to know plotting points in the x-y plane on the graph. The right part of the origin in the horizontal line is the positive x-axis and the left part of the origin in the horizontal line is the negative x-axis. Topside of the origin is the positive y-axis and the bottom part from the origin is the negative y-axis.

## Plotting Points in x-y Plane Problems with Solutions

Example 1.
Plot the Points (12, -4) on the Coordinate Graph in the x-y plane.
Solution:
First Taking the scale 1 cm = 4 on the graph
Let the point P = (12, -4).
Here the x-coordinate = 12 and y-coordinate = -4.
So, move 12 units from origin to towards the positive direction of the x-axis.
From that place to move 4 units towards the negative direction of the y-axis.
Now the position of the point is reached as the coordinates (12, -4)

Example 2.
Plot the points (-11, -8) on the Coordinate Graph in the x-y plane.
Solution:
First taking the scale as your wish
Here I am taking 1 cm = 4 units on the graph.
Let the point Q = (-11, -8).
Here, the x-coordinate = -11 and the y-coordinate = -8.
So, move 11 units from the origin towards the negative direction of the x-axis.
From that place to move 8 units towards the negative direction of the y-axis.
Now the position of the point reached as the coordinates (-11, -8)

Example 3.
Plot the Points (-2.6, -1.6) on the Coordinate Graph in the x-y plane.
Solution:
First taking the scale as your wish
Here I am taking 1 cm = 5 units on the graph.
Let the point E = (-2.6, -1.6).
Here, the x-coordinate = -2.5 and the y-coordinate = -1.5.
So, move 2.6 units from the origin towards the negative direction of the x-axis.
From that place to move 1.6 units towards the negative direction of the y-axis.
Now the position of the point is reached as the coordinates (-2.6, -1.6).

Example 4.
Plot the Points (-6, 0) on the Coordinate Graph in the x-y plane.
Solution:
First taking the scale as your wish
Here I am taking 1 cm = 5 units on the graph.
Let the point B = (-6, 0).
Here, the x-coordinate = -6 and the y-coordinate = 0.
So, the point is on the x-axis. Move 6 units from the origin towards the negative direction of the x-axis. Now the position of the point is reached as the coordinates (-6, 0).

Example 5.
Plot the Points (2, 1.5) on the Coordinate Graph in the x-y plane.
Solution:
First taking the scale as your wish
Here I am taking 1 cm = 3 units on the graph.
Let the point C = (2, 1.5).
Here, the x-coordinate = 2 and the y-coordinate = 1.5.
So, move 2 units from the origin towards the positive direction of the x-axis.
From that place to move 1.5 units towards the positive direction of the y-axis.
Now the position of the point reached has the coordinates (2, 1.5).

Example 6.
Plot the points (12, -8) and (-16, -20) in the x-y plane.
Solution:
Taking 1 cm = 4 as the scale of representation the points are plotted as shown below.
(i) Let P = (12, -8). Here the x-coordinate = 12 and y-coordinate = -8. So, move 12 units from O towards the positive direction of the x-axis. From that place move 8 units towards the negative direction of the y-axis. The position of the point now reached has the coordinates (12, -8).

## Reflection of a Point in the Origin – Definition, Formula, Rules | How to find Reflection of a Point in the Origin?

Hello Students!!! Are you excited and interested to know in-depth about reflection? Don’t worry we are here to help you to give detailed information about the reflection of a point in the origin, x-axis, y-axis, and so on. We know that point reflection is a type of reflection. Know how to find the coordinates of the reflection of the point in the origin from this article. The image of the point (p, q) in the origin is the point (-p, -q).

## Reflection of a Point in the Origin

In order to locate the coordinates in the contiguous figure, the origin represents the plane mirror. Let X be any point in the first coordinates (h, k). M’ is the reflected point in the origin formed by the third quadrant whose coordinates will be (-h, -k). To find the reflection of a point in the origin are have to change the signs of the x-coordinates and y-coordinates.

By seeing the above figure we can say that X'(-h, -k) is the mirror image of X(h, k).

### Rules to find the Reflection of a Point in the Origin

To find the reflection of the point in the origin there are some rules to follow.
i. Change the sign of the x-coordinate i.e., abscissa.
ii. Change the sign of the y-coordinate.

### Reflection of a Point in the Origin Solved Problems

Example 1.
What is the reflection of the point A(2, 5) in origin?
Solution:
Given that A(2,5)
The reflection of the point A (2, 5) is P’ (-2, -5).

Example 2.
What is the reflection of the point B(-6, -9) in the origin?
Solution:
Given that B(-6,-9)
The reflection of point B (-6, -9) is B’ (6, 9).

Example 3.
What is the reflection of the point C(-3,8) in the origin?
Solution:
Given that C(-3,8)
The reflection of point C (-3, 8) is C’ (3, -8).

Example 4.
What is the reflection of the point D(1, -1) in the origin?
Solution:
Given that D(1,-1)
The reflection of the point D(1,-1) is C'(-1, 1)

Example 5.
What is the reflection of the point E(5,-1)
Solution:
Given that E(5,-1)
The reflection of the point S (5, -1) is S’ (-5, 1).

### FAQs on Reflection of a Point in the Origin

1. What does reflection in the origin mean?

When you reflect a point in the origin, both the x-coordinate and the y-coordinate change their signs. In a point reflection in the origin, the image of the point (x,y) is the point (-x,-y).

2. Which functions have graphs that are symmetrical with respect to the origin?

A function that is symmetrical with respect to the origin is called an odd function. f(x).
Since f(−x) = f(x), this function is symmetrical with respect to the y-axis.

3. What is the rule for reflection?

If you reflect a point across the line y = x, the x-coordinate, and y-coordinate change places.
If you reflect over the line y = -x, the x-coordinate and y-coordinate change places.
The line y = x is the point (y, x). the line y = -x is the point (-y, -x).

## Worksheet on Slope and Y-Intercept | Finding Slope and Y Intercept from an Equation Worksheets

Worksheet on Slope and Y-intercept with stepwise solutions are available here. So, the students who are all in search of the concept of slope and Y-intercept in coordinate geometry can make use of this slope and y intercept worksheets with answer key pdf and practice the problems. You can find different types of problems related to the slope (m) and y-intercept. Look into the problems given below in the slope and y intercept worksheet with answers and enhance your math skills.

Do Refer:

## Identifying Slope and Y Intercept Worksheet PDF

Example 1.
Find the slope of the line joining the points (4,−6) and (5,−2).

Solution:

Let A(4,−8) and B(5,−2) be two points.
Slope of the line = y2 – y1/x2 – x1
= -2-(-6)/5 – 4
= -2+6/1
= 4/1
= 4
Therefore the slope of the given points are 4.

Example 2.
If the slope of the line joining the points A(x,3) and B(6,−8) is -6/4, find the value of x.

Solution:

Given that the two points are
A(x,3) and B(6,-8)
x1 = x, y1 = 3, x2 = 6, y2 = -8
Given slope = -6/4
We know that
x2 – x1/y2 – y1
6 – x/-8 – 3 = -6/4
6 – x/-11 = -6/4
24 – 4x = -66
24 + 66 = 4x
90 = 4x
x = 90/4
Hence the value of x = 90/4

Example 3.
The following points are plotted in the x-y plane. Find the slope and y-intercept of the line joining each pair of (1,4) & (-2,3).

Solution:

Given that the points are (1,4) and (-2,3)
x1 = 1, y1 = 4, x2 = -2 and y2 = 3
slope is (y2-y1)/(x2-x1)
(3 – 4)/(-2-1)
= -1/-3
Slope = 1/3
then y=mx+c
you get x-3y+4=0
at y axis x=0
y = 4/3
Therefore y-intercept = 4/3

Example 4.
Find the slope of the line, which makes an angle of 40° with the positive direction of the y-axis measured anticlockwise.

Solution:

If a line makes an angle of 40° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° +40° = 130°
Thus, the slope of the given line is
tan 130°
= tan(180° – 50°)
= −tan50°
= − 1.19175

Example 5.
Determine the slope and y-intercept of the line 4x + 16y + 10 = 0

Solution:

Given that the equation is 4x + 16y + 10 = 0
16y = – 4x – 10
y = -4/16x – 10/16
y = -1/4x – 5/8.
Comparing this with y = mx + c,
Then we get m = – 1/4 and c = – 5/8
Therefore, slope = -1/4 and y-intercept = -5/8

Example 6.
The points (-3, 3) and (1, -4) are plotted in the x-y plane. Find the slope and y-intercept of the line joining the points.

Solution:

Let the line graph obtained by joining the points are (-3, 3) and (1, -4) be the graph of y = mx + c.
So, the given pairs of values of (x, y) obey the relation y = mx + c.
Therefore, 3 = -3m + c …….(i)
-4 = m + c …… (ii)
Subtracting (ii) from (i),
then we get
3 + 3 = -2m – m
9 = -3m
-3m = 9
m = 9/-3
m = -3
Putting m = -3 in (ii),
Then
-4 = -3 + c
c = -1.
Now, m = -3
The slope of the line graph = -3,
c = -1
The y-intercept of the line graph = -1

Example 7.
Find the slope and y-intercept of 3x – √4y = 2√4

Solution:

Given that the equation is 3x – √3y = 2√4
– √4y = -3x + 2√4
√4y = 3x – 2√4
y = 3/√4x – 2√4/√4
y = 3/√4x – 2
Comparing the above equation with y = mx + c,
Then the slope m = 3/√4 and y-intercept = -2.

Example 8.
Find the equation of a line in the form of y = mx + c, having a slope of 10 units and an intercept of -12 units.

Solution:

Given that
The slope of the line, m = 10, and The y-intercept of the line, c = -12.
We know that
The slope-intercept form of the equation of a line is y = mx + c.
From the equation
y = 10x – 12
Therefore the required equation of the line is y = 10x – 12.

Example 9.
Determine the slope and y-intercept of the line 4y + 12 = 0

Solution:

Given that the equation is
4y + 12 = 0
4y = -12
y = -12/4
y = 0 and x = -3
Comparing with y = mx + c,
Then we get m = 0 and c = -3
Therefore, slope = 0 and y-intercept = -3

Example 10.
What is the y-intercept of the graph of 2x + 7y = 6?

Solution:

Given that the equation is 2x + 7y = 6
7y = -2x + 6
y = – 2/7x + 6/7
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 13/4. So, the y-intercept = 13/4
and m = -2/7 therefore slope = -2/7