Vijaya Sree

Worksheet on Graph of Linear Relations in x, y with step-wise solutions is available on this page. So, the students of grade 9 can go through the problems provided here and try to solve them and test how much they have learned from the chapter coordinate geometry. This page is the one-stop for the students who are lagging in graphing linear equations using x, y.

We will learn how to draw the graphs of linear relations in x, y coordinates. Hence practice the problems given below and score better grades in the exams. These concepts will be helpful for you in further classes.

See More:

• Slope of the Graph of y = mx + c
• Worksheet on Plotting Points in the Coordinate Plane
• Worksheet on Slope and Y-intercept

Graphing Linear Relations in x, y Worksheet Answers

You can see different types of equations in the x-y plane in the graph of linear relations worksheet.

Example 1.
Draw the graph of the linear equation in two variables x – 3y + 1 = 0

Solution:

Given equation is x – 3y + 1 = 0
x + 1 = 3y
y = x + 1/3
In the given equation
If x = -1 then y = (-1) + 1/3 = -2/3
If x = 0 then y = (0) + 1/3= 1/3
If x = 1 then y = (1) + 1/3 = 4/3
If x = 2 then y = (2) + 1/3 = 7/3
Plot the graph using the points (-1,-2/3), (0,1/3), (1,4/3), (2,7/3).

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Example 2.
Draw the graph of the linear equation in two variables x + y = 1

Solution:

x + y = 1
– y = x – 1
y = -x + 1
If x = 0 then y = 0 + 1 = 1
If x = 1 then y = -1 + 1 = 0
If x = 2 then y = -2 + 1 = -1
If x = 4 then y = – 4 + 4 = 0
Then plot the following points in the graph (0,1), (1,0), (2,-2) and (4,0)

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Example 3.
Draw the graph of the linear equation in two variables y = 3x + 2.

Solution:

Given equation is y = 3x + 2
In the given equation
If x = -2 then y = 3(-2) + 2 = -6+2 = -4
If x = -1 then y = 3(-1) + 2 = -2
If x = 0 then y = 3(0) + 2 = 2
If x = 1 then y = 3(1) + 2 = 5
If x = 2 then y = 3(2) + 2 = 8
Plot the graph using the points (-2,-4), (-1,-2), (0,2), (1,5) and (2,8).

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Example 4.
Draw the graph of the linear equation in two variables x − y = 6

Solution:

x – y = 6
– y = 6 – x
y = x – 6
If x = 0 then y = 0 – 6 = -6
If x = 1 then y = 1 – 6 = -5
If x = 2 then y = 2 – 6 = -4
If x = 4 then y = 4 – 6 = -2
Then plot the graph using the points (0,-6), (1,-5), (2,-4), (4,-2).

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Example 5.
Draw the graph of the linear equation in two variables y = 3x + 1.

Solution:

Given equation is y = 3x + 1
In the given equation
If x = -5 then y = 3(-5) + 1 = -15+ 1 = -14
If x = -1 then y = 3(-1) + 1 = -2
If x = 0 then y = 3(0) + 1 = 1
If x = 1 then y = 3(1) + 1 = 4
If x = 2 then y = 3(2) + 1 = 7
Plot the graph using the points (-5,-14), (-1,-2), (0,1), (1,4) and (2,7).

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Example 6.
Draw the graph of the linear equation in two variables 4x + 3y + 1 = 0

Solution:

Given equation is 4x + 3y + 1 = 0
-3y = 4x + 1
y = -4x/3 -1/3
In the given equation
If x = -1 then y = -4(-1)/3 – 1/3 = 1
If x = 0 then y = (0)/3 – 1/3= -1/3
If x = 1 then y = -4(1)/3 – 1/3 = -5/3
If x = 2 then y = -4(2)/3 – 1/3 = -9/3
Plot the graph using the points (-1,1), (0,-1/3), (1,-5/3), (2,-9/3).

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Example 7.
Draw the graph of the linear equation in two variables 2x + y = 4

Solution:

2x + y = 4
– y = 2x – 4
y = -2x + 4
If x = 0 then y = -2(0) + 4 = 4
If x = 1 then y = -2(1) + 4 = 2
If x = 2 then y = -2(2) + 4 = 0
If x = 4 then y = -2(4) + 4 = -4
Then plot the following points in the graph (0,4), (1,2), (2,0) and (4,-4)

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Example 8.
Draw the graph of the linear equation in two variables 2x − y = 6

Solution:

2x – y = 6
– y = 6 – 2x
y = 2x – 6
If x = 0 then y = 2(0) – 6 = -6
If x = 1 then y = 2(1) – 6 = -4
If x = 2 then y = 2(2) – 6 = -2
If x = 4 then y = 2(4) – 6 = 2
Then plot the graph using the points (0,-6), (1,-4), (2,-2), (4,2).

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Example 9.
Draw the graph of the linear equation in two variables x – y + 2 = 0

Solution:

Given equation is x – y + 2 = 0
x + 2 = y
In the given equation
If x = -1 then y = (-1) + 2 = 1
If x = 0 then y = (0) + 2 = 2
If x = 1 then y = (1) + 2 = 3
If x = 2 then y = (2) + 2 = 4
Plot the graph using the points (-1,1), (0,2), (1,3), (2,4).

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Example 10.
Draw the graph of the linear equation in two variables x + y = 6

Solution:

x + y = 6
– y = x – 6
y = -x + 6
If x = 0 then y = 0 + 6 = 6
If x = 1 then y = -1 + 6 = 5
If x = 2 then y = -2 + 6 = 4
If x = 4 then y = – 4 + 6 = 2
Then plot the following points in the graph (0,6), (1,5), (2,4) and (4,2)

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In maths, coordinate geometry is one of the most important and interesting concepts at the secondary level. Coordinate geometry is the part of the geometry that uses two or more to specify the point. The position of the point or figure can be determined in a line or three-dimensional space. The concept of coordinate geometry is explained with definitions, formulas, examples here. In addition to this, we also provide worksheets on coordinate geometry to test your knowledge of this chapter.

Topics in Co-ordinate Geometry Grade 9 | Coordinate Geometry List of Contents

• Independent Variables and Dependent Variables
• Coordinates of a Point
• Rectangular Cartesian Coordinates of a Point
• Quadrants and Convention for Signs of Coordinates
• Plotting a Point in Cartesian Plane
• Coordinate Geometry Graph
• Graph of Standard Linear Relations Between x, y
• Slope of the Graph of y = mx + c
• y-intercept of the Graph of y = mx + c
• Drawing Graph of y = mx + c Using Slope and y-intercept
• Problems on Plotting Points in the x-y Plane
• Problems on Slope and Y-intercept
• Worksheet on Plotting Points in the Coordinate Plane
• Worksheet on Graph of Linear Relations in x, y
• Worksheet on Slope and Y-intercept

Introduction to Coordinate Geometry

Coordinate geometry is determined as the study of geometry using coordinate points. With the help of the points, we can find the distance between the two points, divide the lines into m:n ratio, find the midpoint, calculate the area of a triangle in the cartesian plane, and so on.

Definition of Coordinate geometry	It is a part of the geometry where the position of the point is determined using coordinates.
What is meant by coordinate?	Coordinates are nothing but the set of values that helps to show the exact point in the coordinate plane.
Distance Formula	The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).
What is meant by Coordinate Plane?	A coordinate plane is a two-dimensional plane that is formed by the intersection of two perpendicular lines that is x-axis and y-axis.
Section Formula	Section formula is used to divide any line into two parts m:n ratio
Mid-point theorem	Mid-point theorem is used to find the coordinates at which a line is divided into two parts.

What is a Co-ordinate and a Co-ordinate Plane?

Coordinate and coordinate plane sounds the same but they are different. A coordinate is the set of values that is used to show the exact point in the coordinate plane. Whereas a coordinate plane is a 2D plane that is formed by the intersection of two perpendicular lines that is x-axis and y-axis. The point where the two axes intersect is known as the origin. The location of the point (x, y) is known as the coordinates.
There are four quadrants in the graph they are,
Quadrant 1: (+x, +y)
Quadrant 2 : (-x, +y)
Quadrant 3 : (-x, -y)
Quadrant 4 : (+x, -y)

Equation of a Line in Cartesian Plane

Equation of the line can be represented as follows,
i. General Form:
The general form of a line is given as Ax + By + C = 0.
ii. Slope intercept form:
Let x, y be the coordinate of a point that passes through the line, m be the slope of the line, and c be the y-intercept.
y=mx + c
iii. Intercept Form of a line:
Consider a and b be the x-intercept and y-intercept of a line, then the equation of a line is represented as
y=mx + c

Slope of a Line

The general form of the line is Ax + By + C = 0.
The slope can be found by converting this form to the slope-intercept form.
Ax + By + C = 0
By = − Ax – C
By = − Ax – C
y = -A/B x – C/B
Comparing the above equation with y = mx + c,
m = -A/B
By using this formula we can find the slope of a line from the general equation of a line.

Coordinate Geometry Formulas and Theorems

Distance Formula:
The distance formula is used to find the distance between two points A(x1, y1) and B(x2, y2).
d = √(x2 – x1)² + (y2 – y1)²
where (x1, y1) and (x2, y2) are the coordinates of the plane.
Mid-point Theorem:
The mid-point theorem is used to find the coordinates at which a line is divided into two parts. Consider two points A and B having the coordinates (x1, y1) and (x2, y2)
M(x, y) = [(x1+x2)/2, (y1+y2)/2]
Angle Formula:
Consider two lines A and B, having their slopes to be m1 and m2 respectively.
Let “θ” be the angle between these two lines, then the angle between them.
tanθ = (m1-m2)/1+m1m2
Case 1: When the two lines are parallel to each other,
m1 = m2 = m
tanθ = (m1-m2)/1+m1m2 = 0
Thus θ = 0
Case 2: When the two lines are perpendicular to each other,
m1.m2 = -1
tanθ = (m1-m2)/1+m1m2
tanθ = (m1-m2)/1+(-1) = (m1-m2)/0 = undefined
θ = 90°

Coordinate Geometry Example Questions

Example 1.
Find the distance of the point P (7, 8) from the x-axis.
Solution:
We know that,
(x, y) = (7, 8) is a point on the Cartesian plane in the first quadrant.
x = Perpendicular distance from the y-axis
y = Perpendicular distance from x-axis
Therefore, the perpendicular distance from x-axis = y coordinate = 8

Example 2.
Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 4) and B (3, 6).
Solution:
Let P (x, y) be equidistant from the points A (2, 4) and B (3, 7).
Therefore AP = BP …[Given]
AP² = BP² …[Squaring both sides]
(x – 2)² + (y – 4)² = (x – 3)² + (y – 6)²
x² – 4x + 4 + y² – 8y + 16 = x² – 6x + 9 + y² -12y + 36
-4x -8y + 20 = -6x – 12y + 45
-4x -8y + 20 + 6x + 12y – 45 = 0
2x + 4y – 25 = 0
2x + 4y = 25
2x – 25 = -4y
2x – 25 = -4y is the required relation.

Example 3.
Three vertices of a parallelogram taken in order are (1,0) (2,1) and (2,3) respectively. Find the coordinates of the fourth vertex
Solution:
Consider D(x,y) as the fourth vertex.
Let A(1,0) B(2,1) C(2,3) and D(x,y) be the vertices of a parallelogram.
ABCD took each other.
Since the diagonal of a parallelogram bisects each other.
Coordinates of the mid point of AC = coordinates of the midpoint of BD.
(1 + 2/2 , 0 + 3/2) = (2 + x/2 , 1 + y/2)
(3/2 , 3/2) = (2 + x/2 , 1 + y/2)
2 + x/2 + 3/2
= 2 + x = 3
x = 3 – 2
x = 1
And
1 + y/2 = 3/2
1 + y = 3
y = 3 – 1
y = 2
Hence coordinates of the fourth vertex D(1,2)

Example 4.
Find that value(s) of x for which the distance between the points P(x, 3) and Q(7, 10) is 10 units.
Solution:
Given that
PQ = 10
PQ² = 10²
PQ² = 100
(7 – x)² + (10 – 3)² = 100
Using distance formula
(7 – x)² + (10 – 3)² = 100
(7 – x)² = 100 – 49
(7 – x)² = 51
(7 – x) = √51
7 – x = + or – 7.1
7 – x = 7.1 or 7 – x = -7.1
x = 7.1 – 7 or x = -7.1 – 7
x = 0.1 or x = -14.1

Example 5.
Find the centroid of the triangle whose vertices are (1,2) (3,4) (5,6).
Solution:
The centroid of the triangle whose vertices are (1,2) (3,4) and (5,6)
The centroid of the triangle ABC = (x1 + x2 + x3/2, y1 + y2 + y3/2)
The centroid of the triangle whose vertices A(1,2) B(3,4), C(5,6)
= (1+3+5/2 ,2+ 4 + 6/2)
(9/2, 12/2)
(9/2, 6).

FAQs on Coordinate Geometry

1. How can you prepare and complete coordinate geometry?

Pay attention to the classes and practice the problems thoroughly. Make use of the links given on our page and clarify your doubts.

2. How Can You Define the Position of an Object on a Floor?

The students of 9th grade can define the position of an object on the floor by considering two adjacent walls as two coordinate axes.

3. What are the signs of the coordinates in the four quadrants of a cartesian plane?

The signs of the coordinates in the four quadrants of a cartesian plane are (+, +) in the first quadrant, (-, +) in the second quadrant, (-, -) in the third quadrant of a cartesian plane, and (+, +) in the fourth quadrant of a cartesian plane.

In the previous articles, we have discussed about slope and y-intercept, plotting points in the XY plane, independent variables and dependent variables, etc. Let us learn how to solve the slope of the graph y = mx + c from this page. In addition that you can find the problems on the slope of the graph y = mx + c in the below section.

Also Read:

• Problems on Slope and Y-intercept
• Worksheet on Slope and Y-intercept
• y-intercept of the Graph of y = mx + c

Introduction to Slope-Intercept Form | y=mx+c Equation of Straight Line Slope

y = mx + c is an equation of the line having a slope and y-intercept of the line in coordinate geometry. m represents the slope or gradient of the line and c is the y-intercept that cuts the point on the y-axis. The line cuts the y-axis at the point (0, c) which is the distance from the origin to the point c.
m = (y – c)/(x – 0)
m = (y – c)/x
mx = y – c
mx + c = y
y = mx + c

y = mx + c Straight Line Graphs Slope Examples

Example 1.
What is the slope of the equation 6x – 4y + 5 = 0?
Solution:
Given that the equation is
6x – 4y + 5 = 0
4y = 6x + 5
y = 6/4x + 5/4.
y = 3/2x + 5/4
Comparing the equation with y = mx + c,
we have m = 3/2.
Therefore, the slope of the line is 3/2

Example 2.
Find the equation of a line in the form of y = mx + c, having a slope of 4 units and an intercept of -6 units.
Solution:
Given that
The slope of the line, m = 4 and The y-intercept of the line, c = -6.
We know that
The slope-intercept form of the equation of a line is y = mx + c.
From the equation
y = 4x – 6
Therefore the required equation of the line is y = 4x – 6

Example 3.
Convert the equation 5x + 4y = 12 into y = mx + c and find its y-intercept.
Solution:
Given that the equation is
5x + 6y + 12 = 0
6y = 5x + 12
y = 5/6x + 12/6.
y = 5/6x + 2
Comparing the equation with y = mx + c,
we have m = 5/6 and c = 2
Therefore, the slope of the line is 5/6. y-intercept is c = 2

Example 4.
What is the slope of the equation 2x – 4y + 10 = 0?
Solution:
Given that the equation is
2x – 4y + 10 = 0
4y = 2x + 10
y = 2/4x + 10/4.
y = 1/2x + 5/2
Comparing the equation with y = mx + c,
we have m = 1/2.
Therefore, the slope of the line is 1/2.

Example 5.
What is the slope and y-intercept of the line 3x – 6y + 12 = 0?
Solution:
Given that the equation is
3x – 8y + 12 = 0
8y = 3x + 12
y = 3/8x + 12/8.
y = 3/8x + 3/2
Comparing the equation with y = mx + c,
we have m = 3/8 and c = 3/2
Therefore, the slope of the line is ⅜ and y intercept c = 3/2

FAQs on Slope of the graph y = mx + c

1. What is the equation of a straight line?

The equation of the straight line is y = mx +c
where,
m is the slope and c is the y-intercept.

2. How do I find the slope in a graph?

Take two points on the line and determine their coordinates. Determine the difference in y-coordinates of these two points. Find the difference in x-coordinates for these two points. Divide the difference in y-coordinates by the difference in x-coordinates.

3. What is mx in the slope formula?

In the equation of a straight line y = mx + c, the slope is m which is multiplied by x and c is the y-intercept.

y = mx + c is called the slope-intercept form equation of the straight line. In this article, we are going to learn more details about the y = mx + c graph of the straight line. In y = mx + c, m represents the slope value of the equation and c is the y-intercept. The graph y = mx + c cuts the y-axis at the point p the OP is the y-intercept of the graph and O is the origin.

Related Topics:

• Coordinate Geometry
• Problems on Slope and Y-intercept
• Worksheet on Slope and Y-intercept

y-intercept of y = mx + c

The general form of the equation of the straight line is y = mx + c where m is slope or gradient and c is the y-intercept that cuts the y-axis at some point c. It is a linear equation and the variables x and y are the coordinates on the plane. x, y are independent and dependent variables respectively. The y-intercept of the line that is on the graph y = mx + c is c.

Slope y = mx+c Formula

The equation y = mx + c is derived from the slope formula.
m = (y-c)/(x-0)
m = (y-c)/x
mx = y – c
mx + c = y
y = mx + c
Thus the slope-intercept form of the equation of the line is derived.

How to find the y-intercept of Equation of Straight Line y = mx + c?

1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
2. Substitute x = 0 into the equation to find the y-intercept.
3. Declare the coefficient of x.

Slope Intercept Form of Equation y = mx + c Examples

Example 1.
What is the y-intercept of the graph of 6x + 2y = 3?
Solution:
Given that the equation is 6x + 2y = 3
2y = -6x + 3
y = – 6/2x + 3/2
y = -3x + 3/2
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3/2. So, the y-intercept = 3/2

Example 2.
What is the y-intercept of the graph of x + y = 3?
Solution:
Given that the equation is 7x + 2y = 3
y = -x + 3
y = – x + 3
y = -x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3

Example 3.
What is the y intercept of the graph of 5y = 2x + 3
Solution:
Given that the equation is 5y = 2x + 3
y = 2/5x + 3/5
We know that
y = mx + c
Comparing the equation with y = mx + c
We get c = 3/5
Therefore the y-intercept = 3/5

Example 4.
What is the y-intercept of the graph of 4x + 2y = 6?
Solution:
Given that the equation is 4x + 2y = 6
4x + 2y – 6 = 0
2y = -4x + 6
y = – 4/2x + 6/2
y = -2x + 3
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 3. So, the y-intercept = 3

Example 5.
What is the y-intercept of the graph of 7x + 8y = 26?
Solution:
Given that the equation is 7x + 8y = 26
8y = -7x + 26
y = – 7/8x + 26/8
y = -7/8x + 13/4
We know that
y = mx + c,
Comparing the equation with y = mx + c
we get c = 13/4. So, the y-intercept = 13/4

FAQs on y-intercept of the Graph of y = mx + c

1. What is y = mx + c?

The expression y = mx + c is an equation of a line
Here m is the slope of the line and the y-intercept of c.
This equation is formed by knowing the slope of the line and the intercept which the line cuts on the y-axis. This equation y = mx + c is the basic equation of the line.

2. How to use y = mx + c?

1. To find the y-intercept of the line y = mx + c we have to rearrange the equation to make y the subject.
2. Substitute x = 0 into the equation to find the y-intercept.
3. Declare the coefficient of x.

3. What is the y-intercept of the graph 5x + 2y = 5?

5x + 2y = 5
2y = -5x + 5
y = -5/2 x + 5/2
Now compare the equation with y = mx + c
c = 5/2 so, the y-intercept = 5/2.

Plotting a Point in Cartesian Plane is very easy if you know how to draw a cartesian plane in a coordinate geometry graph. A cartesian graph is an intersection of two points at O. They are represented by two numbers in parentheses, separated by a comma that is called coordinates. Learn more about the cartesian plane from here and know how to draw a cartesian plane and then plot points on the cartesian graph. Scroll down this page to learn plotting points on a cartesian plane with solved examples.

Read Articles:

• Worksheet on Plotting Points in the Coordinate Plane
• Quadrants and Convention for Signs of Coordinates

What is Cartesian Plane?

The cartesian plane is a representation of the position of a point by referring to it in terms of x-axis horizontal line and y-axis vertical line respectively. The point of intersection of x-axes and y-axes is known as the origin. The origin is represented by the letter ‘O’. The coordinates have two values the first number represents the x-coordinate and the second number represents the y-coordinate.

How to Plot Points in Cartesian Plan?

1. First, check the signs of the coordinates of a point and then identify the quadrant in which the point is to be plotted.
2. Take the rectangular cartesian frames XOX’ and YOY’ intersecting at the right angles at origin ‘O’.
3. Take a point P on the x-axis on the side of the involved quadrant such that the distance of the point P from the origin O equals the value of the x-coordinate.
4. Draw a perpendicular line PQ on the x-axis. Take a point M on this perpendicular such that PM is equal to the value of the y-coordinate and M in the relevant quadrant. Thus the point M is plotted as per the given coordinate points.

Plotting a Point in Cartesian Plane Examples

Plot the following points in the cartesian plane then identify which quadrant or axis it belongs

Example 1.
Plot the points (3,4) in the Cartesian plane.
Solution:
Given that the points are (3,4)
Here the x-coordinate is 3 and the y-coordinate is 4.
(3, 4) is a point in the first quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (3,4)

Example 2.
Plot the points (5,-6) in the Cartesian plane.
Solution:
Given that the points are (5,-6)
Here the x-coordinate is 5 and the y-coordinate is -6.
(5, -6) is a point in the fourth quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (5,-6)

Example 3.
Plot the points (-7,-8) in the Cartesian plane.
Solution:
Given that the points are (-7,-8)
Here the x-coordinate is -7 and the y-coordinate is -8.
(-7, -8) is a point in the third quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (-7,-8)

Example 4.
Plot the points (-9,2) in the Cartesian plane.
Solution:
Given that the points are (-9,2)
Here the x-coordinate is -9 and the y-coordinate is 2.
(-9, 2) is a point in the second quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (-9,2)

Example 5.
Plot the points (1,1) in the Cartesian plane.
Solution:
Given that the points are (1,1)
Here the x-coordinate is 1 and the y-coordinate is 1.
(1, 1) is a point in the first quadrant because both the coordinates are positive.
Now plot the graph using the coordinates (1,1).

Invariant Points for Reflection in a Line is nothing a point that remains unvaried after the transformation applied to it. Any point on a graph on the line of reflection is an invariant point. Learn more about Invariant Points Under Reflection in a Line from this article. By making this page as a reference you can know about the Invariant Points on the graph on the line of reflection which helps you to score good marks in the exams.

Also Read:

• Reflection of a Point in a Line Parallel to the x-axis
• Reflection of a Point in a Line Parallel to the y-axis

Invariant Points for Reflection in a Line – Definition

Invariant points means the points which lie on the line and when reflected in the line. So, only those points are invariant which lie on the y-axis.
The invariant points must have x-coordinate = 0.
Therefore, only (0, 4) is the invariant point. No points lying outside the line will be an invariant point.

Invariant Points in Reflection in a Line Examples with Answers

Example 1.
Which of the following points (-2, 0), (0, -5), (3, -3) are invariant points when reflected in the x-axis?
Solution:
Given points are (-2,0), (0,-5) and (-3,3)
We know that
The points lying on the line are invariant points. when reflected in the line.
So, only those points are invariant which lie on the x-axis.
Hence, the invariant points must have y-coordinate = 0.
Therefore, only (-2, 0) is the invariant point.

Example 2.
Which of the following points (-2, 3), (0, 6), (4, -3), (-3, 6) are invariant points when reflected in the line parallel to the x-axis at a distance of 4 on the positive side of the y-axis?
Solution:
Given points are (-2,3), (0,6), (4,-3) and (-3,6)
We know that,
The points lying on the line are invariant points when reflected in the line.
So, only those points are invariant which are on the line parallel to the x-axis at a distance of 4 on the positive side of the y-axis.
Hence, the invariant points must have a y-coordinate = 4.
Therefore, (0, 6) and (-3, 6) are the invariant points.

Example 3.
Points A and B have coordinates (4,6) respectively. Find the reflection of A’ of A under reflection in the x-axis and A” of A under reflection in the y axis.
Solution:
Given that the coordinates are (4,6)
A’ = image of A under reflection in the x axis = (4,-6).
A” = image of A under reflection in the y axis = (-4,6).

Example 4.
A point P(a,b) is reflected in the x-axis to P'(6,-5). 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 y-axis such that x = 3.
Solution:
A point P(a,b) is reflected in the x axis P'(6,-5)
We know that Mx(x,y) = (x,-y)
Thus, the coordinates of P are (6,5)
Hence a = 6, and b = 5.
P” = image of P reflected in the y axis = (-6,5)
P”‘ = reflection of P in the line (x = 3) = (-9,5).

Example 5.
Which of the following points (-1, 0), (0, -8), (2, -7) are invariant points when reflected in the x-axis?
Solution:
Given points are (-1,0), (0,-8) and (2,-7)
We know that
The points lying on the line are invariant points. when reflected in the line.
So, only those points are invariant which lie on the x-axis.
Hence, the invariant points must have y-coordinate = 0.
Therefore, only (-1, 0) is the invariant point.

FAQs on Invariant Points Under Reflection in a Line

1. Is a line of invariant points an invariant line?

In linear transformation maps the origin to the origin.
So the origin is always an invariant point under a linear transformation. Every point on the invariant line maps to a point on the line itself.

2. What are invariant points?

Invariant points are points on a line or a shape which do not move when the specific transformation is applied.

3. How do you find invariant points on a graph?

Sketch each graph by using the key points, including invariant points, Determine the image points on the graph by square rooting the points. And locate invariant points on y = f(x) and y = g(x). When graphing the square root of a function, invariant points occur at y = 0 and y = 1.

You must have searched various sites to learn about Rectangular Cartesian Coordinates of a Point. Cartesian coordinates are the pair of numbers that specify the distance on the coordinate axis. Rectangular Cartesian Coordinates of a Point are defined by the ordered pairs x and y that indicate the points on the graph. Refer to our page to gather more information regarding the rectangular cartesian coordinates of a point. In addition to the concept of cartesian coordinates, the students can also find the solved examples on Rectangular Cartesian Coordinates of a Point from this article.

Also, See:

• Worksheet on Plotting Points in the Coordinate Plane
• Problems on Plotting Points in the x-y Plane

Rectangular Cartesian Coordinates of a Point – Definition

In order to find the rectangular cartesian coordinates of a point, you have to take two intersecting lines x-axis and the y-axis. They are perpendicular to each other which cuts at the origin O. Let P be the point in a plane. Draw the perpendicular lines from P to x-axis and P to y-axis. That point is M and N. Measure the lines PA and PB in a and b units. Here (a,b) are called the rectangular Cartesian coordinates of a point. The point P represents the position of the point in a plane.

Rectangular Coordinate System Examples with Answers

Example 1.
Find the new coordinates of the point (3,4) if the origin is shifted to the point (1,2) by translation of the axis.
Solution:
The given points are
(h,k)=(1,2) and (x,y)=(3,4)
Therefore new coordinates (X,Y)
x=X+h and y=Y+k
Therefore 3=X+1 and 4=Y+2
Which gives, X=4 and Y=2.
Hence, the new coordinates are (4,2)

Example 2.
Plot the rectangular coordinates of the points (2,3) (-2,4).
Solution:
Given that
The first point is (2,3)
Here the x coordinate is positive and the y coordinate is also positive, so the point lies in the first coordinate and plots the graph using the points (2,3).
The second point is (-2,4).
Here the x coordinate is negative and the y coordinate is positive, so the points lie in the second quadrant and plot the graph using the points (-2,4).

Example 3.
Find the new coordinates of the point (5,6) if the origin is shifted to the point (3,7) by translation of the axis.
Solution:
The given points are
(h,k)=(5,6) and (x,y)=(3,7)
Therefore new coordinates (X,Y) .
x=X+h and y=Y+k
Therefore 3 = X+5 and 7 = Y+6
Which gives, X=2 and Y= -1
Hence, the new coordinates are (2,-1)

Example 4.
Plot the rectangular coordinates of the points (-1,-2) (-12,8)
Solution:
Given that
The first point is (-1,-2)
Here the x coordinate is negative and the y coordinate is also negative, so the point lies in the third coordinate and plots the graph using the points (-1,-2).
The second point is (-12,8)
Here the x coordinate is negative and the y coordinate is positive, so the points lie in the second quadrant and plot the graph using the points (-12,8).

Example 5.
Plot the rectangular coordinates of the points (7,-6)
Solution:
Given that
The first point is (7,-6)
Here the x coordinate is positive and the y coordinate is negative, so the point lies in the fourth coordinate and plot the graph using the points (2,3).

Frequently Asked Questions

1. What is a rectangular Cartesian coordinate?

Cartesian coordinates are also called rectangular coordinates, it is a method of plotting graphs and pointing the positions of points on a two-dimensional surface or in three-dimensional space. The Cartesian plane consists of two perpendicular axes that cross the origin.

2. What is the point in rectangular coordinates?

In the rectangular coordinate every point is represented as an ordered pair. The first number in the ordered pair is called the x-coordinate, and the second number is called the y-coordinate.

3. How are polar coordinates and rectangular coordinates related?

Rectangular coordinates or cartesian coordinates come in the form of (x,y). Polar coordinates on the opposite hand, are available in the shape (r,θ). rather than moving out from the origin victimization horizontal and vertical lines, we tend to instead choose the angle θ, that is that the direction, then moves out from the origin a particular distance r.