What are the lines which lie on the same plane and do not intersect called a parallel lines B perpendicular lines C intersecting lines D collinear lines?

In geometry, it's a common mistake to say a segment and a line are one and the same.

A segment has a definite beginning and a definite end, with each end represented by a point.

Examples of segments include the length of a table, the distance of a straight road, etc.

On the other hand, a line has no definite beginning or end.

A segment is part of a line, but a line is not part of a segment.

We can see so many straight-line examples around us, edges of a building, roads we use to travel.

Let us study about lines in detail.

A line is a figure formed when two points are connected with minimum distance between them, and both the ends extended to infinity.

A line AB is represented by

While lines have no definite beginning or end, they are represented in our day-to-day lives with examples such as railway tracks or the freeway. 

A ray is a part of a line that has only one fixed point and the other point does not have any end.

A ray AB is represented by

While rays have a fixed beginning and no definite end, they are represented in our day-to-day lives with examples such as the sunlight or the light of a torch.

A segment is a part of a line that has a fixed length or we can say that both the ends of a segment are fixed. Segments, sometimes also referred to as line segments.

A segment AB is represented by

While line segments have both fixed ends, they are represented in our day-to-day lives with examples such as the edge of a table or some wire or pole.

Think Tank

• Let's suppose you tied a rope on one point on the Earth and started walking with the other end of the rope in your hand. You are walking in a straight line, after so many months you somehow managed to return to the same point from where you started. One of your friends points out that while you travelled in a line, the figure obtained by the rope was not a line. What answer do you give your friend?

What Are Parallel Lines?

Two lines are said to be parallel lines if they lie in the same plane and never meet.

Or we can say that if two lines do not have any intersection point they are said to be parallel lines.

Intersecting Lines

If two lines meet at a point then they are said to be interesting lines.

What Are Perpendicular Lines?

If two intersecting lines meet each other at \(90^{\circ}\) angle, then they are said to be perpendicular lines.

Or if the angle between two lines is \(90^{\circ}\) then they are perpendicular to each other.

Explore the types of lines in the below simulation.

Horizontal And Vertical Lines

Lines are classified on the basis of their alignment.

Horizontal Lines

The lines which are drawn horizontally are called horizontal lines.

Vertical Lines

The lines which are drawn vertically are called vertical lines.

Oblique or Slanting Lines

The lines which are drawn in a slanting position or it is forming some angle other than 0, 90, 180, 270, 360 degrees with the horizontal or vertical lines are called oblique or slanting line.

Important Notes

• A line does not have any endpoints.
• A ray has only one fixed point.
• A line segment is a part of a line that has a fixed length.
• Perpendicular lines have a 90-degree angle between them.
• Parallel lines do not meet.

Solved Examples

Sam wants to find out the figures which are made up of line segments only. Help him to pick out the correct figures from the following.

Solution

Let's discuss each option one by one:

Option A is made up of 5 line segments.

Option B is made up of 3 line segments and a curved part.

Option C is a circle hance it is made up of only curved parts.

Option D is made up of 4 line segments.

Find the correct types of lines from the figure given below.

Solution

Option A is a pair of parallel lines

Option B is a pair of non-parallel lines or intersection lines.

Option C is an example of perpendicular lines.

Tom is picking the points of intersection of the lines given in the figure below, he observed that there are 5 points of intersection. Is he right?

Solution

The points at which any of these lines are intersecting are: \(\text{A, B, C, D, E, F}\)

Hence there are total 6 points of intersections.

Interactive Questions

Here are a few activities for you to practice.

Select/Type your answer and click the "Check Answer" button to see the result.

We hope you enjoyed learning about lines with the interactive questions. Now, you will be able to easily solve problems on lines.

About Cuemath

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

FAQs on Lines

The general equation of a straight line can be given as \(ax + by + c= 0\).

Where \(a, b, c\) are constants and \(x, y\) are variables.

2. What is the angle between two perpendicular lines?

The angle between the two perpendicular lines is 90 degrees.

3. What is a secant line?

A secant line is a line that intersects at two points with a circle.

The line AB is the secant to the circle.

4. What is a tangent line?

A tangent line is a line that touches the circle at one point.

The line \(L\) is the tangent to the circle at point \(P\).

5. What are the parallel lines?

Two lines are said to be parallel lines if they lie in the same plane and never meet.

ASSUMED KNOWLEDGE

• Fluency with the arithmetic of the rational numbers
• Knowledge of ratios
• Congruent and similar triangles
• Basic algebraic notation
• Fluency with algebraic expressions and equations
• Basic plotting points in the Cartesian plane including plotting points from a table of values.

MOTIVATION

Coordinate geometry is one of the most important and exciting ideas of mathematics. In particular it is central to the mathematics students meet at school. It provides a connection between algebra and geometry through graphs of lines and curves. This enables geometric problems to be solved algebraically and provides geometric insights into algebra.

The invention of calculus was an extremely important development in mathematics that enabled mathematicians and physicists to model the real world in ways that was previously impossible. It brought together nearly all of algebra and geometry using the coordinate plane. The invention of calculus depended on the development of coordinate geometry.

CONTENT

It is expected that students have met plotting points on the plane and have plotted points from tables of values of both linear and non linear functions.

The number plane (Cartesian plane) is divided into four quadrants by two perpendicular axes called the x-axis (horizontal line) and the y-axis (vertical line). These axes intersect at a point called the origin. The position of any point in the plane can be represented by an ordered pair of numbers (x, y). These ordered pairs are called the coordinates of the point.

Once the coordinates of two points are known the distance between the two points and midpoint of the interval joining the points can be found.

Distance between two points

Distances are always positive, or zero if the points coincide. The distance from A to B is the same as the distance from B to A. We first find the distance between two points that are either vertically or horizontally aligned.

EXAMPLE

Find the distance between the following pairs of points.

a A(1, 2) and B(4, 2)b A(1, −2) and B(1, 3)

Solution

Note: The distance AB is obtained from
the difference of the x-coordinates of
the two points.

Note: The distance AB is obtained from
the difference of the y-coordinates of the
two points.

The example above considered the special cases when the line interval AB is either horizontal or vertical. Pythagoras’ theorem is used to calculate the distance between two points when the line interval between them is neither vertical nor horizontal.

The distance between the points A(1, 2) and B(4, 6) is calculated below.

AC = 4 − 1 = 3 and BC = 6 − 2 = 4.

By Pythagoras’ theorem,

AB2 = 32 + 42 = 25

And so AB = 5

The general case

We can obtain a formula for the length of any interval. Suppose that P(x1, y1) and Q(x2, y2) are two points.

Form the right-angled triangle PQX, where X is the point (x2, y1),

PX = x2 − x1 or x1 − x2 and QX = y2 − y1 or y1 − y2

depending on the positions of P and Q.

By Pythagoras’ theorem:

PQ2 = PX2 + QX2

= (x2 − x1)2 + (y2 − y1)2

Therefore PQ = QP =

Note that (x2 − x1)2 is the same as (x1 − x1)2 and therefore it doesn’t matter whether we go from P to Q or from Q to P − the result is the same.

EXAMPLE

Find the distance between the points A(−4, −3) and B(5, 7).

Solution

In this case, x1 = −4, x2 = 5, y1 = −3 and y2 = 7.

AB2 = (x2 − x1)2 + (y2 − y1)2

= (5 − (−4))2 + (7 − (−3))2

= 92 + 102

= 181

Thus, AB =

Note that we could have chosen x1 = 5, x2 = −4, y1 = 7 and y2 = −3 and still obtained the same result. As long as (x1, y1) refers to one point and (x2, y2) the other point, it does not matter which one is which.

EXERCISE 1

Show that the distance between the points A(a, b) and B(c, d) is the same as the
distance between

• the points P(a, d) and Q(c, b)
• the points U(b, a) and V(d, c)

Illustrate both of these.

EXERCISE 2

The distance between the points (1, a) and (4, 8) is 5. Find the possible values of a and use a diagram to illustrate.

The midpoint of an interval

The coordinates of the midpoint of a line interval can be found using averages as we will see.

We first deal with the situation where the points are horizontally or vertically aligned.

EXAMPLE

Find the coordinates of the midpoint of the line interval AB, given:

aA(1, 2) and B(7, 2) b A(1, −2) and B(1, 3)

Solution

between 1 and 7.

Note: 4 is the average of 1 and 7, that is, 4 =

b The midpoint of AB has coordinates 1,

Note that

When the interval is not parallel to one of the axes we take the average of the x-coordinate and the y-coordinate. This is proved below.

Let M be the midpoint of the line AB. Triangles AMS and MBT are congruent triangles (AAS), and so AS = MT and MS = BT.

Hence the x-coordinate of M is the average of 1 and 5.

x =

The y coordinate of M is the average of 2 and 8.

x =

Thus the coordinates of the midpoint M are (3, 5).

The general case

We can find a formula for the midpoint of any interval. Suppose that P(x1, y1) and Q(x2, y2)are two points and let M(x, y) be the midpoint.

Triangles PMS and MQT are congruent triangles (AAS), and so PS = MT and MS = QT.

Hence the x-coordinate of M is the average of x1 and x2, and y-coordinate of M is the average of y1 and y2. Therefore

x =

Midpoint of an interval

The midpoint of an interval with endpoints P(x1, y1) and Q(x2, y2) is

Take the average of the x-coordinates and the average of the y-coordinates.

EXAMPLE

Find the coordinates of the midpoint of the line interval joining the points (6, 8) and (−3, 2).

Solution

The midpoint has coordinates,

, 5

EXAMPLE

If C(3, 6) is the midpoint of line interval AB and A has coordinates (−1, 1), find the coordinates of B.

Solution

Let the coordinates of B be (x1, y1).

x1 − 1 = 6 y1 + 1 = 12

so x1 = 7 so y1 = 11.

Thus B has coordinates (7, 11).

EXERCISE 3

A square has vertices O(0, 0), A(a, 0), B(a, a) and C(0, a).

The gradient of a line

Gradient of an interval

The gradient is a measure of the steepness of line. There are several ways to measure steepness. In coordinate geometry the standard way to define the gradient of an interval AB is

Given the points A(2, 1) and B(5, 6):

gradient of interval AB =

Notice that as you move from A to B along the interval the y-value increases as the x-value increases. The gradient is positive.

gradient of interval AB =

or

gradient of interval BA =

Notice that in this case as we move from A to B the y value decreases as the x value increases. The gradient is negative. Similarly the gradient of BA = −

gradient of line interval AB =

m =

Note that since

which point we take as the second.

defined. This is shown by interval AB.

If the interval is horizontal, the rise is zero
as shown by interval PQ. The gradient of
the interval is zero.

Gradient of a line

The gradient of a line is defined to be the gradient of any interval within the line.

This definition depends on the fact that two intervals on a line have the same gradient.

Suppose AB and PQ are two intervals on the same straight line. Draw right-angled triangles ABX and PQY with sides AX and PY parallel to the x-axis and sides BX and QY parallel to the y-axis.

Triangle ABX is similar to triangle PQY since the corresponding angles are equal. Therefore:

That is, the intervals have the same gradient.

EXAMPLE

A line passes through the points (1, 2) and (5, 10). Find its gradient.

Solution

=

= 2

EXAMPLE

A line passes through the point (5, 7) and has gradient

Solution

18 = 2(x − 5)

9 = x − 5

x = 14

EXERCISE 4

Find the gradient of the line passing through (a, b) and (0, c)

Intercepts

The x-intercept of a line is the point at which it crosses the x-axis.

The y-intercept of a line is the point at which it crosses the y-axis.

In the diagram to the left the y-intercept is at A and the x-intercept at B.

The second diagram shows a line parallel to the x-axis and it has a y-intercept at C.

The third diagram shows a line parallel to the y-axis and it has an x-intercept at D.

Equation of a straight line

Can we find the equation of the line given suitable geometric information about the line? The following shows that this can be done given the gradient of the line and the y-intercept.

The line y = 3x + 2

Consider the line with gradient 3 and y-intercept 2. This passes through the point A(0, 2). Let B(x, y) be any point on this line.

	Gradient of interval AB	=
		=
		=

The gradient of the line is 3.

So,		= 3
Rearranging	y − 2	= 3x
	y	= 3x + 2

So the coordinates of B(x, y) satisfies y = 3x + 2. This is called the equation of the line.

Conversely suppose that B(x, y) satisfies the equation y = 3x + 2, then

We summarise this by saying that the equation of the line is y = 3x + 2.

The equation y = mx + c

Consider the line with gradient m and y-intercept c. If passes through the point A(0, c).
Let B(x, y) be any point on this line.

	Gradient of interval AB	=
		=

We know the gradient of the line is m.

	Therefore		= m
		y − c	= mx
		y	= mx + c

That is, the line in the cartesian plane with gradient m and y-intercept c has equation
y = mx + c. Conversely, the points whose coordinates satisfy the equation y = mx + c always lie on the line with gradient m and y-intercept c.

Vertical and horizontal lines

Vertical lines

In a vertical line all points have the same x-coordinate, but the y-coordinate can take any value. The equation of the vertical line through the point (6, 0) is x = 6. The x-axis intercept is 6. All the points on this line have x-coordinate 6.

In general, the equation of the vertical line through P(a, b) is x = a.

Because this line does not have a gradient it cannot be written in the form

y = mx + b.

Horizontal lines

A horizontal line has gradient 0. In a horizontal line all points have the same y-coordinate, but the x-coordinate can take any value. The equation of the horizontal line through the point (0, 5) is y = 5. The equation of the horizontal line through the point (9, 5) is y = 5.

In general, the equation of the horizontal line through P(a, b) is y = b.

EXAMPLE

Write down the gradient and y-intercept of the line with equation y = 3x − 4.

Solution

The gradient of the line is 3 and the y-intercept is −4.

Sometimes an equation needs to be rearranged before the gradient and y-intercept can be determined. Consider the following example.

EXAMPLE

Rewrite the equation 2x + 3y = 6 in the form y = mx + c and hence find the value of the gradient and y-intercept.

Solution

	2x +	3y	= 6
so		3y	= 6 − 2x
that is,		y	= 2 −
Thus		y	= − + 2

The gradient of the line is −

Equation of a line given its gradient and a point on the line

Let P(x, y) be any point with x ≠ x1 on the line
passing through the point A(x1, y1) and let m
be the gradient of this line.

Using gradient,	m =
and	y − y1 = m(x − x1)

This is the equation of the straight line with gradient m passing through the point A(x1, y1).

EXAMPLE

Find the equation of the line that passes through the point (−2, 3) with gradient −4.

Solution

The equation for this line is:

Note that it is usual to give the answer in the form y = mx + c

Equation of a straight line given two points

Given two points A(x1, y1) and B(x2, y2) the equation of the line passing through the two points can be found.

The gradient m of the line passing through A(x1, y1) and B(x2, y2) =

Substituting into y − y1 = m(x − x1) gives

y − y1 =

Graphing straight lines

Using the equation to sketch the line

If you are given an equation of a straight line and asked to draw its graph all you need to do is find two points whose coordinates satisfy the equation and plot the points. There are two commonly used methods to find two points.

Using the y-intercept and one other point

Using the y-intercept and a second point the equation can be found .

EXAMPLE

Draw the graph of y = 2x + 3.

Solution

Substitute x = 1, so y = 5 giving the point (1, 5)
lies on the line.

Plot the two points and draw the line through them.

This method does not work if the line is parallel to the y-axis.

EXAMPLE

The gradient of a line is −6 and the y-intercept is 2. Find the equation of the line and sketch it.

Solution

Using the y = mx + c form for the equation of a straight line.

The point (0, 2) lies on the line.

The point (1, −4) lies on the line. The graph is shown.

x-intercept y-intercept method

In this method both intercepts are found.

• The x-intercept is found by substituting y = 0 and
• The y-intercept is found by substituting x = 0.

This method does not work if the line is parallel to an axis or passes through the origin.

EXAMPLE

Using the x-intercept y-intercept method sketch the graph of:

Solution

a	When x = 0,	y = −4
	When y = 0,	3x − 4	= 0
		3x	= 4
		x	=

The general form for the equation of a straight line

The equations y = 2x − 3, x = 6 and 2x − 3y = 6 can be written as −2x + y + 3, x − 6 = 0 and 2x − 3y − 6 = 0 respectively.

The general form for the equation of a line is ax + by + c = 0 where a, b and c are constants and a ≠ 0 or b ≠ 0. The equation of every line can be put in general form. The general form is not unique. The equation x + 2y +1 = 0 is the same straight line as
2x + 4y + 2 = 0.

EXERCISE 5

An equilateral triangle ABC has coordinates O(0, 0), B(a, 0) and C(c, d).

a Find c and d in terms of a by using the fact that OB = BC = CO.

b Find the equation of the lines OB, BC and CO.

Parallel and perpendicular lines

Parallel lines

If two lines l1 and l2 are parallel then corresponding angles are equal.

Conversely, if corresponding angles are equal then the lines are parallel.

Theorem

Two lines are parallel if they have the same gradient and conversely,
two lines with the same gradient are parallel.

Proof

In the diagram, two lines are drawn and the right-angled triangles PQX and ABY are added with QX = BY.

If the lines are parallel then

The two triangles are congruent by the AAS test.

Therefore PX = PY and

That is, the gradients are equal.

Conversely. If the gradients are equal

Now QX = BY and therefore PX = AY.

Hence the triangles QPX and ABY are congruent by the SAS test.

Hence the corresponding angles PQX and ABY are equal and the lines are parallel.

EXAMPLE

Show that the line passing through the points A(6, 4) and B(7, 11) is parallel to the line passing through P(0, 0) and Q(2, 14).

Solution

Gradient of AB = = 7

Gradient of PQ = = 7

The two lines have the same gradient and so are parallel.

EXAMPLE

Find the equation of the line that is parallel to the line y = −2x + 6 and passing through the point A(1, 10).

Solution

The gradient of the line y = −2x + 6 is −2.

Therefore the line through the point A(1, 10) parallel to y = −2x + 6 has equation:

Perpendicular lines

When we draw y = 3x and draw a line perpendicular to it passing through the origin then it is clear that y = −ax where a is a small positive number. We will show that the equation is y = −

We are now going to show the surprising result that ‘if two lines are perpendicular then the product of their gradients is −1’ (or if one is vertical and the other horizontal). The converse is also true. That is, ‘ If the product of the gradients of two lines is −1 then they are perpendicular’.

We first consider the case when both lines pass through the origin.

Draw two lines passing through the origin with one of the lines having positive gradient and the other negative gradient.

Form right-angled triangles OPQ and OAB with OQ = OB.

Gradient of the line OA	=
Gradient of the line through OP	=	−
The product	=	− ×
	=	− ×	(since OQ = OB)
	=	−

If the lines are perpendicular,

Therefore

so PQ = AB and the product, −

Conversely If the product is −1, then AB = PQ,

so

Therefore

We have now proved the result for lines through the origin.

If we are given two lines anywhere in the plane, we can draw lines through the origin parallel to the given two lines. The gradient of each new line is the same as the gradient of the corresponding original line. So the result holds for lines that do not necessarily pass through the origin.

EXAMPLE

Show that the line through the points A(6, 0) and B(0, 12) is perpendicular to the line through P(8, 10) and Q(4, 8).

Solution

Gradient of AB	=
	= −2
Gradient of PQ	=
	=
	=

(Gradient of AB) × (gradient of PQ)	= −2 ×
	= −1

Hence the lines are perpendicular.

Parallel and perpendicular lines

• If two non-vertical lines are parallel then they have the same gradient.
  Conversely if two non-vertical lines have the same gradient then they are parallel.
• If two non-vertical lines are perpendicular then the product of their gradients is −1. Conversely if the product of the gradients of two lines is −1 then they are perpendicular.

EXAMPLE

Find the equation of the line which passes through the point (1, 3) and is perpendicular to the line whose equation is y = 2x + 1.

Solution

Gradient of the line y = 2x + 1 is 2.

Gradient of a line perpendicular to this line is −

The required equation is y − 3	=	− (x − 1)
2(y − 3)	=	−(x − 1)
2y + x	=	7

Thus the equation of the required line is 2y + x = 7.

Proofs with coordinate geometry

Coordinate geometry can be used to prove results in Euclidean Geometry. An important aspect of doing this is placing objects on the Cartesian plane in a way that minimises calculations.

EXAMPLE

Prove that the midpoints of a parallelogram bisect each other using coordinate geometry.

Solution

Let the coordinates of the vertices be O(0, 0), A(a, 0), B(a + c, d) and C(c, d).

There is no loss in generality in placing the vertices of the parallelogram on the Cartesian plane in this way.

The midpoint M of OB	=	,
The midpoint N of AC	=	,

M = N and so the midpoints coincided which means that the diagonals bisect each other.

EXAMPLE

Prove that the diagonals of a rhombus bisect each other at right angles using coordinate geometry.

Solution

Let the coordinates of the vertices be O(0, 0), A(a, 0), B(a + c, d) and C(c, d).

Because it is a rhombus all the sides are of equal length.

	OA = AB = BC = CO
	Gradient of OB	= and,
	Gradient of AC	=

The product of the gradients of the diagonals =

OA2 = a2 and by Pythagoras’ theorem, AB2 = c2 + d2

so a2 = c2 + d2.

Hence d2 = −(c2 − a2)

Thus the product of the gradients of the diagonals = −1.

EXERCISE 6

In any triangle ABC prove that AB2 + AC2 = 2(AD2 + DC2)

Where D is the midpoint of BC.

EXERCISE 7

Prove that set of points equidistant from two given points is a straight line.

EXERCISE 8

Prove that the lines joining the midpoints of opposite sides of a quadrilateral and the lines joining the midpoints of its diagonals meet in a point and bisect each other.

LINKS FORWARD

Coordinate geometry leads into many other topics in school mathematics. The techniques of coordinate geometry are used in calculus, functions, statistics and many other important areas.

HISTORY

There were three facets of the development of coordinate geometry.

• The invention of a system of coordinates
• The recognition of the correspondence between geometry and algebra
• The graphic representation of relations and functions

The Greek mathematician Menaechmus (380−320 BC) proved theorems by using a method that was very close to using coordinates and it has sometimes been maintained that he had introduced coordinate geometry.

Apollonius of Perga (262−190 BC) dealt with problems in a manner that may be called an coordinate geometry of one dimension; with the question of finding points on a line that were in a ratio to the others. The results and ideas of the ancient Greeks provided a motivation for the development of coordinate geometry.

Coordinate geometry has traditionally been attributed to René Descartes (1599−1650) and Pierre de Fermat (1601−1665) who independently provided the beginning of the subject as we know it today.

ANSWERS TO EXERCISES

EXERCISE 1

AB2 = PQ2= UV2 = (d − b)2 + (c − a)2

EXERCISE 2

a = 4 or a = 12

EXERCISE 3

a

,

b

diagonal = a, radius =

c

x − 2 + y − 2 =

EXERCISE 4

EXERCISE 5

a

c = , d =

b

OA: y = 0, OC: y = , OB: y = (x − a)

EXERCISE 6

Place the triangle so D is at the origin. Then let the coordinates of B and C be (−a, 0) and (a, 0) respectively. Let the coordinates of A be (d, c).

EXERCISE 7

Let P(x, y) be a point equidistant from A(a, b) and C(c, d)

PA = PC

2(c − a)x + 2(d − b)y = d2 + c2 − a2 − b2

EXERCISE 8

Let the coordinates of the vertices be O(0, 0), A(a, c), B(m, n) and D(b, 0)

Show that the midpoint of all the required line segments has coordinates

The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.

The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.

© The University of Melbourne on behalf of the International Centre of Excellence for Education in Mathematics (ICE-EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
https://creativecommons.org/licenses/by-nc-nd/3.0/