Class 9 Chapter 6 (Lines and Angles) Class Notes

Lines and Angles

(1) Line Segment : It is a line with two end points.It is denoted by \(\overline {AB} \).

(2) Ray: It is a line with one end point.
It is denoted by \(\overrightarrow {AB} \) .

(3) Collinear Points : If three or more points lie on the same line, they are called collinear points; otherwise they are called non – collinear points. 

For the figure shown above, A, B and C are collinear points.

 

For the figure shown above, A, B and C are non - collinear points.

(4) Angle : It is formed when two rays originate from the same end point. The rays which form an angle are called its arms and the end point is called the vertex of the angle.

(5) Types of Angles:
(i) Acute angle: It is the angle whose measure is between 0ᵒ and 90ᵒ.

(ii) Right angle: It is the angle whose measure is equal to 90ᵒ.

(iii) Obtuse angle: It is the angle whose measure is greater 90ᵒ than but less than 180ᵒ.

(iv) Straight angle: It is the angle whose measure is equal to 180ᵒ.

(v) Reflex angle: It is the angle whose measure is greater 180ᵒ than but less than 360ᵒ.

(vi) Complementary angles: The two angles whose sum is 90ᵒ are known as complementary angles. 

For the figure shown above, the sum of angles a & b is 90ᵒ, hence these two angles are complementary angles.

(vii) Supplementary angles: The two angles whose sum is 180ᵒ are known as supplementary angles. 

For the figure shown above, sum of the angles 45ᵒ and 135ᵒ is 180ᵒ, hence these two angles are supplementary angles.

(viii) Adjacent angles: Two angles are said to be adjacent if they have a common vertex, a common arm and their non-common arms are on different side of the common arm.

When two angles are adjacent, then their sum is always equal to the angle formed by the two non common arms.

 

For the figure shown above, ∠ ABD and ∠ DBC are adjacent angles. Here, ray BD is the common arm and B is the common vertex. And ray BA and BC are non common arms.
Here, ∠ ABC = ∠ ABD + ∠ DBC.

(ix) Linear pair of angles: Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The linear pair of angles must add up to 180ᵒ.

 

For the figure shown above, ∠ ABD and ∠ DBC are called linear pair of angles.

(x) Vertically Opposite angles: These are the angles opposite each other when two lines cross. 

For the figure shown above, ∠ AOD and ∠ BOC are vertically opposite angles. Also, ∠ AOC and ∠ BOD are vertically opposite angles.

(xi) Intersecting lines: These are the lines which cross each other.

 

For the figure shown above, lines PQ and RS are the intersecting lines.

(xii) Non-intersecting lines: These are the lines which do not cross each other. 

For the figure shown above, lines PQ and RS are the non-intersecting lines.

(6) Pair of Angles:

Axiom 1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.
Axiom 2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
Theorem 1: If two lines intersect each other, then the vertically opposite angles are equal.
Proof:

  

Suppose AB and CD are two lines intersecting each other at point O.
Here, the pair of vertically opposite angles formed are (i) ∠ AOC and ∠ BOD (ii) ∠ AOD and ∠BOC And we need to prove that ∠ AOC = ∠ BOD and ∠ AOD = ∠ BOC.
Here, ray OA stands on line CD. Hence, ∠ AOC + ∠ AOD = 180° as per linear pair axiom. Similarly, ∠ AOD + ∠ BOD = 180°.
On equating both, we get, ∠ AOC + ∠ AOD = ∠ AOD + ∠BOD
Thus, ∠ AOC = ∠BOD
Similarly, it can be proved that ∠AOD = ∠BOC.

(7) Some Examples:
For Example: ∠ PQR = ∠ PRQ, then prove that ∠ PQS = ∠ PRT.

 

From the figure, we can see that ∠ PQS and ∠ PQR forms a linear pair.
Hence, ∠ PQS +∠ PQR = 180° i.e. ∠ PQS = 180° - ∠ PQR - (i)
Also, from the figure, we can see that ∠ PRQ and ∠ PRT forms a linear pair.
Hence, ∠ PRQ +∠ PRT = 180° i.e. ∠ PRT = 180° - ∠ PRQ
Given, ∠ PQR = ∠ PRQ
Therefore, ∠ PRT = 180° - ∠ PQR - (ii)
From (i) and (ii),
∠ PQS = ∠ PRT = 180° - ∠ PQR
Thus, ∠ PQS = ∠ PRT

For Example: OP, OQ, OR and OS are four rays. Prove that ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.

 

Firstly, let us make ray OT as shown in figure below to make a line TOQ.

 

From the above figure, we can see that, ray OP stands on line TOQ.

Hence, as per linear pair axiom, ∠ TOP + ∠ POQ = 180° - (i)
Similarly, from the figure, we can see that, ray OS stands on line TOQ.
Hence, as per linear pair axiom, ∠ TOS + ∠ SOQ = 180°
But, from the figure, ∠ SOQ = ∠ SOR + ∠ QOR
So, ∠ TOS + ∠ SOR + ∠ QOR = 180° - (ii)
On adding (i) & (ii), we get,
∠ TOP + ∠ POQ + ∠ TOS + ∠ SOR + ∠ QOR = 360°
From the figure, ∠ TOP + ∠ TOS = ∠ POS
Therefore, ∠ POQ + ∠ QOR + ∠ SOR + ∠ POS = 360°.

(8) Parallel lines and a Transversal:
Transversal: It is a line which intersects two or more lines at distinct points. 

Here, line l intersects lines m and n at P and Q respectively. Thus, line l is transversal for lines m and n.
(a) Exterior angles : These are the angles outside the parallel lines.
Here, ∠ 1, ∠ 2, ∠ 7 and ∠ 8 are exterior angles.
(b) Interior angles : These are the angles inside the parallel lines.
Here, ∠ 3, ∠ 4, ∠ 5 and ∠ 6 are interior angles.
(c) Corresponding angles : These are angles in the matching corners.
Here, (i) ∠ 1 and ∠ 5 (ii) ∠ 2 and ∠ 6 (iii) ∠ 4 and ∠ 8 (iv) ∠ 3 and ∠ 7 are corresponding angles.
(d) Alternate interior angles : The angles that are formed on opposite sides of the transversal and inside the two lines are alternate interior angles.
Here, (i) ∠ 4 and ∠ 6 (ii) ∠ 3 and ∠ 5 are alternate interior angles.
(e) Alternate exterior angles : The angles that are formed on opposite sides of the transversal and outside the two lines are alternate exterior angles.
Here, (i) ∠ 1 and ∠ 7 (ii) ∠ 2 and ∠ 8 are alternate exterior angles.

(f) Interior angles on the same side of the transversal : (i) ∠ 4 and ∠ 5 (ii) ∠ 3 and ∠ 6.They are also known as consecutive interior angles or allied angles or co-interior angles.

Axiom 1: If a transversal intersects two parallel lines, then each pair of corresponding angles is equal.
Axiom 2: If a transversal intersects two lines such that a pair of corresponding angles is equal, then the two lines are parallel to each other.
Theorem 1: If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.
Theorem 2: If a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.
Theorem 3: If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.
Theorem 4: If a transversal intersects two lines such that a pair of interior angles on the same side of the transversal is supplementary, then the two lines are parallel.

(9) Lines Parallel to the Same Line:
Theorem 1: Lines which are parallel to the same line are parallel to each other.

For Example: If AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE. 

From the figure, we can see that, ∠ AGE and ∠ GED forms alternate interior angles.
Therefore, ∠ AGE = ∠ GED = 126°
From the figure, we can see that, ∠ GEF = ∠ GED - ∠ FED = 126° - 90° = 36°
Again from the figure, we can see that, ∠ FGE and ∠ AGE forms linear pair.
Therefore, ∠ FGE + ∠ AGE = 180°
∠ FGE = 180° - 126° = 54°.

For Example: AB || CD and CD || EF. Also, EA ⊥ AB. If ∠ BEF = 55°, find the values of x, y and z. 

From the figure, we can see that, ∠ y and ∠ DEF forms interior angles on the same side of the transversal ED.
Therefore, y + 55° = 180° => y = 180° - 55° = 125°
From the figure, we can see that, AB || CD, so as per corresponding angles axiom x = y.
So, x = 125°
From the figure, we can see that, AB || CD and CD || EF, hence, AB || EF.
Therefore, ∠ EAB + ∠ FEA = 180° - (i)
From the figure, ∠ FEA = ∠ FEB + ∠ BEA.
Substituting in (i), we get,
∠ EAB + ∠ FEB + ∠ BEA = 180°
90° + z + 55° = 180° i.e. z = 35°.

(10) Angle Sum Property of a Triangle:

Theorem 1: The sum of the angles of a triangle is 180°.
Theorem 2: If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Proof:

 

For the given triangle PQR, we need to prove that ∠ 1 + ∠ 2 + ∠ 3 = 180°.
Firstly, we will draw line XPY parallel to QR passing through P as shown in figure below. 

From the figure, we can see that ∠ 4 + ∠ 1 + ∠ 5 = 180° - (1)
Here, XPY || QR and PQ, PR are transversals. So, ∠ 4 = ∠ 2 and ∠ 5 = ∠ 3 (Pairs of alternate angles).
Substituting ∠ 4 and ∠ 5 in (1), we get, ∠ 1 + ∠ 2 + ∠ 3 = 180°.
Hence, the sum of the angles of a triangle is 180°.

For Example: The side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠ QTR = 1/2 ∠ QPR.

 

We know that, the exterior angle of triangle is equal to the sum of the two interior angles.
So, ∠ TRS = ∠ TQR + ∠ QTR i.e. ∠ QTR = ∠ TRS - ∠ TQR – (i)
Similarly, ∠ SRP = ∠ QPR + ∠ PQR – (ii)
From the figure, ∠ SRP = 2 ∠ TRS and ∠ PQR = 2 ∠ TQR
Hence, equation (ii) becomes,
2 ∠ TRS = ∠ QPR + 2 ∠ TQR
∠ QPR = 2 ∠ TRS - 2 ∠ TQR => ½ ∠ QPR = ∠ TRS - ∠ TQR – (iii)
On equating (i) and (iii), we get,
∠ QTR = ½ ∠ QPR.

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Class 9 Linear Equations in two variables questions and answers

Questions on Linear Equations in two variables class 9

Attempt these to score better.

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Class 9 Chapter 5 (Introduction to Euclid's Geometry) Class Notes

 

Introduction to Euclid’s Geometry

 

Introduction to Euclid’s Geometry

Euclid was a teacher of mathematics at Alexandria in Egypt, popularly known as ‘Father of Geometry”.

He introduced the method of proving mathematical results by using deductive logical reasoning and the previously proved result.

He collected all his work in a book called “Elements”. This book is divided into thirteen chapters and each chapter is called a book.

Definitions of Euclid’s 

Euclid thought that the geometry is an abstract model of the world which we can see around us. Like the notions of line, plane, surface etc.

He had given these notions in the form of definitions-

1. Anything which has no component is called Point.

2. A length without breadth is called Line.

3. The endpoints of any line are called Points which make it line segment.

4. If a line lies evenly with the points on itself then it is called A Straight Line.

5. Any object which has length and breadth only is called Surface.

6. The edges of a surface are lines.

7. A plane surface is a surface which lies evenly with the straight lines on itself.

Euclid’s Axioms and Postulates

Euclid assumed some properties which were actually ‘obvious universal truth’. He had bifurcated them in two types: Axioms and postulates.

Axioms

Some common notions which are used in mathematics but not directly related to mathematics are called Axioms.

Some of the Axioms are-

1. If the two things are equal to a common thing then these are equal to one another.

If p = q and s = q, then p = s.

2. If equals are added to equals, the wholes are equal.

If p = q and we add s to both p and q then the result will also be equal.

p + s = q + s

3. If equals are subtracted from equals, the remainders are equal.

This is same as above, if p = q and we subtract the same number from both then the result will be the same.

p – s = q - s

4. Things which coincide with one another are equal to one another.

If two figures fit into each other completely then these must be equal to one another.

5. The whole is greater than the part.

This circle is divided into four parts and each part is smaller than the whole circle. This shows that the whole circle will always be greater than any of its parts.

6. Things which are double of the same things are equal to one another.

This shows that this is the double of the two semicircles, so the two semicircles are equal to each other.

7. Things which are halves of the same things are equal to one another. This is the vice versa of the above axiom.

Postulates

The assumptions which are very specific in geometry are called Postulates.

There are five postulates by Euclid-

1. A straight line may be drawn from any one point to any other point.

This shows that a line can be drawn from point A to point B, but it doesn’t mean that there could not be other lines from these points.

2. A terminated line can be produced indefinitely.

This shows that a line segment which has two endpoints can be extended indefinitely to form a line.

3. A circle can be drawn with any centre and any radius.

This shows that we can draw a circle with any line segment by taking one of its points as a centre and the length of the line segment as the radius. As we have AB line segment, in which we took A as the centre and the AB as the radius of the circle to form a circle.

4. All right angles are equal to one another.

As we know that a right angle is equal to 90° and all the right angles are congruent because if any angle is not 90° then it is not a right angle.

As in the above figure ∠DCA =∠DCB =∠HE =∠HGF= 90°

5. Parallel Postulate

If there is a line segment which passes through two straight lines while forming two interior angles on the same side whose sum is less than 180°, then these two lines will definitely meet with each other if extended on the side where the sum of two interior angles is less than two right angles.

And if the sum of the two interior angles on the same side is 180° then the two lines will be parallel to each other.

Equivalent Versions of Euclid’s Fifth Postulate

1. Play fair’s Axiom

This says that if you have a line ‘l’ and a point P which doesn’t lie on line ‘l’ then there could be only one line passing through point P which will be parallel to line ‘l’. No other line could be parallel to line ‘l’ and passes through point P.

2. Two distinct intersecting lines cannot be parallel to the same line.

This also states that if two lines are intersecting with each other than a line parallel to one of them could not be parallel to the other intersecting line.

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Class 9 Chapter 4 (Linear Equations in Two Variables) Class Notes

 

Linear Equations in Two Variables

 

Linear Equations

The equation of a straight line is the linear equation. It could be in one variable or two variables.

Linear Equation in One Variable

The equation with one variable in it is known as a Linear Equation in One Variable.

The general form is px + q = s, where p, q and s are real numbers and p ≠ 0.

Example

x + 5 = 10

y – 3 = 19

These are called Linear Equations in One Variable because the highest degree of the variable is one.

Graph of the Linear Equation in One Variable

We can mark the point of the linear equation in one variable on the number line.

x = 2 can be marked on the number line as follows -

Linear Equation in Two Variables

An equation with two variables is known as a Linear Equation in Two Variables. The general form of the linear equation in two variables is

ax + by + c = 0

where a and b are coefficients and c is the constant. a ≠ 0 and b ≠ 0.

Example

6x + 2y + 5 = 0, etc.

Slope Intercept form

Generally, the linear equation in two variables is written in the slope-intercept form as this is the easiest way to find the slope of the straight line while drawing the graph of it.

The slope-intercept form is

Where m represents the slope of the line and b tells the point of intersection of the line with the y-axis.

Remark: If b = 0 i.e. if the equation is y = mx then the line will pass through the origin as the y-intercept is zero.

Solution of a Linear Equation

• There is only one solution in the linear equation in one variable but there are infinitely many solutions in the linear equation in two variables.

• As there are two variables, the solution will be in the form of an ordered pair, i.e. (x, y).

• The pair which satisfies the equation is the solution of that particular equation.

Example:

Find the solution for the equation 2x + y = 7.

Solution:

To calculate the solution of the given equation we will take x = 0

2(0) + y = 7

y = 7

Hence, one solution is (0, 7).

To find another solution we will take y = 0

2x + 0 = 7

x = 3.5

So another solution is (3.5, 0).

Graph of a Linear Equation in Two Variables

To draw the graph of linear equation in two variables, we need to draw a table to write the solutions of the given equation, and then plot them on the Cartesian plane.

By joining these coordinates, we get the line of that equation.

• The coordinates which satisfy the given Equation lies on the line of the equation.

• Every point (x, y) on the line is the solution x = a, y = b of the given Equation.

• Any point, which does not lie on the line AB, is not a solution of Equation.

Example:

Draw the graph of the equation 3x + 4y = 12.

Solution:

To draw the graph of the equation 3x + 4y = 12, we need to find the solutions of the equation.

Let x = 0

3(0) + 4y = 12

y = 3

Let y = 0

3x + 4(0) = 12

x = 4

Now draw a table to write the solutions.

x	0	4
y	3	0

Now we can draw the graph easily by plotting these points on the Cartesian plane.

Equations of Lines Parallel to the x-axis and y-axis

When we draw the graph of the linear equation in one variable then it will be a point on the number line.

x - 5 = 0

x = 5

This shows that it has only one solution i.e. x = 5, so it can be plotted on the number line.

But if we treat this equation as the linear equation in two variables then it will have infinitely many solutions and the graph will be a straight line.

x – 5 = 0 or x + (0) y – 5 = 0

This shows that this is the linear equation in two variables where the value of y is always zero. So the line will not touch the y-axis at any point.

x = 5, x = number, then the graph will be the vertical line parallel to the y-axis.

All the points on the line will be the solution of the given equation.

Similarly if y = - 3, y = number then the graph will be the horizontal line parallel to the x-axis.

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