Welcome dear students! Today we are going to learn about Lines and Angles from Class 9 Maths.
In Chapter 5, you studied that a minimum of two points are required to draw a line. You also studied some axioms and, with the help of these axioms, proved some other statements. In this chapter, you will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Further you will use these properties to prove some statements using deductive reasoning. You have already verified these statements through some activities in the earlier classes. In your daily life, you see different types of angles formed between the edges of plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles. For instance, suppose you want to make a model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you would make it? You would keep some of the sticks parallel to each other, and some sticks would be kept slanted. Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles. Without the knowledge of the properties of these lines and angles, do you think she can draw the layout of the building? In science, you study the properties of light by drawing the ray diagrams. For example, to study the refraction property of light when it enters from one medium to the other medium, you use the properties of intersecting lines and parallel lines. When two or more forces act on a body, you draw the diagram in which forces are represented by directed line segments to study the net effect of the forces on the body. At that time, you need to know the relation between the angles when the rays or line segments are parallel to or intersect each other. To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle formed between the horizontal and the line of sight. Plenty of other examples can be given where lines and angles are used. In the subsequent chapters of geometry, you will be using these properties of lines and angles to deduce more and more useful properties. Let us first revise the terms and definitions related to lines and angles learnt in earlier classes.
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Recall that a part or portion of a line with two end points is called a line segment and a part of a line with one end point is called a ray. Note that the line segment AB is denoted by AB, and its length is denoted by AB. The ray AB is denoted by AB with an arrow, and a line is denoted by AB with arrows on both ends. However, we will not use these symbols, and will denote the line segment AB, ray AB, length AB and line AB by the same symbol, AB. The meaning will be clear from the context. Sometimes small letters l, m, n will be used to denote lines. If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points. Recall that an angle is formed when two rays originate from the same end point. The rays making an angle are called the arms of the angle and the end point is called the vertex of the angle. You have studied different types of angles, such as acute angle, right angle, obtuse angle, straight angle and reflex angle in earlier classes. In Figure 6.1, we see five types of angles. An acute angle measures between 0° and 90°, written as 0° < x < 90°. A right angle is exactly equal to 90°, written as y = 90°. An obtuse angle measures greater than 90° but less than 180°, written as 90° < z < 180°. A straight angle is equal to 180°, written as s = 180°. A reflex angle measures greater than 180° but less than 360°, written as 180° < t < 360°.
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Two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles. You have also studied about adjacent angles in the earlier classes. In Figure 6.2, we see two angles sharing a vertex and an arm. Two angles are adjacent, if they have a common vertex, a common arm and their non-common arms are on different sides of the common arm. In Figure 6.2, ∠ABD and ∠DBC are adjacent angles. Ray BD is their common arm and point B is their common vertex. Ray BA and ray BC are non common arms. Moreover, when two angles are adjacent, then their sum is always equal to the angle formed by the two non-common arms. So, we can write ∠ABC = ∠ABD + ∠DBC. Note that ∠ABC and ∠ABD are not adjacent angles. Why? Because their non-common arms BD and BC lie on the same side of the common arm BA. If the non-common arms BA and BC in Figure 6.2 form a line then it will look like Figure 6.3. In this case, ∠ABD and ∠DBC are called linear pair of angles. You may also recall the vertically opposite angles formed when two lines, say AB and CD, intersect each other, say at the point O, as shown in Figure 6.4. There are two pairs of vertically opposite angles. One pair is ∠AOD and ∠BOC. The other pair is ∠AOC and ∠BOD.
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Draw two different lines PQ and RS on a paper. You will see that you can draw them in two different ways as shown in Figure 6.5. In Figure 6.5 part one, the lines cross each other. These are intersecting lines. In Figure 6.5 part two, the lines never meet. These are non-intersecting or parallel lines. Recall the notion of a line, that it extends indefinitely in both directions. Lines PQ and RS in Figure 6.5 part one are intersecting lines and in Figure 6.5 part two are parallel lines. Note that the lengths of the common perpendiculars at different points on these parallel lines is the same. This equal length is called the distance between two parallel lines. In Section 6.2, you have learnt the definitions of some of the pairs of angles such as complementary angles, supplementary angles, adjacent angles, linear pair of angles. Can you think of some relations between these angles? Now, let us find out the relation between the angles formed when a ray stands on a line. Draw a figure in which a ray stands on a line as shown in Figure 6.6. Name the line as AB and the ray as OC. What are the angles formed at the point O? They are ∠AOC, ∠BOC and ∠AOB. Can we write ∠AOC + ∠BOC = ∠AOB? Yes, because they are adjacent angles. What is the measure of ∠AOB? It is 180°, because it is a straight angle. From these, can you say that ∠AOC + ∠BOC = 180°? Yes. From the above discussion, we can state the following Axiom: Axiom 6.1: If a ray stands on a line, then the sum of two adjacent angles so formed is 180°.
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Recall that when the sum of two adjacent angles is 180°, then they are called a linear pair of angles. In Axiom 6.1, it is given that a ray stands on a line. From this given, we have concluded that the sum of two adjacent angles so formed is 180°. Can we write Axiom 6.1 the other way? That is, take the conclusion of Axiom 6.1 as given and the given as the conclusion. So it becomes: If the sum of two adjacent angles is 180°, then a ray stands on a line, meaning the non-common arms form a line. Now you see that Axiom 6.1 and this statement are in a sense the reverse of each others. We call each as converse of the other. We do not know whether the converse is true or not. Let us check. Draw adjacent angles of different measures as shown in Figure 6.7. Keep the ruler along one of the non-common arms in each case. Does the other non-common arm also lie along the ruler? You will find that only in Figure 6.7 part three, both the non-common arms lie along the ruler, that is, points A, O and B lie on the same line and ray OC stands on it. Also see that ∠AOC + ∠COB = 125° + 55° = 180°. From this, you may conclude that the converse is true. So, you can state in the form of an axiom as follows: Axiom 6.2: If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line. For obvious reasons, the two axioms above together is called the Linear Pair Axiom. Let us now examine the case when two lines intersect each other. Recall, from earlier classes, that when two lines intersect, the vertically opposite angles are equal. Let us prove this result now.
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Theorem 6.1: If two lines intersect each other, then the vertically opposite angles are equal. Proof: In the statement above, it is given that two lines intersect each other. So, let AB and CD be two lines intersecting at O as shown in Figure 6.8. They lead to two pairs of vertically opposite angles, namely, ∠AOC and ∠BOD, and ∠AOD and ∠BOC. We need to prove that ∠AOC = ∠BOD and ∠AOD = ∠BOC. Now, ray OA stands on line CD. Therefore, ∠AOC + ∠AOD = 180° by the Linear pair axiom. Can we write ∠AOD + ∠BOD = 180°? Yes, because ray OB stands on line CD. From these two equations, we can write ∠AOC + ∠AOD = ∠AOD + ∠BOD. This implies that ∠AOC = ∠BOD by Axiom 3 from Section 5.2. Similarly, it can be proved that ∠AOD = ∠BOC. Now, let us do some examples based on Linear Pair Axiom and Theorem 6.1.
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Example 1: In Figure 6.9, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7, find all the angles. Solution: ∠POR + ∠ROQ = 180° because they form a linear pair of angles. But ∠POR : ∠ROQ = 5 : 7. Therefore, ∠POR = 5/12 × 180° = 75°. Similarly, ∠ROQ = 7/12 × 180° = 105°. Now, ∠POS = ∠ROQ = 105° because they are vertically opposite angles. And ∠SOQ = ∠POR = 75° because they are vertically opposite angles. Example 2: In Figure 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ, respectively. If ∠POS = x, find ∠ROT. Solution: Ray OS stands on the line POQ. Therefore, ∠POS + ∠SOQ = 180°. But ∠POS = x. Therefore, x + ∠SOQ = 180°. So, ∠SOQ = 180° – x. Now, ray OR bisects ∠POS, therefore, ∠ROS = 1/2 × ∠POS = 1/2 × x = x/2. Similarly, ∠SOT = 1/2 × ∠SOQ = 1/2 × (180° – x) = 90° – x/2. Now, ∠ROT = ∠ROS + ∠SOT = x/2 + 90° – x/2 = 90°.
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Example 3: In Figure 6.11, OP, OQ, OR and OS are four rays. Prove that ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°. Solution: In Figure 6.11, you need to produce any of the rays OP, OQ, OR or OS backwards to a point. Let us produce ray OQ backwards to a point T so that TOQ is a line, as shown in Figure 6.12. Now, ray OP stands on line TOQ. Therefore, ∠TOP + ∠POQ = 180° by the Linear pair axiom. Similarly, ray OS stands on line TOQ. Therefore, ∠TOS + ∠SOQ = 180°. But ∠SOQ = ∠SOR + ∠QOR. So, the second equation becomes ∠TOS + ∠SOR + ∠QOR = 180°. Now, adding the first and third equations, you get ∠TOP + ∠POQ + ∠TOS + ∠SOR + ∠QOR = 360°. But ∠TOP + ∠TOS = ∠POS. Therefore, the equation becomes ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°.
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Now we move to Exercise 6.1. Question 1: In Figure 6.13, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE. Solution: Since AB and CD intersect at O, ∠AOC and ∠BOD are vertically opposite angles. Therefore, ∠AOC = ∠BOD = 40°. We are given ∠AOC + ∠BOE = 70°. Substituting 40° for ∠AOC, we get 40° + ∠BOE = 70°. Therefore, ∠BOE = 30°. To find reflex ∠COE, note that angles on a straight line sum to 180°. So ∠AOC + ∠COE + ∠BOE = 180°. Substituting known values, 40° + ∠COE + 30° = 180°. So ∠COE = 110°. The reflex ∠COE is 360° – 110° = 250°. Question 2: In Figure 6.14, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c. Solution: Since XY is a straight line, ∠POX + ∠POY = 180°. But ∠POY is 90°, so ∠POX is 90°. The angles a and b together make up ∠POX, so a + b = 90°. Given a : b = 2 : 3, let a = 2x and b = 3x. So 2x + 3x = 90°, which gives 5x = 90°, so x = 18°. Therefore, a = 36° and b = 54°. Since MN is a straight line, b + c = 180°. Substituting b = 54°, we get 54° + c = 180°, so c = 126°.
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Question 3: In Figure 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT. Solution: Since SQR is a straight line, ∠PQS + ∠PQR = 180°. So ∠PQS = 180° – ∠PQR. Similarly, since PRT is a straight line, ∠PRT + ∠PRQ = 180°. So ∠PRT = 180° – ∠PRQ. Given ∠PQR = ∠PRQ, substituting this into the equations shows that 180° – ∠PQR = 180° – ∠PRQ. Therefore, ∠PQS = ∠PRT. Question 4: In Figure 6.16, if x + y = w + z, then prove that AOB is a line. Solution: The sum of all angles around point O is 360°. So x + y + w + z = 360°. We are given x + y = w + z. Substituting w + z with x + y in the first equation gives x + y + x + y = 360°, which simplifies to 2(x + y) = 360°. Therefore, x + y = 180°. Since x and y are adjacent angles and their sum is 180°, by the converse of the Linear Pair Axiom, their non-common arms form a straight line. Hence, AOB is a line. Question 5: In Figure 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS). Solution: Since OR is perpendicular to PQ, ∠POR = 90° and ∠QOR = 90°. From the figure, ∠QOS = ∠QOR + ∠ROS, so ∠QOS = 90° + ∠ROS. Also, ∠POS + ∠ROS = ∠POR, which is 90°. So ∠POS = 90° – ∠ROS. Now consider ∠QOS – ∠POS. Substituting the expressions, we get (90° + ∠ROS) – (90° – ∠ROS), which simplifies to 2∠ROS. Therefore, ∠ROS = 1/2(∠QOS – ∠POS).
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Question 6: It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP. Solution: When XY is produced to P, XYP is a straight line. So ∠XYZ + ∠ZYP = 180°. Given ∠XYZ = 64°, we get 64° + ∠ZYP = 180°, so ∠ZYP = 116°. Ray YQ bisects ∠ZYP, so ∠ZYQ = ∠QYP, each being half of 116°, which is 58°. Now, ∠XYQ = ∠XYZ + ∠ZYQ, which is 64° + 58° = 122°. The reflex ∠QYP is 360° – 58° = 302°. Now let us move to Section 6.5. If two lines are parallel to the same line, will they be parallel to each other? Let us check it. See Figure 6.18 in which line m is parallel to line l and line n is parallel to line l. Let us draw a line t transversal for the lines l, m and n. It is given that line m is parallel to line l and line n is parallel to line l. Therefore, ∠1 = ∠2 and ∠1 = ∠3 by the Corresponding angles axiom. So, ∠2 = ∠3. But ∠2 and ∠3 are corresponding angles and they are equal. Therefore, you can say that Line m is parallel to Line n by the Converse of corresponding angles axiom. This result can be stated in the form of the following theorem: Theorem 6.6: Lines which are parallel to the same line are parallel to each other. Note that the property above can be extended to more than two lines also.
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Now, let us solve some examples related to parallel lines. Example 4: In Figure 6.19, if PQ is parallel to RS, ∠MXQ = 135° and ∠MYR = 40°, find ∠XMY. Solution: Here, we need to draw a line AB parallel to line PQ, through point M as shown in Figure 6.20. Now, AB is parallel to PQ and PQ is parallel to RS. Therefore, AB is parallel to RS. Now, ∠QXM + ∠XMB = 180° because AB is parallel to PQ and they are interior angles on the same side of the transversal XM. But ∠QXM = 135°. So, 135° + ∠XMB = 180°. Therefore, ∠XMB = 45°. Now, ∠BMY = ∠MYR because AB is parallel to RS and they are alternate angles. Therefore, ∠BMY = 40°. Adding these two results, you get ∠XMB + ∠BMY = 45° + 40°. That is, ∠XMY = 85°. Example 5: If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel. Solution: In Figure 6.21, a transversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of ∠ABQ and ray CG is the bisector of ∠BCS; and BE is parallel to CG. We are to prove that PQ is parallel to RS. It is given that ray BE is the bisector of ∠ABQ. Therefore, ∠ABE = 1/2 ∠ABQ. Similarly, ray CG is the bisector of ∠BCS. Therefore, ∠BCG = 1/2 ∠BCS. But BE is parallel to CG and AD is the transversal. Therefore, ∠ABE = ∠BCG by the Corresponding angles axiom. Substituting the first two equations into the third, you get 1/2 ∠ABQ = 1/2 ∠BCS. That is, ∠ABQ = ∠BCS. But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal. Therefore, PQ is parallel to RS by the Converse of corresponding angles axiom.
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Example 6: In Figure 6.22, AB is parallel to CD and CD is parallel to EF. Also EA is perpendicular to AB. If ∠BEF = 55°, find the values of x, y and z. Solution: y + 55° = 180° because they are interior angles on the same side of the transversal ED. Therefore, y = 180° – 55° = 125°. Again x = y because AB is parallel to CD and they are corresponding angles. Therefore x = 125°. Now, since AB is parallel to CD and CD is parallel to EF, therefore, AB is parallel to EF. So, ∠EAB + ∠FEA = 180° because they are interior angles on the same side of the transversal EA. Therefore, 90° + z + 55° = 180°. Which gives z = 35°. Now we move to Exercise 6.2. Question 1: In Figure 6.23, if AB is parallel to CD, CD is parallel to EF and y : z = 3 : 7, find x. Solution: Since AB is parallel to CD and CD is parallel to EF, then AB is parallel to EF. Angles y and z are interior angles on the same side of the transversal, so y + z = 180°. Given y : z = 3 : 7, let y = 3k and z = 7k. So 3k + 7k = 180°, giving 10k = 180°, so k = 18°. Thus, y = 54° and z = 126°. Since AB is parallel to CD, angle x and angle y are interior angles on the same side of the transversal, so x + y = 180°. Therefore, x + 54° = 180°, giving x = 126°.
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Question 2: In Figure 6.24, if AB is parallel to CD, EF is perpendicular to CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE. Solution: Since EF is perpendicular to CD, ∠FED = 90°. ∠GED is given as 126°. So ∠GEF = ∠GED – ∠FED = 126° – 90° = 36°. Since AB is parallel to CD and GE is a transversal, ∠AGE and ∠GED are alternate interior angles. Therefore, ∠AGE = 126°. In triangle GEF, the sum of angles is 180°. We know ∠GEF is 36° and ∠EFG is 90° because EF is perpendicular to CD and AB is parallel to CD, making EF perpendicular to AB as well. So ∠FGE = 180° – (90° + 36°) = 54°. Question 3: In Figure 6.25, if PQ is parallel to ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. Hint: Draw a line parallel to ST through point R. Solution: Let us draw a line UV through R parallel to PQ and ST. Since PQ is parallel to UV, ∠PQR and ∠QRU are interior angles on the same side, so they sum to 180°. Thus, 110° + ∠QRU = 180°, giving ∠QRU = 70°. Since UV is parallel to ST, ∠RST and ∠SRV are interior angles on the same side, so they sum to 180°. Thus, 130° + ∠SRV = 180°, giving ∠SRV = 50°. Now, ∠QRS, ∠QRU, and ∠SRV lie on a straight line UV, so they sum to 180°. Therefore, ∠QRS + 70° + 50° = 180°. So ∠QRS = 60°.
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Question 4: In Figure 6.26, if AB is parallel to CD, ∠APQ = 50° and ∠PRD = 127°, find x and y. Solution: Since AB is parallel to CD and PQ is a transversal, ∠APQ and ∠PQR are alternate interior angles. Therefore, x = 50°. Now, consider transversal PR. Since AB is parallel to CD, the exterior angle ∠PRD equals the sum of the interior opposite angles ∠APQ and ∠QPR. So 127° = 50° + y. Therefore, y = 127° – 50° = 77°. Question 5: In Figure 6.27, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB is parallel to CD. Solution: Draw normals BE at B and CF at C perpendicular to the mirrors PQ and RS respectively. Since PQ is parallel to RS, the normals BE and CF are also parallel. By the law of reflection, the angle of incidence equals the angle of reflection. So ∠ABE = ∠EBC. Let this angle be α. Similarly, ∠BCF = ∠FCD. Let this angle be β. Since BE is parallel to CF and BC is a transversal, the alternate interior angles are equal. So ∠EBC = ∠BCF. This means α = β. Now, consider the lines AB and CD with transversal BC. The alternate interior angles are ∠ABC and ∠BCD. ∠ABC = ∠ABE + ∠EBC = α + α = 2α. ∠BCD = ∠BCF + ∠FCD = β + β = 2β. Since α = β, 2α = 2β. Therefore, ∠ABC = ∠BCD. Since these alternate interior angles are equal, AB is parallel to CD.
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In this chapter, you have studied the following points. First, if a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and vice-versa. This property is called as the Linear pair axiom. Second, if two lines intersect each other, then the vertically opposite angles are equal. Third, lines which are parallel to a given line are parallel to each other. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]