LINES AND ANGLES-Notes

A line segment is a part of a line that has two definite end points. It is the shortest distance between these two points and contains all the points lying between them.

Line segment AB is denoted by \(\mathrm{\overline{AB}}\).

A line segment is finite, measurable, and forms the foundation for constructing more complex geometric

A ray is a part of a line that starts from a fixed point and extends endlessly in one direction. The fixed point is called the initial point of the ray, while the other end shows the direction in which it continues infinitely.

The ray AB is denoted by \(\mathrm{\overrightarrow{AB}}\)

A ray has one fixed starting point and no end point, making it infinite in length but directional in nature. It is used to represent light rays, paths, and directions in geometry and real life.

Collinear points are points that lie on the same straight line.

For example, if points A, B, and C lie on the same line, they are collinear points. However, if no single line can pass through all of them, they are non-collinear.

Non-collinear points are points that do not lie on the same straight line.

An angle is formed when two rays originate from the same endpoint. The common starting point of these rays is called the vertex of the angle, and the two rays themselves are known as the arms of the angle.
For example, if two rays OA and OB start from a common point O, then the figure formed is an angle AOB \((\theta)\). Here, O is the vertex, and OA and OB are the arms of the angle.

An acute angle is an angle that measures less than 90°. It is smaller than a right angle.
In other words, if the measure of an angle is greater than 0° but less than 90°, it is called an acute angle.

A right angle is an angle that measures exactly 90°. It represents a perfect quarter turn and forms a shape like the corner of a square or rectangle.

An obtuse angle is an angle that measures more than 90° but less than 180°. It is wider than a right angle but smaller than a straight angle.

A straight angle is an angle that measures exactly 180°. It represents a straight line, where the two rays lie in opposite directions from the vertex.

A reflex angle is an angle that measures more than 180° but less than 360°. It is larger than a straight angle but smaller than a complete angle.

Two angles are said to be complementary when the sum of their measures is 90°.

Two angles are said to be supplementary when the sum of their measures is 180°.
In simple terms, if two angles together form a straight angle, they are called supplementary angles.

Two angles are said to be adjacent if they share a common vertex, a common arm, and their non-common arms lie on opposite sides of the common arm.
In simple words, adjacent angles are next to each other, sitting side by side, sharing a corner (vertex) and one arm.

Two angles are said to form a linear pair if they are adjacent and their non-common arms form a straight line.
In simple words, when two angles lie side by side on a straight line, they make a linear pair.

When two straight lines intersect each other, they form two pairs of opposite angles that are called vertically opposite angles.

Intersecting Lines

Two lines are said to be intersecting lines if they meet or cross each other at a single point. The point where they meet is called the point of intersection.

When two lines share exactly one common point, they are said to intersect. At the point of intersection, angles are formed, which can be acute, obtuse, right, or vertically opposite.

Non-Intersecting Lines

Two lines are said to be non-intersecting lines if they never meet or cross each other, no matter how far they are extended in either direction.

Non-intersecting lines maintain the same distance between them and do not have any common point. If they are in the same plane and never meet, they are called parallel lines.

Axiom-1

If a ray stands on a line, then the sum of two adjacent angles so formed is \(180^\circ\).

Axiom-2

If the sum of two adjacent angles is \(180^\circ\), then the non-common arms of the angles form a line.

If two lines intersect each other, then the vertically opposite angles are equal.

Given that: Two lines intersect each other
Let AB and CD intersect each other at point \(O\) they lead two pair of vertically opposite angles namely $$\begin{aligned}\left( i\right) \angle AOC\text{ and }\angle BOD\\ \left( ii\right) \angle AOD\text{ and }\angle BOC\end{aligned}$$ We need to prove that $$\begin{aligned}\angle AOC=\angle BOD\\ \angle AOD=\angle BOC\end{aligned}$$ Now ray OA stand on line CD, therefore, \(\angle AOC \text{ and } \angle AOD\) forms linear pair, From Axion-1 $$\angle AOC+\angle AOD=180^{\circ }\tag{1}$$ ray OD stands on line A B, therefore we can write $$\angle AOD+BOD=180^{\circ }\tag{2}$$ as both angles are linear pair

From Equation (1) and (2), we can write $$\scriptsize\require{cancel}\begin{aligned}\angle AOC+\angle AOD&=\angle AOD+\angle BOD=180^\circ\\\\ \angle AOC+\cancel{\angle AOD}&=\cancel{\angle AOD}+\angle BOD\\\\ \Rightarrow \angle AOC&=\angle BOD\end{aligned}$$ similarly we can prove that $$\angle AOD=\angle BOC$$

lines PQ and RS intersect each other at point O. If \(\angle POR : \angle ROQ = 5 : 7\), find all the angles.

Solution:
Given: $$\begin{aligned}\angle POR:\angle ROQ=5:7\end{aligned}$$ \(\angle POR\) and \(\angle ROQ\) are Linear Pair of Angles, therefore $$\begin{aligned}\angle POR:\angle ROQ=5:7\\\angle POR+\angle ROQ=180^{\circ}\end{aligned}$$ $$\begin{aligned}\angle POR&=\dfrac{5}{12}\times 180^\circ\\\\ &=75^{\circ}\\\\ \angle ROQ&=\dfrac{7}{12}\times 180^\circ\\\\ &=105^{\circ }\\ \end{aligned}$$ \(\angle POR=\angle SOQ\) (vertically opposite angle) $$\therefore \angle SOQ=75^{\circ}$$ Similarly $$\begin{aligned}\angle POS&=\angle ROQ\\ \therefore \angle POS&=105^{\circ}\end{aligned}$$

Ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of \(\angle POS\) and \(\angle SOQ,\) respectively. If \(\angle POS = x\), find \(\angle ROT\).

To find $$\angle ROT$$ (Given) $$\angle POS=x$$ $$\angle SOQ+\angle POS=180^{\circ }$$ (linear pair) $$\begin{align}x+\angle POS&=180^{\circ}\\ \angle POS&=180^{\circ }-x\tag{1}\end{align}$$ OT is angle bisector $$\begin{aligned}\therefore \angle SOT=\dfrac{1}{2}\angle POS\\ =\dfrac{1}{2}\left( 180^{\circ}-x\right) \end{aligned}$$ OR is angle bisector of \(\angle POS\) $$\begin{align}\therefore \angle ROS=\dfrac{x}{2}\tag{2}\\ \angle ROT=\angle ROS+\angle SOT\end{align}$$ From Eqn (1) and (2) $$\begin{aligned} \angle ROT&=\dfrac{x}{2}+\dfrac{1}{2}\left( 180^{\circ}-x\right) \\\\ &=\dfrac{x+180-x}{2}\\\\ &=\dfrac{180^{\circ}}{2}\\\\ &=90^{\circ }\end{aligned}$$

OP, OQ, OR and OS are four rays. Prove that \(\angle POQ + \angle QOR + \angle SOR + \angle POS = 360°\).

Solution:

Prove that $$\angle POQ+\angle QOR+\angle SOR+\angle POS=360^{\circ }$$ Construction: Line QO is extended to point T

\(\angle TOP \text{ and }\angle POQ\) are Linear Pair, therefore, $$\angle TOP + \angle POQ=180^{\circ }\tag{1}$$ \(\angle TOS\) and \(\angle SOQ\) are also Linear pair, therfore, $$\angle TOS+\angle SOQ=180^\circ\tag{2}$$ But, $$\angle SOQ=\angle SOR+\angle QOR\\$$ Substituting Value of \(\angle SOQ\) in equation(2) $$\begin{aligned}\angle TOS+\angle SOR+\angle ROQ=180^{0}\end{aligned}\tag{3}$$ Adding equations (1) + (3) $$\scriptsize\begin{aligned}\angle TOP+\angle POQ+\angle TOS+\angle SOR+\angle QOR&=360^{0}\\ (\angle TOP+\angle TOS&=\angle SOP)\\ \angle TOP+\angle TOS+\angle POQ+\angle TOS+\angle SOR+\angle QOR&=360^{0}\\ \angle POS+\angle POQ+\angle SOR+\angle QOR&=360^{0}\end{aligned}$$ Hence, Proved

If \(PQ \parallel RS\), \(\angle MXQ = 135^\circ\) and \(\angle MYR = 40^\circ\), find \(\angle XMY\)

To find $$\angle XMY$$

Construction: Draw line \(AB \parallel PQ\)

$$\begin{aligned} AB\parallel PQ\\PQ\parallel RS\\\text{ therefore } AB\parallel RS\end{aligned}$$ Line parallel to the same line are parallel to each other $$\begin{aligned}\angle QXM=135^{0}\\ \angle PXM+\angle QXM=180^{\circ }\end{aligned}$$ \(\left(\angle PXM \text{ and }\angle QXM \text{ are linear pair}\right)\) $$\begin{aligned}\angle PXM&=180^{\circ}-135^{\circ }\\ &=45^{\circ}\\\\ PQ\parallel AB\end{aligned}$$ XM is transversal to parallel lines PQ and AB, therefore $$\begin{aligned}\angle PXM=\angle XMB\\ \Rightarrow \angle XMB=45^{\circ}\end{aligned}$$ (Alternate Angles) MY is transversal to || lines AB and RS $$\begin{aligned}\angle RYM&=\angle YMB=40^{\circ}\quad \text{ (Alternate Angles)}\\ \therefore \angle XMY&=\angle XMB+\angle YMB\\ &=45^{\circ}+40^{\circ}\\ &=85^{0}\end{aligned}$$