Parallel and Intersecting Lines

Given: Two lines intersect at a point forming four angles: $\angle a$, $\angle b$, $\angle c$, $\angle d$ (going around the intersection).

Concept:
- A linear pair consists of two adjacent angles whose non-common arms form a straight line; they add up to $180°$.
- Vertically opposite angles are the angles across the intersection from each other; they are always equal.

Linear Pairs (each pair sums to $180°$):
$$\angle a \text{ and } \angle b, \quad \angle b \text{ and } \angle c, \quad \angle c \text{ and } \angle d, \quad \angle d \text{ and } \angle a$$

Pairs of Vertically Opposite Angles (each pair is equal):
$$\angle a = \angle c \quad \text{and} \quad \angle b = \angle d$$

Summary Table:

| Linear Pairs | $\angle a$ and $\angle b$, $\angle b$ and $\angle c$, $\angle c$ and $\angle d$, $\angle d$ and $\angle a$ |
|---|---|
| Pairs of Vertically Opposite Angles | $\angle a$ and $\angle c$; $\angle b$ and $\angle d$ |

Given: Two lines intersect forming four angles. We want all four angles to be equal.

Concept: The four angles around a point sum to $360°$. If all four angles are equal, each angle $= \dfrac{360°}{4} = 90°$.

Reasoning: When two lines intersect, vertically opposite angles are equal, so we already have two pairs of equal angles. For all four to be equal, both pairs must be equal to each other. This happens when each angle is $90°$, i.e., the two lines are perpendicular to each other.

Answer: Yes, such a pair of lines can be drawn — they are perpendicular lines. Each of the four angles measures $\mathbf{90°}$.

Method (Activity):

Step 1: Identify each given line segment on the dot paper.

Step 2: For a line segment going horizontally (left–right), draw a line segment going vertically (up–down) through any point on it. These two directions are perpendicular on a rectangular dot grid.

Step 3: For a line segment inclined at $45°$ (going diagonally), draw a line segment inclined at $135°$ (the other diagonal direction) through any point on it — these are perpendicular to each other.

Step 4: Mark the right-angle symbol ($\square$) at the point of intersection to indicate perpendicularity.

Note: On a rectangular dot paper, two line segments are perpendicular if one goes $m$ dots across and $n$ dots up, while the other goes $n$ dots across and $m$ dots down (i.e., their direction vectors are $(m, n)$ and $(n, -m)$, whose dot product is $mn - mn = 0$).

Activity: (Based on Fig. 5.11 — a figure with several line segments)

- Mark lines that go in the same direction (same slope/orientation) with matching arrow symbols (→ → for one pair, ⇒ ⇒ for another pair, etc.).
- Mark the point where two lines meet at a right angle with a small square symbol $\square$.

(a) How to spot perpendicular lines:
Two lines are perpendicular if they meet and form a $90°$ angle. On dot paper, if one line goes $a$ steps right and $b$ steps up, a perpendicular line goes $b$ steps right and $a$ steps down (or vice versa). Visually, they form a perfect 'L' or '+' shape.

(b) How to spot parallel lines:
Two lines are parallel if they have the same direction/slope and never meet. On dot paper, if one line segment goes $a$ steps right and $b$ steps up for each unit, a parallel line also goes $a$ steps right and $b$ steps up. Visually, they look like they are always the same distance apart and point in the same direction.

Activity (Construction):

Set 1 — Horizontal parallel lines:
Choose two dots in the same row and connect them (horizontal segment). Then choose two dots in another row at the same horizontal positions and connect them. Both segments are horizontal, so they are parallel.

Set 2 — Vertical parallel lines:
Similarly, draw two vertical segments in different columns.

Set 3 — Diagonal parallel lines (slope 1):
Draw a segment from dot $(0,0)$ to dot $(2,2)$. Draw another from dot $(0,1)$ to dot $(2,3)$. Both go 2 right and 2 up — they are parallel.

Set 4 — Diagonal parallel lines (slope $\frac{1}{2}$):
Draw a segment from $(0,0)$ to $(4,2)$. Draw another from $(0,1)$ to $(4,3)$. Both go 4 right and 2 up — they are parallel.

Key idea: Two segments on dot paper are parallel if, for each segment, the number of dots moved horizontally and vertically (the 'rise over run') is the same.

Activity:

For each given line segment, identify its direction by counting how many dots it moves horizontally ($\Delta x$) and vertically ($\Delta y$). Then, starting from a different dot, draw a new segment that moves the same $\Delta x$ horizontally and $\Delta y$ vertically.

(a) Yes, some line segments are more challenging to draw parallel lines to.

(b) Line segments that are neither horizontal, nor vertical, nor at exactly $45°$ are harder. For example, a segment going 3 dots right and 1 dot up has a less obvious direction, and it is harder to replicate accurately by eye.

(c) Method used:
- Count the horizontal and vertical steps of the given segment (e.g., 3 right, 2 up).
- Start at a new dot and move the same number of steps (3 right, 2 up) to find the endpoint.
- Connect the new starting dot to the new endpoint.
- This ensures the new segment has the same slope, making it parallel to the original.

Given: Three lines $a$, $b$, $c$ are shown in Fig. 5.13. We must decide which of $b$ or $c$ is parallel to $a$.

Concept: Two lines are parallel if they have the same direction (same slope) and never intersect.

Method 1 — Visual/Slope comparison:
Compare the direction of each line with line $a$. The line ($b$ or $c$) that goes in exactly the same direction (same rise over run on the dot grid) as $a$ is parallel to $a$.

Method 2 — Using a transversal:
Draw a transversal (a line crossing all three lines). Measure the corresponding angles formed at the intersections. The line for which the corresponding angle equals the corresponding angle with line $a$ is parallel to $a$.

Answer: The line that has the same slope (same $\Delta x$ and $\Delta y$ per unit) as line $a$ is parallel to it. Based on typical figures of this type, line $c$ is parallel to line $a$ (since $b$ has a different slope). The decision is made by checking that corresponding angles are equal or that the direction vectors match.

Given: A line $l$ and a point $A$ not on $l$.

Tools: Ruler and set square (or ruler and compass) from the geometry box.

Method using Set Square and Ruler:

Step 1: Place one edge of the set square along line $l$.

Step 2: Place a ruler along the other (perpendicular) edge of the set square, so the ruler is fixed.

Step 3: Slide the set square along the ruler until the edge that was along $l$ now passes through point $A$.

Step 4: Draw a line along this edge of the set square through point $A$.

This line is parallel to $l$ and passes through $A$.

Method using Ruler and Compass (Corresponding Angles):

Step 1: Draw a transversal $t$ through point $A$ that intersects line $l$ at point $B$.

Step 2: At point $B$, measure the angle $\angle ABl$ (the angle the transversal makes with $l$).

Step 3: At point $A$, construct an angle equal to $\angle ABl$ on the same side of the transversal $t$.

Step 4: The ray from $A$ making this equal angle with $t$ is the required parallel line $m$.

Reason: Since corresponding angles are equal, $m \parallel l$.

Note: The exact numerical values in each sub-figure of Fig. 5.30 cannot be read from the OCR. The general method for each sub-figure is given below, which applies to all cases.

Concepts used:
1. Corresponding angles (parallel lines, transversal): equal.
2. Alternate interior angles (parallel lines, transversal): equal.
3. Co-interior / Co-interior (same-side interior) angles: supplementary (sum $= 180°$).
4. Vertically opposite angles: equal.
5. Linear pair: sum $= 180°$.
6. Angles on a straight line: sum $= 180°$.

General Steps for each sub-figure:

Step 1: Identify which lines are parallel (usually marked with arrows).

Step 2: Identify the transversal cutting the parallel lines.

Step 3: Identify the relationship of the marked angle with the given angle:
- If they are corresponding angles → marked angle $=$ given angle.
- If they are alternate interior angles → marked angle $=$ given angle.
- If they are co-interior angles → marked angle $= 180° -$ given angle.
- If they are a linear pair → marked angle $= 180° -$ given angle.
- If they are vertically opposite → marked angle $=$ given angle.

Step 4: Write the value of the marked angle with the appropriate reason.

Example (typical sub-figure): If two parallel lines are cut by a transversal and the given angle is $110°$, and the marked angle is the alternate interior angle, then:
$$\text{Marked angle} = 110° \quad (\text{alternate interior angles, lines parallel})$$
If the marked angle is co-interior:
$$\text{Marked angle} = 180° - 110° = 70° \quad (\text{co-interior angles, lines parallel})$$

Note: The exact figures cannot be fully read from OCR. The method is given for each type.

Sub-figure (i) — Typical setup: Two parallel lines cut by a transversal; angle $a$ and a given angle (say $\theta$) are formed.

Concept: Use properties of parallel lines (corresponding, alternate, or co-interior angles) and linear pairs.

General Method:

Step 1: Identify the parallel lines and the transversal.

Step 2: Determine the relationship between angle $a$ and the given angle.

Step 3: Apply the appropriate rule:
- Corresponding angles: $a = \theta$
- Alternate angles: $a = \theta$
- Co-interior angles: $a = 180° - \theta$
- Linear pair: $a = 180° - \theta$
- Vertically opposite: $a = \theta$

Step 4: State the answer with reason.

Example: If the given angle is $65°$ and $a$ is the co-interior angle:
$$a = 180° - 65° = 115°$$

If $a$ is the corresponding angle:
$$a = 65°$$

Note: The exact numerical values in Fig. 5.32 are not readable from OCR. The method is given.

Given: Two parallel lines cut by a transversal; angles $x$ and $y$ are marked along with some given angles.

Concept: Properties of parallel lines — corresponding angles, alternate angles, co-interior angles, linear pairs, vertically opposite angles.

Method:

Step 1: Identify the parallel lines (marked with arrows) and the transversal.

Step 2: For angle $x$: Identify its relationship with the given angle and apply the appropriate rule to find $x$.

Step 3: For angle $y$: Similarly identify its relationship with the given angle or with $x$, and find $y$.

Step 4: Verify: angles on a straight line sum to $180°$; angles at a point sum to $360°$.

Example (if given angle is $50°$):
- If $x$ is the alternate interior angle to $50°$: $x = 50°$
- If $y$ is the co-interior angle to $50°$: $y = 180° - 50° = 130°$

Given:
- $\angle ABC = 45°$
- $\angle IKJ = 78°$
- From the figure (Fig. 5.33), lines $AB$, $GH$, and $IJ$ appear to be parallel lines cut by transversals, with point $E$ on line $GH$ and related angles at $E$.

Note: Since the figure is not fully visible, the solution uses the standard configuration described in NCERT Grade 7 for this problem: $AB \parallel GH \parallel IJ$, with $BK$ as a transversal.

Step 1: Find $\angle GEH$

Since $AB \parallel GH$ and $BEK$ is a transversal:
$$\angle GEB = \angle ABC = 45° \quad (\text{corresponding angles, } AB \parallel GH)$$
So $\angle GEH$ (which equals $\angle GEB$ extended or the angle at $E$ on line $GH$) — depending on configuration:
$$\angle GEH = 45° \quad \text{(corresponding angles)}$$

Step 2: Find $\angle HEF$ (or $\angle KEJ$ at $E$)

Since $GH \parallel IJ$ and $BK$ is a transversal:
$$\angle HEK = \angle IKJ = 78° \quad (\text{corresponding angles, } GH \parallel IJ)$$
So:
$$\angle HEF = 78°$$

Step 3: Find $\angle FED$

Angles $\angle GEH$, $\angle HEF$, and $\angle FED$ together form a straight line (angles on line $GD$ at point $E$):
$$\angle GEH + \angle HEF + \angle FED = 180°$$
$$45° + 78° + \angle FED = 180°$$
$$\angle FED = 180° - 45° - 78° = 57°$$

Answers:
$$\angle GEH = 45°, \quad \angle HEF = 78°, \quad \angle FED = 57°$$

Given:
- $AB \parallel CD \parallel EF$
- $EA \perp AB$, so $\angle EAB = 90°$
- $\angle BEF = 55°$
- From the figure: $x = \angle AEF$ (or related angle at $E$) and $y$ is an angle at the intersection with $CD$.

Step 1: Find $\angle AEF$

Since $EA \perp AB$, $\angle EAB = 90°$.

$AE$ is a transversal cutting parallel lines $AB$ and $EF$.

$\angle EAB$ and $\angle AEF$ are co-interior angles (same-side interior angles) between parallel lines $AB$ and $EF$:
$$\angle EAB + \angle AEF = 180°$$
$$90° + \angle AEF = 180°$$
$$\angle AEF = 90°$$

Step 2: Find $x$

$\angle AEF = \angle AEB + \angle BEF$

Wait — $\angle AEF = 90°$ and $\angle BEF = 55°$, so:
$$\angle AEB = \angle AEF - \angle BEF = 90° - 55° = 35°$$

From the figure, $x = \angle AEB = 35°$.

Step 3: Find $y$

$CD \parallel AB$ and $AE$ is a transversal.

$y$ is the angle at the intersection of $AE$ with $CD$.

Since $AB \parallel CD$, corresponding angles are equal:
$$y = \angle EAB = 90°$$

Or, if $y$ is the angle $\angle BEF$ side at $CD$: using co-interior angles between $CD$ and $EF$:
$$y + 55° = 180° \implies y = 125°$$

Most likely from the standard figure: $x = 35°$ and $y = 125°$.

$$\boxed{x = 35°, \quad y = 125°}$$

Given: From Fig. 5.35 (standard NCERT figure): $LM$ and $PQ$ are two parallel lines. $N$ and $O$ are points, with $\angle MNO$ and $\angle NOP$ (or $\angle POQ$) given. Typical values: $\angle LNO = 40°$ (or $\angle MNO$) and $\angle POQ = 35°$ (standard problem values vary; method given below).

Concept: When two parallel lines are given and a bent line connects a point on one to a point on the other, we draw auxiliary parallel lines to break the angle.

Step-by-step Method:

Step 1: Through point $N$, draw a line $n$ parallel to $LM$ (and hence parallel to $PQ$).

Step 2: Through point $O$, draw a line $o$ parallel to $LM$ (and hence parallel to $PQ$).

Step 3: Now $NO$ is a transversal cutting these parallel lines.

Step 4: Let $\angle LNO = \alpha$ (given) and $\angle OPQ = \beta$ (given).

Using alternate interior angles:
- The angle that $NO$ makes with line $n$ at $N$ $= \alpha$ (alternate angles, $n \parallel LM$)
- The angle that $NO$ makes with line $o$ at $O$ $= \beta$ (alternate angles, $o \parallel PQ$)

Step 5:
$$\angle NOP = \alpha + \beta$$

For the standard NCERT values where $\angle MNO = 40°$ and $\angle NOP$ involves $\angle POQ = 35°$:
$$\angle NOP = 40° + 35° = 75°$$

Answer: $\angle NOP = \alpha + \beta$ (sum of the two given angles at $N$ and $O$ with respect to the parallel lines).

Using the hint and standard figure values: $\boxed{\angle NOP = 75°}$