Class 9 Mathematics Chapter 6 Lines And Angles

Key Concept: Collinear Points

a) Yes

[Solution Description]

To determine whether the three points are collinear, we calculate the slope between A and B, and between A and C. The slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 – y_1}{x_2 – x_1}$.

Slope of AB: $m_{AB} = \frac{4 – 2}{3 – 1} = \frac{2}{2} = 1$

Slope of AC: $m_{AC} = \frac{6 – 2}{5 – 1} = \frac{4}{4} = 1$

Since the slopes are equal, the points A, B, and C are collinear.

Key Concept: Intersecting and Non-Intersecting Lines

b) Intersecting lines

[Solution Description] Two lines that meet at a point are called intersecting lines. The point at which they meet is called the point of intersection.

Key Concept: Lines Parallel to the Same Line

c) Lines $l$ and $n$ are parallel

[Solution Description] According to Theorem 6.6, if two lines are parallel to the same line, they are parallel to each other. Since $l \parallel m$ and $m \parallel n$, it follows that $l \parallel n$.

Key Concept: Parallel Lines Theorem

b) $70^\circ$

[Solution Description]

Since line $m \parallel$ line $l$ and a transversal $t$ intersects them, $\angle 1$ and $\angle 2$ are corresponding angles. According to the corresponding angles axiom, $\angle 1 = \angle 2$. Given $\angle 1 = 70^\circ$, it follows that $\angle 2 = 70^\circ$.

Key Concept: Definition of an Angle

a) Vertex

[Solution Description]

The question asks for the name of the common endpoint where two rays meet to form an angle. According to the definition, this point is called the vertex.

Key Concept: Lines and Angles

d) 360 degrees

[Solution Description]

When two lines intersect, they form four angles. The sum of all these angles is 360 degrees. This is because the angles around a point add up to 360 degrees.

Key Concept: Line Segment

c) 8 units

[Solution Description]

Since points A, B, and C are collinear, the length of AC is the sum of the lengths of AB and BC. Therefore, AC = AB + BC = 5 units + 3 units = 8 units.

Key Concept: Adjacent Angles

b) $∠ ABD$ and $∠ DBC$

[Solution Description]

Adjacent angles share a common vertex, a common arm, and their non-common arms lie on opposite sides of the common arm. From the given options, only one pair satisfies this condition.

Key Concept: Vertically Opposite Angles

c) It measures $70^\circ$

[Solution Description]

According to Theorem 6.1, vertically opposite angles are equal. Thus, if one angle measures $70^\circ$, its vertically opposite angle will also measure $70^\circ$.

Key Concept: Angles formed by Parallel Lines and a Transversal

b) $45°$

[Solution Description]

Since AB || CD and a transversal intersects them, the interior angles on the same side of the transversal are supplementary. Given $∠ QXM = 135°$, we can find $∠ XMB$ as follows:

$∠ QXM + ∠ XMB = 180°$

Substituting the given value:

$135° + ∠ XMB = 180°$

Solving for $∠ XMB$:

$∠ XMB = 180° – 135° = 45°$

Therefore, the measure of $∠ XMB$ is $45°$.

Key Concept: Adjacent Angles

b) A common vertex and a common arm

[Solution Description]

Two angles are adjacent if they share a common vertex, a common arm, and their non-common arms are on different sides of the common arm.

Key Concept: Intersecting Lines

b) They cross each other at one point

[Solution Description]

When two lines intersect, they cross each other at a single point, called the point of intersection. The lines are called intersecting lines.

Key Concept: Vertically opposite angles, Linear pair

c) 30°

[Solution Description]

Vertically opposite angles are equal. Therefore, $∠ AOC = ∠ BOD$. Given $∠ AOC = 3x + 20°$ and $∠ BOD = 5x – 40°$, set them equal: $3x + 20° = 5x – 40°$. Subtract $3x$ from both sides: $20° = 2x – 40°$. Add $40°$ to both sides: $60° = 2x$. Divide by 2: $x = 30°$.

Key Concept: Alternate Interior Angles

c) The lines are parallel

[Solution Description]

If the alternate interior angles are equal when two lines are intersected by a transversal, then the lines are parallel according to the alternate interior angles theorem.

Key Concept: Supplementary Angles

b) 60°

[Solution Description] Supplementary angles add up to 180°. If one angle is 120°, the other angle can be found by subtracting 120° from 180°. So, $180° – 120° = 60°$. The other angle measures 60°.

Key Concept: Linear Pair Axiom, Vertically Opposite Angles

b) 35°

[Solution Description]

Given that lines MN and OP intersect at point Q, and $∠MQO = 4y – 15°$ and $∠NQP = 3y + 25°$. Since $∠MQO$ and $∠NQP$ are vertically opposite angles, they are equal. Therefore, $4y – 15° = 3y + 25°$. Solving for $y$, we subtract $3y$ from both sides to get $y – 15° = 25°$. Adding $15°$ to both sides gives $y = 40°$. Now, $∠OQP$ is adjacent to $∠MQO$, so $∠OQP = 180° – (4y – 15°) = 180° – 145° = 35°$.

Key Concept: Lines Parallel to the Same Line

c) They are parallel to each other

[Solution Description]

According to Theorem 6.6, lines which are parallel to the same line are parallel to each other. Since both line $m$ and line $n$ are parallel to line $l$, it follows that line $m \parallel$ line $n$.

Key Concept: Converse of Linear Pair Axiom

c) The non-common arms form a line

[Solution Description]

According to Axiom 6.2, if the sum of two adjacent angles is $180^\circ$, then the non-common arms of the angles form a line.

Key Concept: Lines Parallel to the Same Line

c) They are parallel

[Solution Description]

According to Theorem 6.6, if two lines are parallel to the same line, then they are parallel to each other. This is derived from the corresponding angles axiom and its converse. Therefore, the two lines will also be parallel to each other.

Key Concept: Adjacent Angles

b) Angles whose non-common arms lie on the same side of the common arm

[Solution Description]

Adjacent angles must have their non-common arms on different sides of the common arm. If both non-common arms lie on the same side of the common arm, they are not adjacent.

Key Concept: Linear Pair Axiom

c) $180^\circ$

[Solution Description] According to the Linear Pair Axiom (Axiom 6.1), if a ray stands on a line, then the sum of the two adjacent angles so formed is $180^\circ$.

Key Concept: Linear Pair Axiom, Vertically Opposite Angles

b) $85^\circ$

[Solution Description]

According to the Linear Pair Axiom, the sum of two adjacent angles should be $180^\circ$. Here, the sum is given as $170^\circ$, which contradicts the axiom. However, assuming this scenario, if two adjacent angles sum up to $170^\circ$, then each angle would be $\frac{170^\circ}{2} = 85^\circ$. Since vertically opposite angles are equal, both pairs of vertically opposite angles would also measure $85^\circ$.

Key Concept: Distance Between Parallel Lines

b) 3 units

[Solution Description]

The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:

$\text{Distance} = \frac{|C_2 – C_1|}{\sqrt{A^2 + B^2}}$

Here, $A = 3$, $B = 4$, $C_1 = -10$, and $C_2 = 5$. Plugging these values into the formula:

$\text{Distance} = \frac{|5 – (-10)|}{\sqrt{3^2 + 4^2}} = \frac{15}{5} = 3$

So, the distance between the two parallel lines is 3 units.

Key Concept: Properties of parallel lines and a transversal

b) 70$^\circ$

[Solution Description]

When a transversal intersects two parallel lines, the sum of the interior angles on the same side of the transversal is $180^\circ$. This is because these angles are supplementary. Given one interior angle as $110^\circ$, the other interior angle can be calculated as $180^\circ – 110^\circ = 70^\circ$.

Key Concept: Angles formed by intersecting lines

b) 130°

[Solution Description]

When two lines intersect, they form two pairs of adjacent angles that are supplementary (i.e., their measures add up to 180°). Given one angle as 50°, its adjacent angle can be calculated as follows:

$\text{Adjacent Angle} = 180° – 50° = 130°$

Key Concept: Parallel beams in architecture

a) $75^{\circ}$

[Solution Description] Corresponding angles formed by a transversal with two parallel lines are equal. Since the angles are corresponding, their measures will be identical regardless of beam orientation. Direct application of this property gives the answer.

Key Concept: Linear Pair Axiom

c) 180°

[Solution Description]

According to the Linear Pair Axiom, if a ray stands on a line, the sum of the two adjacent angles formed is $180°$.

Key Concept: Linear Pair Axiom

c) $110°$

[Solution Description]

Since $∠ AOC$ and $∠ AOD$ form a linear pair, their sum is $180°$. Given that $∠ AOC = 70°$, we can find $∠ AOD$ by subtracting $70°$ from $180°$. Thus, $∠ AOD = 180° – 70° = 110°$.

Key Concept: Lines Parallel to the Same Line

a) $70^\circ$

[Solution Description]

Draw a line parallel to ST through point R. Since PQ || ST, the angles $\angle PQR$ and $\angle QRS$ are co-interior angles. Therefore, $\angle QRS = 180^\circ – \angle PQR = 180^\circ – 110^\circ = 70^\circ$.

Key Concept: Types of Angles

a) Acute angle

[Solution Description]

An acute angle is defined as an angle measuring between $0°$ and $90°$.