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Geometry and Measurement - Properties of Parallel and Perpendicular Lines

Grade 7IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Parallel lines are lines in the same plane that never intersect, no matter how far they are extended. Perpendicular lines are lines that intersect at a right angle, which is exactly 90∘90^\circ. A transversal is a line that crosses at least two other lines.

Two parallel lines L1 and L2 intersected by a transversal line t.
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When two parallel lines are intersected by a transversal, Corresponding Angles are equal. These angles occupy the same relative position at each intersection where a straight line crosses two others.

Corresponding angles a and b are equal when the horizontal lines are parallel.
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Alternate Interior Angles are located between the parallel lines and on opposite sides of the transversal. These angles are equal when the lines are parallel. Co-interior angles are on the same side of the transversal and between the parallel lines; they sum to 180∘180^\circ (supplementary).

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Perpendicularity is a special relationship where two lines meet to form four right angles. If a line is perpendicular to one of two parallel lines, it is also perpendicular to the other.

📐Formulae

If L1∣∣L2L_1 || L_2, then ∠Corresponding1=∠Corresponding2\angle \text{Corresponding}_1 = \angle \text{Corresponding}_2

If L1∣∣L2L_1 || L_2, then ∠Alternate Interior1=∠Alternate Interior2\angle \text{Alternate Interior}_1 = \angle \text{Alternate Interior}_2

If L1∣∣L2L_1 || L_2, then ∠Co-interior1+∠Co-interior2=180∘\angle \text{Co-interior}_1 + \angle \text{Co-interior}_2 = 180^\circ

For L1⊥L2L_1 \perp L_2, the angle of intersection is 90∘90^\circ

∠Linear Pair1+∠Linear Pair2=180∘\angle \text{Linear Pair}_1 + \angle \text{Linear Pair}_2 = 180^\circ

∠Vertically Opposite1=∠Vertically Opposite2\angle \text{Vertically Opposite}_1 = \angle \text{Vertically Opposite}_2

💡Examples

Problem 1:

In the diagram, line ABAB is parallel to line CDCD. A transversal line EFEF intersects ABAB at point PP and CDCD at point QQ. If ∠APQ=(2x+10)∘\angle APQ = (2x + 10)^\circ and ∠DQP=(3x−20)∘\angle DQP = (3x - 20)^\circ are alternate interior angles, find the value of xx and the measure of each angle.

Solution:

  1. Since AB∣∣CDAB || CD, alternate interior angles are equal: (2x+10)=(3x−20)(2x + 10) = (3x - 20).
  2. Subtract 2x2x from both sides: 10=x−2010 = x - 20.
  3. Add 2020 to both sides: x=30x = 30.
  4. Substitute xx back into the expressions: ∠APQ=2(30)+10=70∘\angle APQ = 2(30) + 10 = 70^\circ and ∠DQP=3(30)−20=70∘\angle DQP = 3(30) - 20 = 70^\circ.

Explanation:

Because the lines are parallel, we can use the property that alternate interior angles (the 'Z' shape) are equal in measure to set up an algebraic equation and solve for the unknown.

Problem 2:

Given two parallel lines intersected by a transversal, two co-interior angles are represented by xx and 4x4x. Find the measure of the larger angle.

Solution:

  1. Co-interior angles between parallel lines are supplementary: x+4x=180∘x + 4x = 180^\circ.
  2. Combine like terms: 5x=180∘5x = 180^\circ.
  3. Divide by 55: x=1805=36∘x = \frac{180}{5} = 36^\circ.
  4. Find the larger angle: 4x=4×36=144∘4x = 4 \times 36 = 144^\circ.

Explanation:

We use the co-interior angle property (the 'C' shape), which states that these angles sum to 180∘180^\circ, to create a linear equation and solve for the variable.

Problem 3:

In the following figure, line mm is parallel to line nn. If ∠1=115∘\angle 1 = 115^\circ, find the measure of ∠2\angle 2.

Two parallel lines m and n with a transversal. Angle 1 is exterior-top-right and Angle 2 is interior-bottom-left.

Solution:

∠1+adjacent angle on straight line=180∘\angle 1 + \text{adjacent angle on straight line} = 180^\circ Adjacent angle=180∘−115∘=65∘\text{Adjacent angle} = 180^\circ - 115^\circ = 65^\circ Since m∣∣n,∠2=65∘ (Corresponding Angles)\text{Since } m || n, \angle 2 = 65^\circ \text{ (Corresponding Angles)} Alternatively, ∠1 and ∠2 are co-exterior angles on the same side, or ∠2 is vertically opposite to the co-interior partner of ∠1.\text{Alternatively, } \angle 1 \text{ and } \angle 2 \text{ are co-exterior angles on the same side, or } \angle 2 \text{ is vertically opposite to the co-interior partner of } \angle 1. ∠2=180∘−115∘=65∘\angle 2 = 180^\circ - 115^\circ = 65^\circ

Explanation:

Angles 1 and 2 are related through parallel line properties. We can first find the linear pair for angle 1, then use the property that corresponding angles are equal, or use the property that co-interior angles sum to 180∘180^\circ.

Problem 4:

Lines ABAB and CDCD are parallel. A third line XYXY is perpendicular to ABAB at point PP. Does XYXY also intersect CDCD at a 90∘90^\circ angle? Calculate the angle if ∠BPY=90∘\angle BPY = 90^\circ.

Horizontal parallel lines AB and CD with a vertical line XY crossing both at 90 degrees.

Solution:

Given AB∣∣CD and XY⊥AB\text{Given } AB || CD \text{ and } XY \perp AB ∠BPY=90∘\angle BPY = 90^\circ Let the intersection of XY and CD be point Q.\text{Let the intersection of } XY \text{ and } CD \text{ be point } Q. ∠DQX=∠BPY (Corresponding Angles)\angle DQX = \angle BPY \text{ (Corresponding Angles)} ∠DQX=90∘\angle DQX = 90^\circ Therefore, XY⊥CD\text{Therefore, } XY \perp CD

Explanation:

Using the property of corresponding angles, if a transversal (XY) intersects one of two parallel lines at 90∘90^\circ, it must intersect the other at the same angle.