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Practical Geometry - Construction of a Line Parallel to a Given Line

Grade 7CBSE

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

🔑Concepts

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The fundamental concept of constructing parallel lines relies on the properties of a transversal intersecting two lines. If a transversal intersects two lines such that a pair of alternate interior angles are equal, then the lines must be parallel.

Diagram showing line l parallel to m with alternate interior angles x and y marked as equal.
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The step-by-step construction involves: 1. Taking a line ll and a point AA outside it. 2. Taking any point BB on ll and joining ABAB. 3. Constructing an angle at AA equal to ∠ABC\angle ABC on the opposite side of the transversal ABAB to create alternate interior angles.

Construction diagram showing a point A outside line l, a transversal AB, and the parallel line m.
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Corresponding angles property can also be used for construction. If a transversal intersects two lines such that a pair of corresponding angles are equal, the lines are parallel.

Two parallel lines with a transversal and equal corresponding angles marked.
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The distance between two parallel lines remains constant throughout their length. This 'perpendicular distance' can be used to construct a parallel line at a specific offset from the original line.

📐Formulae

If ∠1=∠2\angle 1 = \angle 2 (Alternate Interior Angles), then l∥ml \parallel m

If ∠1=∠3\angle 1 = \angle 3 (Corresponding Angles), then l∥ml \parallel m

Sum of interior angles on the same side of transversal: ∠1+∠4=180∘\angle 1 + \angle 4 = 180^\circ (Co-interior angles)

Distance dd between lines l1l_1 and l2l_2 is constant: d(P,l2)=d(Q,l2)d(P, l_2) = d(Q, l_2) for any points P,QP, Q on l1l_1

💡Examples

Problem 1:

Draw a line XYXY. Take a point AA outside it. Through AA, draw a line mm parallel to XYXY using the concept of alternate interior angles.

Solution:

  1. Draw a line XYXY and mark a point AA outside the line.
  2. Mark any point BB on the line XYXY and join the points AA and BB. Now, ABAB is the transversal.
  3. With BB as the center and any convenient radius, draw an arc cutting XYXY at point CC and ABAB at point DD.
  4. With AA as the center and the same radius as in step 3, draw an arc EFEF cutting ABAB at point GG.
  5. Place the compass pointer at CC and adjust the opening to measure the distance to DD.
  6. With the same opening and GG as the center, draw an arc to cut the arc EFEF at point HH.
  7. Draw a line mm passing through points AA and HH.

Explanation:

This construction replicates the angle ∠ABC\angle ABC at point AA such that ∠BAH=∠ABC\angle BAH = \angle ABC. Since these are alternate interior angles and are made equal, line mm becomes parallel to line XYXY.

Problem 2:

Given a line ll and a point PP at a distance of 44 cm from it, construct a line mm parallel to ll passing through PP.

Solution:

  1. Draw a line ll.
  2. Take any point XX on line ll and draw a perpendicular line XYXY using a protractor or compass at 90∘90^\circ.
  3. With XX as the center and a radius of 44 cm on the compass, draw an arc cutting the perpendicular line at point PP.
  4. At point PP, draw another perpendicular line to the segment XPXP.
  5. Extend this line on both sides to name it line mm.

Explanation:

Since line mm is perpendicular to XPXP and line ll is also perpendicular to XPXP, line mm and line ll are parallel because they are both perpendicular to the same transversal line at a distance of 44 cm.

Problem 3:

Draw a line ABAB. Mark a point CC outside it. Using a ruler and compass, construct a line PQPQ passing through CC such that PQ∥ABPQ \parallel AB.

Construction of line PQ parallel to AB passing through C using alternate angles.

Solution:

  1. Draw line ABAB and mark point CC outside.
  2. Take any point DD on ABAB and join CDCD.
  3. With DD as center and a convenient radius, draw an arc cutting ABAB at XX and CDCD at YY.
  4. With CC as center and the same radius, draw an arc EFEF cutting CDCD at GG.
  5. Adjust the compass to the width of arc XYXY. With GG as center and this width, cut arc EFEF at point HH.
  6. Join CHCH and extend it to form line PQPQ. Line PQPQ is parallel to ABAB.

Explanation:

This construction uses the principle of Alternate Interior Angles. By making ∠DCH=∠ADC\angle DCH = \angle ADC, we ensure the lines are parallel.

Problem 4:

Draw a line mm. Construct a line nn parallel to mm at a distance of 33 cm from it.

Construction of a parallel line at a distance of 3cm using perpendiculars.

Solution:

  1. Draw a line mm.
  2. Take any point XX on mm and construct a perpendicular XYXY of any length.
  3. Using a compass, mark a point ZZ on XYXY such that XZ=3XZ = 3 cm.
  4. At point ZZ, construct another perpendicular nn to the line XYXY.
  5. Line nn is parallel to line mm and is at a distance of 33 cm.

Explanation:

Since both lines mm and nn are perpendicular to the same line XYXY, they are parallel to each other. The distance XZ=3XZ = 3 cm defines the separation.