Topic 2 - Parallel and Perpendicular Lines EOC Review

What is m∠2 + m∠3?

(Just type the numerical answer, do not include degrees.)

If m∠4 = 35, find m∠2. Explain.

• 55; ∠2 and ∠4 are complementary angles, so m∠2 = 90 - m∠4.

• 35; ∠2 and ∠4 are corresponding angles, so m∠2 = m∠4.

• 145; ∠2 and ∠4 are supplementary angles, so m∠2 = 180 - m∠4.

• 35; ∠2 and ∠4 are alternate interior angles, so m∠2 = m∠4.

If m∠4 = 35, find m∠3. Explain.

• 55; ∠3 and ∠4 are complementary angles, so m∠3 = 90 - m∠4.

• 35; ∠3 and ∠4 are corresponding angles, so m∠3 = m∠4.

• 145; ∠3 and ∠4 are supplementary angles, so m∠3 = 180 - m∠4.

• 35; ∠3 and ∠4 are alternate interior angles, so m∠3 = m∠4.

If ∠1 ≅ ∠2, can you conclude that any of the lines are parallel? Explain.

• Yes; lines n and p are parallel because corresponding angles are congruent.

• No; ∠1 and ∠2 show no relationship.

• Yes; lines ℓ and m are parallel because corresponding angles are congruent.

• No; neither angle is formed by the transversal, line q.

If m∠3 + m∠4 = 180, which lines can you conclude are parallel? Explain.

• Lines n and p are parallel because alternate interior angles are congruent.

• Lines n and p are parallel because same-side interior angles are supplementary.

• Lines ℓ and m are parallel because same-side interior angles are supplementary.

• Lines ℓ and m are parallel because alternate interior angles are congruent.

What is m∠1?

Which of the following is true?

• m∠BCD − m∠DBC = 10°

• m∠BCD − m∠DBC = 20°

• m∠BCD − m∠DBC = 60°

• m∠BCD − m∠DBC = 80°

Find the slope of each line in the figure.

Slope of p = ​ (a)  

Slope of q =​ (b)  

Slope of r = ​ (c)  

Slope of m = ​ (d)  

Slope of n = ​ (e)  

• -2.5
• 2.5
• 0.4
• -0.4

Select all the true statements.

• m ∥ p

• p ∥ q

• p ⊥ m

• m ∥ n

• q ⊥ n

What must be true to prove that q ⊥ n?

• The product of the slopes of q and n is 1.

• The product of the slopes of q and n is -1.

• The slope of q must be the slope of n multiplied by -1.

• The slope of q must be equal to the slope of n.