Geometry Parallel Lines and Transversals Test Solutions and Scoring (PDF)

1. Same-side interior angles (supplementary).
2. Alternate exterior angles (congruent).

Scoring: 1 point per relationship.

1. \(5x – 3 = 3x + 23\), \(2x = 26\), \(x = 13\)
2. \(5(13) – 3 = 62^{\circ}\)
3. \(180^{\circ} – 62^{\circ} = 118^{\circ}\)

(a) x = 13; (b) 62°; (c) 118°

Scoring: 1 point per part.

Adjacent angles form a linear pair: \(2x + 20 + 3x – 10 = 180\), \(5x = 170\), \(x = 34\).

\(2(34) + 20 = 88^{\circ}\) and \(3(34) – 10 = 92^{\circ}\).

The vertical angles repeat: 88°, 92°, 88°, 92°.

x = 34; 88°, 92°, 88°, 92°

Scoring: 1 point for the equation, 1 point for x and all four angles.

Corresponding angles of parallel lines are congruent. Since 80° is not 82°, the lines are not parallel.

Scoring: 1 point for the answer, 1 point for the reason.

\(7x + 3 + 5x + 9 = 180\)

\(12x = 168\)

\(x = 14\)

\(7(14) + 3 = 101^{\circ}\) and \(5(14) + 9 = 79^{\circ}\)

x = 14; 101° and 79°

Scoring: 1 point for the equation, 1 point for x and the angles.

1. \(3x + 4 = 5x – 8\), \(x = 6\), \(PA = 22\)
2. \(AM = 12\), \(PM = \sqrt{20^{2} – 12^{2}} = \sqrt{256} = 16\)

(a) x = 6, PA = 22; (b) PM = 16

Scoring: (a) 1 point for the equation, 1 point for x and PA. (b) 1 point.

Midpoint \((4, 4)\). Slope of the segment \(\frac{6 – 2}{8 – 0} = \frac{1}{2}\), so the perpendicular slope is \(-2\).

\(y – 4 = -2(x – 4)\) gives \(y = -2x + 12\).

y = −2x + 12

Scoring: 1 point for the midpoint and slope, 1 point for the equation.

Alternate interior angles are congruent: 73°.

Same-side interior angles are supplementary: \(180^{\circ} – 73^{\circ} = 107^{\circ}\).

73° and 107°

Scoring: 1 point each.

Call the vertical angles a and c, and let b be the angle adjacent to both.

\(a + b = 180^{\circ}\) and \(c + b = 180^{\circ}\).

So \(a + b = c + b\), and subtracting b gives \(a = c\).

Scoring: 1 point for the two linear pairs, 1 point for the conclusion.