Honors Geometry Unit 1 Test Review - Practice Flashcards

Flashcards covering key concepts and problem-solving steps from the notes, including midpoint definitions, segment congruence, angle relationships, distance formulas, coordinate geometry steps, and angle-bisection reasoning.

What condition must hold for B to be the midpoint of AC when A, B, C are collinear?

B lies on segment AC and AB = BC.

Are PQ and RS congruent in the collinear setup with PQ = 3x, PR = 7x + 4, QS = 5x + 7, RS = 5, with Q between P and R and R between Q and S?

No. Using QS = QR + RS and PR = PQ + QR leads to x = 2; then PQ = 6 and RS = 5, so they are not congruent.

If ∠ADB = 3x^2 − 2x, ∠BDC = 10x + 20, and ∠ADC = 4x^2 + 7x + 8 are adjacent with ∠ADB + ∠BDC = ∠ADC, what is x and m∠ADB?

Set 3x^2 − 2x + 10x + 20 = 4x^2 + 7x + 8. Solve to x = 4 (positive). Then m∠ADB = 3(4^2) − 2(4) = 40 degrees.

Find the length of JK with J(2,−5) and K(−6,−1).

JK = sqrt[(−6−2)^2 + (−1−(−5))^2] = sqrt(64 + 16) = sqrt(80) = 4√5.

Are JK and AB congruent if AM = 2√5 and M is the midpoint of AB?

Yes, because AB would be AM + MB = 2√5 + 2√5 = 4√5, which equals JK (assuming MB = AM).

What is the midpoint M of PQ for P(−6, 2) and Q(4, 8)?

M = ((−6 + 4)/2, (2 + 8)/2) = (−1, 5).

What is the midpoint S of PM given P(−6, 2) and M(−1, 5)?

S = ((−6 + (−1))/2, (2 + 5)/2) = (−7/2, 7/2).

Point F is located 2/3 of the way from S to Q. What are its coordinates?

F = S + (2/3)(Q − S) = (−7/2, 7/2) + (2/3)( (4,8) − (−7/2,7/2) ) = (3/2, 13/2) = (1.5, 6.5).

A point is chosen at random on PQ. What is the probability that the point lies on SF?

SF is a subsegment of PQ from t = 0.25 to t = 0.75 along PQ, so its length is half of PQ. Probability = 1/2.

Does QF bisect ∠AQN given m∠MQB = 136°, m∠AQF = 5x + 6, and m∠NQF = 8x − 26?

Yes, if x satisfies 5x + 6 = 8x − 26. Solving gives x = 32/3, so ∠AQF = ∠FQN and QF bisects ∠AQN.