Module 02 β Transformations and Congruence
By the end you can
A whole figure is easier when you track one vertex at a time.
x is sideways. y is height. Negative values mean left or down from zero.
For y=k draw a horizontal line at height k. For x=k draw a vertical line at sideways position k.
Write P β Pβ β Pβ. Do not apply the second move to the original point again.
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First work through the short cards. Then try the practice quiz. Each answer includes a reason.
Your course says Part One uses multiple-choice questions. The questions here are new practice across the module; they do not predict the exact questions on your test.
In (x, y), x is the sideways position and y is the height. Positive x means right; negative x means left. Positive y means up; negative y means down.
Aβ² is read βA prime.β It names the point after a transformation. It does not mean a different kind of coordinate.
Blue A(β3, β2) becomes purple Aβ²(1, 0). The changes are +4 in x and +2 in y: right 4 and up 2.
Check B too: (β1, β2) β (3, 0). Same slide. The rule is (x, y) β (x + 4, y + 2). A translation does not turn or flip a figure.
y = 1 is horizontal, one unit above the x-axis. P(2, β2) is 3 units below it, so its reflection is 3 units above it: Pβ²(2, 4). The x-coordinate stays 2.
The y-axis is vertical and has equation x = 0. Do not confuse a line such as y = 1 with the y-axis.
From y=β2 to y=1 is 3 grid spaces upward.
From y=1 go up 3 more spaces to y=4. Keep x=2.
Got it now? Scroll back up and re-read the section β it should land differently.
Across the x-axis: (x, y) β (x, βy). Across the y-axis: (x, y) β (βx, y). Across y = x: (x, y) β (y, x).
For a horizontal or vertical mirror that is not an axis, count to the mirror and then the same distance beyond it. Points already on the mirror stay in place.
These rules use the origin as the center: 90Β° counterclockwise sends (x, y) to (βy, x); 90Β° clockwise sends it to (y, βx); 180Β° sends it to (βx, βy).
Here A(1, 1) turns to Aβ²(β1, 1). Also check B(3, 1) β Bβ²(β1, 3). A 270Β° clockwise turn equals a 90Β° counterclockwise turn.
Apply the moves in their stated order. For P(1, β2), moving left 2 and up 1 gives Pβ(β1, β1).
Then reflect across y = 1. Pβ is 2 below the mirror, so Pβ is 2 above it: Pβ(β1, 3). Check every vertex when deciding whether the entire figure matches.
Translations, reflections, and rotations preserve lengths and angle measures. The image is congruent to the original. A reflection reverses clockwise/counterclockwise vertex order; translations and rotations preserve it.
For symmetry, the figure must land exactly on itself. A non-square rectangle does this after a 180Β° turn about its center, but not a 90Β° turn. A regular hexagonβs smallest positive turn is 360Β° Γ· 6 = 60Β°.
Valid tests are SSS, SAS, ASA, AAS, and HL. HL requires right triangles, a matching hypotenuse, and a matching leg.
βIncludedβ means between: SAS needs the angle between the two known sides. ASA needs the side between the two known angles. AAA proves the same shape, not necessarily the same size. SSA is not a general congruence test.
A shared side matches itself by the Reflexive Property. An angle bisector makes two equal angles; a midpoint splits a segment into two equal parts.
First prove the triangles congruent. Then use CPCTC for their corresponding parts. Do not use CPCTC as the reason that the triangles are congruent.
Triangle angles total 180Β°. Equal sides have equal opposite angles; the converse starts with equal angles and concludes equal opposite sides.
For sides 6 and 9, a third side x must satisfy 3 < x < 15. The endpoints are excluded. A midsegment joins two side midpoints, is parallel to the third side, and is half as long. Here MN = 5 because AB = 10.
An exterior angle equals the sum of the two nonadjacent interior angles. Within a triangle, the larger side is opposite the larger angle.
Why are 3 and 15 excluded when two sides are 6 and 9?
At either endpoint the three lengths make a flat shape. The Triangle Inequality bounds are strict.
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A median joins a vertex to the midpoint of the opposite side; medians meet at the centroid. An altitude is perpendicular to the opposite side or its extension; altitudes meet at the orthocenter.
Angle bisectors meet at the incenter. Perpendicular bisectors of sides meet at the circumcenter. The centroid and incenter are always inside; the other two can be inside, on, or outside depending on the triangle.
A centroid divides each median in a 2:1 ratio. The vertex-to-centroid part is twice the centroid-to-midpoint part. If a whole median is 12, those parts are 8 and 4.
A point on a segmentβs perpendicular bisector is equally far from its endpoints. The three perpendicular bisectors meet at the circumcenter, equally far from all three vertices.
The angle bisectors meet at the incenter, equally far from the three sides. Distance to a side is measured along a perpendicular segment.
Got it now? Scroll back up and re-read the section β it should land differently.
In a parallelogram, opposite sides and angles match, consecutive angles total 180Β°, and the diagonals bisect each other. AO = OC and BO = OD. This does not automatically mean AC = BD.
A rectangle adds congruent diagonals. A rhombus adds four equal sides, perpendicular diagonals, and diagonals that bisect opposite angles. A square has both sets of properties.
A kite has two pairs of adjacent congruent sides; its diagonals are perpendicular. Only the symmetry diagonal necessarily bisects the other diagonal.
Your course uses a trapezoid with exactly one pair of parallel sides. An isosceles trapezoid also has congruent legs, congruent base angles, and congruent diagonals.
Try the full set, then revisit the explanations for any misses. The score is practice feedback, not a school grade.
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