Justify the slope criteria for parallel and perpendicular lines and use coordinates to prove theorems about triangles, quadrilaterals and circles

Learning goal: Justify the slope criteria for parallel and perpendicular lines and use them to solve geometric problems, such as finding the equation of a line parallel or perpendicular to a given line through a given point; and use coordinates to prove simple geometric theorems algebraically and to verify geometric relationships, including properties of special triangles, quadrilaterals and circles, for example deciding whether four given points form a rectangle or whether a point lies on a given circle.

Slope as a rise-over-run triangle. A non-vertical line with slope m rises m units for every 1 unit of run; any two points on it, a units apart horizontally, are ma units apart vertically. Draw the right triangle with horizontal leg a and vertical leg ma between them: the slope triangle. Every slope triangle on the same line is similar to every other, which is why the ratio rise/run is the same no matter which two points are chosen. Vertical lines have no slope; they are handled separately below.

Justifying the parallel criterion. Claim: two distinct non-vertical lines are parallel if and only if they have the same slope. Suppose the lines have equations y = mx + b₁ and y = mx + b₂ with b₁ ≠ b₂. The translation (x, y) → (x, y + (b₂ − b₁)) maps the first line onto the second, because it adds b₂ − b₁ to every y-value. A translation maps a line to a parallel line (it preserves direction), so the lines are parallel. Directly: a common point would need mx + b₁ = mx + b₂, so b₁ = b₂, which is false. Conversely, if the slopes differ, say m₁ ≠ m₂, then m₁x + b₁ = m₂x + b₂ has the solution x = (b₂ − b₁)/(m₁ − m₂), so the lines meet at a point and are not parallel. Two vertical lines are parallel as well; a vertical and a non-vertical line always meet.

Justifying the perpendicular criterion. Claim: two non-vertical lines are perpendicular if and only if the product of their slopes is −1. Take the slope triangle of a line with slope m = b/a, with run a and rise b, placed with its corner at a point O on the line, so it goes from O to O + (a, b). Rotate the plane 90° counterclockwise about O. The point O + (a, b) goes to O + (−b, a), and the line goes to a line through O that is perpendicular to the original, because a 90° rotation turns every line by 90°. The rotated line has run −b and rise a, so its slope is a/(−b) = −a/b, and (b/a)(−a/b) = −1. Conversely, if two lines through O have slopes b/a and −a/b, the second passes through O + (−b, a), which is the 90° rotation of O + (a, b), so the lines are perpendicular. A horizontal line (slope 0) is perpendicular to a vertical line (no slope); the product rule does not apply to that pair, and it must be stated as its own case.

Using the criteria. Find the line through (2, −1) parallel to 3x − 2y = 6. Solving for y gives y = (3/2)x − 3, slope 3/2. The parallel line has the same slope: y + 1 = (3/2)(x − 2), so y = (3/2)x − 4. Check the point: (3/2)(2) − 4 = −1. The perpendicular line through (2, −1) has slope −2/3: y + 1 = −(2/3)(x − 2), so y = −(2/3)x + 1/3. Check: −(2/3)(2) + 1/3 = −4/3 + 1/3 = −1.

Coordinate proofs. To prove a geometric statement with coordinates, translate each geometric property into an algebraic test: equal lengths become equal distances (distance formula), parallel sides become equal slopes, right angles become slope products of −1 (or a horizontal–vertical pair), a midpoint becomes an average of coordinates, and "on the circle" becomes "satisfies the circle's equation." Then compute. A coordinate proof must check every property the conclusion needs, not just the convenient ones.

A worked model inside the lesson: is ABCD a rectangle? A(0, 0), B(6, 3), C(4, 7), D(−2, 4). A rectangle is a quadrilateral with four right angles, or equivalently a parallelogram with one right angle. Slopes: AB = (3 − 0)/(6 − 0) = 1/2; BC = (7 − 3)/(4 − 6) = 4/(−2) = −2; CD = (4 − 7)/(−2 − 4) = (−3)/(−6) = 1/2; DA = (0 − 4)/(0 − (−2)) = −4/2 = −2. AB ∥ CD and BC ∥ DA (equal slopes), so ABCD is a parallelogram. At every vertex the slope product is (1/2)(−2) = −1, so all four angles are right angles: ABCD is a rectangle. Lengths: AB = √(36 + 9) = √45 = 3√5 ≈ 6.71 and BC = √(4 + 16) = √20 = 2√5 ≈ 4.47, so it is not a square. Confirmation by diagonals: AC = √(16 + 49) = √65 and BD = √(64 + 1) = √65; the diagonals are congruent, as a rectangle's must be.

A second model: the midpoint of a hypotenuse. Theorem: the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices. Place the right angle at the origin: A(0, 0), B(8, 0), C(0, 6). The hypotenuse BC has midpoint M(4, 3). MA = √(16 + 9) = 5, MB = √((8 − 4)² + 3²) = 5, MC = √(4² + (6 − 3)²) = 5. All three equal 5, so M is equidistant from A, B and C, which is the same as saying the circle centered at M with radius 5 passes through all three: the hypotenuse is a diameter of the circumscribed circle. With general legs 2p and 2q the same computation gives the distance √(p² + q²) three times, so the theorem holds for every right triangle.

Points on circles. Does (5, 3) lie on the circle centered at (3, −2) with radius 5? The equation is (x − 3)² + (y + 2)² = 25, and (5, 3) gives 4 + 25 = 29 ≠ 25, so it does not; (7, 1) gives 16 + 9 = 25, so it does. This is the algebraic verification of a geometric statement, and it is exact: no measuring is involved.

The misconception: checking only what the conclusion says. Showing that the diagonals of ABCD are congruent does not by itself prove a rectangle; an isosceles trapezoid also has congruent diagonals. Showing that two sides are parallel proves only a trapezoid. A proof has to establish a set of properties that is known to characterize the figure: for a rectangle, four right angles, or a parallelogram with congruent diagonals, or a parallelogram with one right angle. A second misconception is that one numerical example proves a theorem; it verifies the theorem for that figure. To prove a theorem in general, use coordinates with letters, placed to keep the algebra simple, as in the hypotenuse example.

Worked parallel and perpendicular lines: Line L: 4x + 5y = 20, so y = −(4/5)x + 4, slope −4/5. The line parallel to L through (−5, 2): y − 2 = −(4/5)(x + 5), so y = −(4/5)x − 2; check: −(4/5)(−5) − 2 = 4 − 2 = 2. The line perpendicular to L through (−5, 2) has slope 5/4: y − 2 = (5/4)(x + 5), so y = (5/4)x + 33/4; check: (5/4)(−5) + 33/4 = −25/4 + 33/4 = 2. Product of slopes: (−4/5)(5/4) = −1.

Worked isosceles triangle proof: Show that the triangle with vertices P(−3, 0), Q(3, 0), R(0, 5) is isosceles and that the median from R is perpendicular to PQ. PR = √(9 + 25) = √34 and QR = √(9 + 25) = √34, so PR = QR and the triangle is isosceles. The midpoint of PQ is (0, 0), so the median from R is the segment from (0, 5) to (0, 0), which is vertical; PQ is horizontal (both endpoints have y = 0). A vertical and a horizontal line are perpendicular, so the median is perpendicular to the base.

Worked "is it a rhombus?": A(1, 1), B(4, 5), C(9, 5), D(6, 1). Sides: AB = √(9 + 16) = 5, BC = 5, CD = √(9 + 16) = 5, DA = 5. All four sides are congruent, so ABCD is a rhombus. Slopes of the diagonals: AC = (5 − 1)/(9 − 1) = 1/2, BD = (1 − 5)/(6 − 4) = −2; product −1, so the diagonals are perpendicular, as a rhombus's must be. It is not a square: slope AB = 4/3 and slope BC = 0, product 0 ≠ −1, so angle B is not a right angle.

Worked point-on-circle: The circle has center (−1, 2) and passes through (2, 6), so r² = 3² + 4² = 25. Is (3, −1) on it? (3 + 1)² + (−1 − 2)² = 16 + 9 = 25: yes. Is (4, 2)? 25 + 0 = 25: yes. Is (2, −2)? 9 + 16 = 25: yes. Is (1, 5)? 4 + 9 = 13 ≠ 25: no, it is inside.

Criteria-and-proof record: Write the two slope criteria with their justifications in your own words, including the special case of vertical lines. Then, for the quadrilateral with vertices (−1, 2), (3, 4), (5, 0), (1, −2), compute all four side slopes, all four side lengths and both diagonal lengths, and decide, with reasons naming the properties used, which of parallelogram, rectangle, rhombus, square applies.

Independent performance: Write two coordinate proofs. First, place a rectangle with vertices (0, 0), (a, 0), (a, b), (0, b), with a > 0 and b > 0, and prove its diagonals are congruent and bisect each other. Second, choose four specific points that you claim form a rectangle that is not aligned with the axes, prove it, and find the equation of the line through one vertex perpendicular to a side. Success criteria: each property is translated into the correct algebraic test; the general proof uses letters and does not rely on a picture; the specific proof checks all four angles (or parallelogram plus one right angle, with a sentence saying why that suffices); and the perpendicular line's equation is checked by substituting the point and by the slope product −1.