Distance and midpoint formulas, equations of lines and circles, coordinate geometry proofs, locus problems, and dividing a segment — the complete analytic geometry reference for precalculus.
Derived from the Pythagorean theorem. The horizontal and vertical differences form the legs of a right triangle; the segment is the hypotenuse.
The midpoint averages the x-coordinates and the y-coordinates. It is always equidistant from both endpoints.
| Points | Distance | Midpoint |
|---|---|---|
| (0, 0) and (3, 4) | 5 | (1.5, 2) |
| (1, 1) and (4, 5) | 5 | (2.5, 3) |
| (−2, 3) and (4, 11) | 10 | (1, 7) |
| (0, −5) and (12, 0) | 13 | (6, −2.5) |
The slope of a line measures steepness and direction. Given two points (x₁, y₁) and (x₂, y₂):
y = mx + b
m = slope, b = y-intercept. Best for graphing and reading slope instantly.
y − y₁ = m(x − x₁)
Use when you know one point and the slope. Fastest path to a line equation.
(y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁)
Encodes both points directly. Derive it by substituting the slope formula into point-slope.
a is the x-intercept, b is the y-intercept. Useful when both intercepts are known. Does not work for lines through the origin.
Distance from point (x₀, y₀) to this line: d = |Ax₀ + By₀ + C| / √(A² + B²). Essential for locus and distance-to-line problems.
Parallel lines have equal slopes and different y-intercepts — they never intersect.
Example: y = 5x + 2 ∥ y = 5x − 7 (equal slope 5, different intercepts)
Perpendicular lines meet at 90°. Their slopes are negative reciprocals.
Example: slope 3/4 ⊥ slope −4/3. Product: (3/4)(−4/3) = −1 ✓
| Given Slope | Parallel Slope | Perpendicular Slope |
|---|---|---|
| m = 3 | m = 3 | m = −1/3 |
| m = −2/5 | m = −2/5 | m = 5/2 |
| m = 7 | m = 7 | m = −1/7 |
| m = 0 (horizontal) | m = 0 | undefined (vertical) |
| undefined (vertical) | undefined | m = 0 (horizontal) |
A circle is the locus of all points in a plane equidistant from a fixed center point. That constant distance is the radius.
Center (h, k), radius r. The signs inside the parentheses are opposite to the coordinates of the center. If the equation reads (x + 3)², then h = −3.
Convert to standard by completing the square for x and y separately. The circle is real when (D/2)² + (E/2)² − F > 0, a point when it equals 0, and imaginary when it is negative.
Group and move constant
Move F to the right side and group x-terms and y-terms: (x² + Dx) + (y² + Ey) = −F
Complete the square for x
Take half of D, square it, add (D/2)² to both sides: (x² + Dx + (D/2)²) + (y² + Ey) = −F + (D/2)²
Complete the square for y
Take half of E, square it, add (E/2)² to both sides: (x² + Dx + (D/2)²) + (y² + Ey + (E/2)²) = −F + (D/2)² + (E/2)²
Rewrite as perfect squares
Factor each group: (x + D/2)² + (y + E/2)² = r², where r² = (D/2)² + (E/2)² − F
Coordinate proofs translate geometric statements into algebraic equations. Place figures strategically to minimize computation, then verify properties using the distance, slope, and midpoint formulas.
Method 1 (Slopes): Show slope AB = slope CD and slope BC = slope AD — both pairs of opposite sides are parallel. Method 2 (Distances): Show AB = CD and BC = AD using the distance formula. Method 3 (Diagonals): Find midpoints of AC and BD — if they are the same point, the diagonals bisect each other, proving a parallelogram.
First prove it is a parallelogram (equal opposite slopes). Then show one interior angle is 90° by confirming two adjacent sides have slopes that are negative reciprocals. Alternatively, show the diagonals are equal in length (distance formula) in addition to bisecting each other.
Use the distance formula to show two sides are equal. Then optionally show the altitude to the third side bisects it (using the midpoint formula), confirming the symmetry of an isosceles triangle.
Calculate the lengths of all three sides using the distance formula. Verify the Pythagorean theorem: a² + b² = c² where c is the longest side. Alternatively, show two sides have slopes whose product is −1.
A locus is the complete set of points satisfying a given geometric condition. Common loci you must recognize:
Perpendicular bisector of the segment joining the two points
Set PA = PB, square both sides, simplify — the y² terms cancel leaving a linear equation.
A line parallel to both, halfway between them
Average the equations of the two lines to find the locus directly.
Circle centered at that point with the given radius
PA = r → (x − a)² + (y − b)² = r². The definition of a circle.
Two lines parallel to the given line, one on each side
Use the point-to-line distance formula |Ax + By + C|/√(A² + B²) = d.
Name the moving point
Let P(x, y) represent any point on the locus.
Translate the condition
Write the geometric condition as an algebraic equation in x and y.
Simplify
Expand, cancel like terms, and rearrange to recognize the curve (line, circle, parabola, etc.).
State restrictions
Note any points to exclude or domain restrictions implied by the original condition.
The section formula locates the point that divides segment AB in a given ratio m:n.
P lies between A and B, dividing AB in ratio m:n from A to B.
The midpoint is the special case m = n = 1: P = ((x₁ + x₂)/2, (y₁ + y₂)/2).
P lies on the extension of AB beyond B (or beyond A), dividing AB externally in ratio m:n.
External division is undefined when m = n (that ratio would create a point at infinity).
The centroid G is the point where all three medians meet. Each median divides internally in ratio 2:1 from vertex to midpoint. For triangle with vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃):
Example: A(0, 0), B(6, 0), C(3, 9) → G = ((0 + 6 + 3)/3, (0 + 0 + 9)/3) = (3, 3).
A = (−3, 2) and B = (5, −4)
Distance: d = √[(5 − (−3))² + (−4 − 2)²]
= √[(8)² + (−6)²] = √[64 + 36] = √100 = 10
Midpoint: M = ((−3 + 5)/2, (2 + (−4))/2) = (1, −1)
Distance = 10, Midpoint = (1, −1)
Verify: distance from (1, −1) to A = √[(1−(−3))² + (−1−2)²] = √[16 + 9] = 5 ✓ (half of 10)
x² + y² − 8x + 6y + 16 = 0
Step 1: Group and move constant.
(x² − 8x) + (y² + 6y) = −16
Step 2: Complete the square for x: (−8/2)² = 16. Add 16 both sides.
Step 3: Complete the square for y: (6/2)² = 9. Add 9 both sides.
(x² − 8x + 16) + (y² + 6y + 9) = −16 + 16 + 9
Step 4: Factor.
(x − 4)² + (y + 3)² = 9
Center (4, −3), radius 3
Quadrilateral ABCD: A(0,0), B(a,0), C(b,c), D(d,e)
Let P, Q, R, S = midpoints of AB, BC, CD, DA.
P = (a/2, 0), Q = ((a+b)/2, c/2)
R = ((b+d)/2, (c+e)/2), S = (d/2, e/2)
Slope PQ = (c/2 − 0)/((a+b)/2 − a/2) = (c/2)/(b/2) = c/b
Slope SR = ((c+e)/2 − e/2)/((b+d)/2 − d/2) = (c/2)/(b/2) = c/b
PQ ∥ SR. By similar argument QR ∥ PS. Therefore PQRS is a parallelogram. ∎
Let P(x, y) be on the locus.
Distance to (0, 4): √[x² + (y − 4)²]
Distance to x-axis: |y|
Set equal and square: x² + (y − 4)² = y²
Expand: x² + y² − 8y + 16 = y²
Simplify: x² − 8y + 16 = 0
Solve for y: y = x²/8 + 2
The locus is the parabola y = x²/8 + 2 (a classic parabola definition).
A(2, −1), B(7, 9), ratio 3:2
P_x = (3·7 + 2·2)/(3 + 2) = (21 + 4)/5 = 25/5 = 5
P_y = (3·9 + 2·(−1))/(3 + 2) = (27 − 2)/5 = 25/5 = 5
P = (5, 5)
Verify: AP/PB = √[(5−2)² + (5+1)²] / √[(7−5)² + (9−5)²] = √45 / √20 = 3/2 ✓
The distance formula gives the length of the segment joining two points (x₁, y₁) and (x₂, y₂): d = √[(x₂ − x₁)² + (y₂ − y₁)²]. It is derived directly from the Pythagorean theorem — the horizontal leg has length |x₂ − x₁| and the vertical leg has length |y₂ − y₁|, so the hypotenuse (the segment connecting the two points) satisfies d² = (x₂ − x₁)² + (y₂ − y₁)². Example: distance from (1, 2) to (4, 6) = √[(4−1)² + (6−2)²] = √[9 + 16] = √25 = 5.
The midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is M = ((x₁ + x₂)/2, (y₁ + y₂)/2). You average the x-coordinates and average the y-coordinates. Example: midpoint of (−3, 4) and (7, −2) is ((−3 + 7)/2, (4 + (−2))/2) = (4/2, 2/2) = (2, 1). To verify, check that M is equidistant from both endpoints using the distance formula.
The standard form of a circle with center (h, k) and radius r is (x − h)² + (y − k)² = r². This comes directly from the definition of a circle as the set of all points exactly r units from the center. Example: a circle centered at (3, −1) with radius 5 has equation (x − 3)² + (y + 1)² = 25. To find the center and radius from a given equation, identify h, k from the subtracted values and take the square root of the right side for r.
The general form is x² + y² + Dx + Ey + F = 0. To convert to standard form, complete the square for both x and y. Step 1: Group x-terms and y-terms. Step 2: Complete the square — for x² + Dx, add (D/2)² to both sides; for y² + Ey, add (E/2)² to both sides. Step 3: Write each group as a perfect square binomial. Example: x² + y² − 6x + 4y + 4 = 0 → (x² − 6x + 9) + (y² + 4y + 4) = −4 + 9 + 4 → (x − 3)² + (y + 2)² = 9. Center (3, −2), radius 3.
There are several coordinate proof strategies for a parallelogram. Method 1 (slopes): Show both pairs of opposite sides have equal slopes, proving they are parallel. Method 2 (distances): Show both pairs of opposite sides are equal in length using the distance formula. Method 3 (diagonals): Show the diagonals bisect each other by finding both midpoints — if the midpoints are the same point, the diagonals bisect each other. Any one of these three methods is sufficient to prove ABCD is a parallelogram.
A locus is the set of all points satisfying a given geometric condition. To find the locus equation: Step 1 — let P(x, y) be a general point on the locus. Step 2 — translate the geometric condition into an algebraic equation involving x and y. Step 3 — simplify the equation. Example: the locus of points equidistant from A(2, 0) and B(8, 0) is the perpendicular bisector. Set PA = PB: √[(x−2)² + y²] = √[(x−8)² + y²]. Square both sides: (x−2)² + y² = (x−8)². Simplify: x² − 4x + 4 = x² − 16x + 64 → 12x = 60 → x = 5. The locus is the vertical line x = 5.
The section formula finds point P dividing segment AB from A(x₁, y₁) to B(x₂, y₂) in ratio m:n internally: P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). For external division (ratio m:n but outside the segment): P = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)). Example: point dividing from A(1, 3) to B(7, 9) in ratio 2:1 internally is P = ((2·7 + 1·1)/3, (2·9 + 1·3)/3) = (15/3, 21/3) = (5, 7).
Step 1: Find the slope m of the given line by rearranging to slope-intercept form y = mx + b. Step 2: The perpendicular slope is −1/m (the negative reciprocal). Step 3: Use point-slope form y − y₀ = (−1/m)(x − x₀) with the given point (x₀, y₀). Step 4: Simplify to the desired form. Example: line 3x − 4y = 8 has slope 3/4. A line perpendicular to it through (6, 2) has slope −4/3: y − 2 = −(4/3)(x − 6) → y = −(4/3)x + 10.
Three points A, B, C are collinear (lie on the same line) if and only if the slope from A to B equals the slope from B to C. Equivalently, use the area method: compute area of triangle ABC = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. If the area equals 0, the points are collinear. Example: A(1, 2), B(3, 6), C(5, 10). Slope AB = (6−2)/(3−1) = 2. Slope BC = (10−6)/(5−3) = 2. Equal slopes confirm the three points are collinear.
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