This lesson is the core of the Analytic Geometry category (10% of the CLEP exam). Its skills also leak into other categories: completing the square is pure algebra (20% category), and matching a conic equation to a graph is a representations task (30% category). Most of these items appear in Section 2, where no calculator is available — everything here should be automatic by hand.
For points (x₁, y₁) and (x₂, y₂):
distance: d = √[(x₂ − x₁)² + (y₂ − y₁)²]
midpoint: M = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
The distance formula is the Pythagorean theorem in disguise — the horizontal and vertical differences are the legs, the segment is the hypotenuse. The midpoint averages the coordinates; it does not subtract them. A sign discipline worth adopting: compute each difference in parentheses before squaring, so (4 − (−2)) becomes 6 on paper rather than 2 in your head.
Slope through two points: m = (y₂ − y₁)/(x₂ − x₁), rise over run.
Perpendicularity requires both moves — flip the fraction and flip the sign. Doing only one is the misconception nearly every distractor on these items is built from. Special cases: horizontal lines (m = 0) are perpendicular to vertical lines (undefined slope).
To write a line through a known point with a known slope, point-slope form is fastest: y − y₁ = m(x − x₁). Verify by substituting the point back in — a ten-second check that catches most errors.
Standard form displays everything:
(x − h)² + (y − k)² = r² center (h, k), radius r
Two persistent traps: the signs (center coordinates are what make each parenthesis zero, so (x − 3)² + (y + 2)² has center (3, −2)), and the right side (r², not r — a right side of 25 means radius 5).
General form hides everything: x² + y² + Dx + Ey + F = 0. Recover the standard form by completing the square in x and in y:
x² + y² − 6x + 4y − 12 = 0
(x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x − 3)² + (y + 2)² = 25 center (3, −2), radius 5
Half each linear coefficient, square it, add it to both sides.
The CLEP exam asks this stem nearly verbatim. An equation graphs as a circle exactly when:
Compare with the impostors: unequal positive coefficients (x² + 4y² = 16) give an ellipse; opposite signs (x² − y² = 9) give a hyperbola; only one squared variable (y = x² + 4) gives a parabola.
y = a(x − h)² + k vertex (h, k), axis of symmetry x = h
x²/a² + y²/b² = 1 (centered at the origin)
The graph crosses the x-axis at (±a, 0) and the y-axis at (0, ±b) — set y = 0 and x = 0 respectively to see why. The major axis lies along whichever denominator is larger, and the vertices are its endpoints. For x²/25 + y²/9 = 1: a² = 25 under x², so the major axis is horizontal and the vertices are (±5, 0); the minor axis endpoints are (0, ±3). Note that the denominators are the squares — read off 5 and 3, not 25 and 9.
x²/a² − y²/b² = 1 opens left/right, vertices (±a, 0)
y²/a² − x²/b² = 1 opens up/down, vertices (0, ±a)
The minus sign between the squared terms is the identifying mark. A hyperbola's two branches approach a pair of crossed asymptotes through the center; for x²/a² − y²/b² = 1 they are
y = ±(b/a)x
— the ratio is (number under y²'s square root)/(number under x²'s square root). For x²/9 − y²/4 = 1: a = 3, b = 2, asymptotes y = ±(2/3)x, vertices (±3, 0).
[GRAPH: Four small panels in the xy-plane, one conic each. Panel 1: circle (x − 3)² + (y + 2)² = 25, center marked (3, −2), radius segment labeled 5. Panel 2: parabola y = 2(x − 3)² + 1, vertex (3, 1), dashed vertical axis of symmetry x = 3. Panel 3: ellipse x²/25 + y²/9 = 1 with vertices (±5, 0) and endpoints (0, ±3) labeled. Panel 4: hyperbola x²/9 − y²/4 = 1, branches opening left and right from vertices (±3, 0), dashed asymptotes y = ±(2/3)x crossing at the origin.]
| equation shape | conic |
|---|---|
| x² and y², equal coefficients, same sign | circle (if r² > 0 after completing the square) |
| x² and y², unequal positive coefficients | ellipse |
| x² and y² with opposite signs | hyperbola |
| exactly one squared variable | parabola |
| no squared variables | line |
1. C — Correct: d = √[(4 − (−2))² + (−3 − 5)²] = √[6² + (−8)²] = √[36 + 64] = √100 = 10. A) adds the differences before rooting: √(6 + 8) = √14, skipping the squares. B) mishandles the sign, computing (4 − 2)² + (−3 − 5)² = 68 = (2√17)² — the −2 was treated as 2. D) adds the absolute differences 6 + 8 = 14 without squaring or rooting (taxicab distance). E) stops at 6² + 8² = 100, forgetting the square root. Fix: write each coordinate difference in parentheses, square both, add, then take the root — the formula is the Pythagorean theorem, so it must end with a √.
2. D — Correct: M = ((−6 + 4)/2, (2 + (−8))/2) = (−1, −3). A) subtracts instead of adds, then halves: ((−6 − 4)/2, (2 + 8)/2) = (−5, 5) — that computes half the difference, which belongs to the distance/radius world, not the midpoint. B) adds the coordinates but forgets to divide by 2. C) drops both negative signs. E) halves the differences in the other order. Fix: the midpoint is the average of the endpoints — add each pair of coordinates and divide by 2; no subtraction appears anywhere.
3. A — Correct: the perpendicular slope is the negative reciprocal of 3, namely −1/3, and through (1, 2): y − 2 = −(1/3)(x − 1) → y = −(1/3)x + 7/3. Check: −(1/3)(1) + 7/3 = 2 ✓. B) uses slope 3 — that line is parallel to ℓ (though it does pass through the point). C) takes the reciprocal without negating (slope 1/3). D) negates without taking the reciprocal (slope −3). E) has the right slope but pastes the point's y-coordinate in as the y-intercept — the line y = −(1/3)x + 2 passes through (0, 2), not (1, 2). Fix: perpendicular = flip AND negate the slope, then use point-slope form and verify the given point satisfies your final equation.
4. B — Correct: complete the square: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4 → (x − 3)² + (y + 2)² = 25, so center (3, −2) and radius √25 = 5. A) flips the center's signs — (x − 3)² is zero at x = +3, and (y + 2)² at y = −2. C) reports r² = 25 as the radius. D) makes both errors at once. E) reads the centers straight off the coefficients −6 and 4 without halving, and treats the constant 12 as r². Fix: half each linear coefficient gives the center (with signs opposite the parentheses), and the right side after completing the square is r² — root it before answering.
5. E — Correct: dividing 2x² + 2y² − 8x + 12y = 0 by 2 gives x² + y² − 4x + 6y = 0, equal unit coefficients; completing the square gives (x − 2)² + (y + 3)² = 13 > 0 — a genuine circle. A) has opposite signs on the squared terms: a hyperbola. B) has unequal positive coefficients (1 and 4): an ellipse. C) has only one squared variable: a parabola. D) passes the coefficient test but fails the radius test: completing the square gives (x + 4)² + (y − 1)² = 16 + 1 − 20 = −3, and no points satisfy it — the tempting trap for students who check coefficients only. Fix: circle = equal same-sign squared coefficients (after dividing out common factors), no xy term, AND a positive right side after completing the square — run all three checks.
6. C — Correct: vertex form y = a(x − h)² + k has axis of symmetry x = h; here h = 3, so the axis is the vertical line x = 3. A) flips the sign of h — but (x − 3)² is zero at x = +3. B) reports the horizontal line through the vertex's y-coordinate; axes of symmetry of these parabolas are vertical. D) grabs the constant k = 1 as if it were h. E) combines both confusions. Fix: the axis of symmetry is x = (the value that zeroes the squared parenthesis) — always an x = equation for a parabola of the form y = a(x − h)² + k.
7. A — Correct: the larger denominator (25) sits under x², so the major axis is horizontal; setting y = 0 gives x = ±√25 = ±5, vertices (±5, 0). B) puts the vertices on the wrong axis. C) reports the minor axis endpoints' distance √9 = 3 on the major axis. D) gives the actual minor axis endpoints — right points, wrong axis of the question. E) forgets to take the square root of the denominator. Fix: vertices lie along the larger-denominator axis at ±(square root of that denominator) — root the denominators before plotting anything.
8. B — Correct: for x²/a² − y²/b² = 1, the asymptotes are y = ±(b/a)x with a = √9 = 3 and b = √4 = 2: y = ±(2/3)x. A) inverts the ratio to a/b. C) uses the denominators 4/9 without taking square roots. D) uses the denominators inverted. E) confuses the vertices x = ±3 with asymptotes — a hyperbola's asymptotes are slanted lines through the center, not vertical lines. Fix: root both denominators first, then slope = (y-root)/(x-root) for an x-opening hyperbola — asymptotes always pass through the center as a crossed pair ±.
9. B — Correct: ℓ has slope (0 − 4)/(6 − 0) = −2/3, so any perpendicular line has slope 3/2; y = (3/2)x − 5 qualifies (the question asks only for perpendicularity, not for a particular point). A) has slope −2/3 — parallel to ℓ, not perpendicular. C) negates the sign but never flips the fraction. D) flips the fraction but keeps the negative sign: (−2/3)(−3/2) = +1, not −1. E) a vertical line is perpendicular only to horizontal lines, and ℓ is not horizontal. Fix: compute the given line's slope first, then demand the product of slopes be exactly −1 — check the product rather than eyeballing "flipped-looking" slopes.
10. D — Correct: both squared terms are present with positive but unequal coefficients (4 and 9), which after dividing by 36 gives x²/9 + y²/4 = 1 — an ellipse. A) would require equal coefficients on x² and y². B) would require exactly one squared variable. C) would require opposite signs between the squared terms. E) would require no squared terms at all. Fix: classify conics by the squared terms alone: equal coefficients → circle, unequal positive → ellipse, opposite signs → hyperbola, one square → parabola.
11. E — Correct: the center is the midpoint of the diameter, ((−1 + 5)/2, (3 + (−5))/2) = (2, −1); the diameter's length is √(6² + 8²) = 10, so r = 5 and r² = 25: (x − 2)² + (y + 1)² = 25. A) squares the full diameter (10² = 100) instead of the radius. B) writes the center's coordinates with the signs copied directly into the parentheses instead of reversed. C) uses the diameter 10 itself as r². D) computes the "center" from half-differences ((5 − (−1))/2, (−5 − 3)/2) = (3, −4) instead of the midpoint. Fix: diameter endpoints → center = midpoint, radius = half the distance between them; then write standard form remembering the sign flips and the square on r.
12. A — Correct: (x − h)² + (y − k)² = r² is the standard form — every point (x, y) at distance r from (h, k). B) reverses the center's signs; (x + h)² is zero at x = −h, not h. C) forgets to square the radius. D) has a subtraction between the squared terms — that is a hyperbola's shape. E) drops the squares entirely, leaving a line's equation. Fix: standard form is the distance formula squared: (difference in x)² + (difference in y)² = r² — both squares and the squared radius are non-negotiable.
1. C — Correct: d = √[(4 − (−2))² + (−3 − 5)²] = √[6² + (−8)²] = √[36 + 64] = √100 = 10. A) adds the differences before rooting: √(6 + 8) = √14, skipping the squares. B) mishandles the sign, computing (4 − 2)² + (−3 − 5)² = 68 = (2√17)² — the −2 was treated as 2. D) adds the absolute differences 6 + 8 = 14 without squaring or rooting (taxicab distance). E) stops at 6² + 8² = 100, forgetting the square root. Fix: write each coordinate difference in parentheses, square both, add, then take the root — the formula is the Pythagorean theorem, so it must end with a √.
2. D — Correct: M = ((−6 + 4)/2, (2 + (−8))/2) = (−1, −3). A) subtracts instead of adds, then halves: ((−6 − 4)/2, (2 + 8)/2) = (−5, 5) — that computes half the difference, which belongs to the distance/radius world, not the midpoint. B) adds the coordinates but forgets to divide by 2. C) drops both negative signs. E) halves the differences in the other order. Fix: the midpoint is the average of the endpoints — add each pair of coordinates and divide by 2; no subtraction appears anywhere.
3. A — Correct: the perpendicular slope is the negative reciprocal of 3, namely −1/3, and through (1, 2): y − 2 = −(1/3)(x − 1) → y = −(1/3)x + 7/3. Check: −(1/3)(1) + 7/3 = 2 ✓. B) uses slope 3 — that line is parallel to ℓ (though it does pass through the point). C) takes the reciprocal without negating (slope 1/3). D) negates without taking the reciprocal (slope −3). E) has the right slope but pastes the point's y-coordinate in as the y-intercept — the line y = −(1/3)x + 2 passes through (0, 2), not (1, 2). Fix: perpendicular = flip AND negate the slope, then use point-slope form and verify the given point satisfies your final equation.
4. B — Correct: complete the square: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4 → (x − 3)² + (y + 2)² = 25, so center (3, −2) and radius √25 = 5. A) flips the center's signs — (x − 3)² is zero at x = +3, and (y + 2)² at y = −2. C) reports r² = 25 as the radius. D) makes both errors at once. E) reads the centers straight off the coefficients −6 and 4 without halving, and treats the constant 12 as r². Fix: half each linear coefficient gives the center (with signs opposite the parentheses), and the right side after completing the square is r² — root it before answering.
5. E — Correct: dividing 2x² + 2y² − 8x + 12y = 0 by 2 gives x² + y² − 4x + 6y = 0, equal unit coefficients; completing the square gives (x − 2)² + (y + 3)² = 13 > 0 — a genuine circle. A) has opposite signs on the squared terms: a hyperbola. B) has unequal positive coefficients (1 and 4): an ellipse. C) has only one squared variable: a parabola. D) passes the coefficient test but fails the radius test: completing the square gives (x + 4)² + (y − 1)² = 16 + 1 − 20 = −3, and no points satisfy it — the tempting trap for students who check coefficients only. Fix: circle = equal same-sign squared coefficients (after dividing out common factors), no xy term, AND a positive right side after completing the square — run all three checks.
6. C — Correct: vertex form y = a(x − h)² + k has axis of symmetry x = h; here h = 3, so the axis is the vertical line x = 3. A) flips the sign of h — but (x − 3)² is zero at x = +3. B) reports the horizontal line through the vertex's y-coordinate; axes of symmetry of these parabolas are vertical. D) grabs the constant k = 1 as if it were h. E) combines both confusions. Fix: the axis of symmetry is x = (the value that zeroes the squared parenthesis) — always an x = equation for a parabola of the form y = a(x − h)² + k.
7. A — Correct: the larger denominator (25) sits under x², so the major axis is horizontal; setting y = 0 gives x = ±√25 = ±5, vertices (±5, 0). B) puts the vertices on the wrong axis. C) reports the minor axis endpoints' distance √9 = 3 on the major axis. D) gives the actual minor axis endpoints — right points, wrong axis of the question. E) forgets to take the square root of the denominator. Fix: vertices lie along the larger-denominator axis at ±(square root of that denominator) — root the denominators before plotting anything.
8. B — Correct: for x²/a² − y²/b² = 1, the asymptotes are y = ±(b/a)x with a = √9 = 3 and b = √4 = 2: y = ±(2/3)x. A) inverts the ratio to a/b. C) uses the denominators 4/9 without taking square roots. D) uses the denominators inverted. E) confuses the vertices x = ±3 with asymptotes — a hyperbola's asymptotes are slanted lines through the center, not vertical lines. Fix: root both denominators first, then slope = (y-root)/(x-root) for an x-opening hyperbola — asymptotes always pass through the center as a crossed pair ±.
9. B — Correct: ℓ has slope (0 − 4)/(6 − 0) = −2/3, so any perpendicular line has slope 3/2; y = (3/2)x − 5 qualifies (the question asks only for perpendicularity, not for a particular point). A) has slope −2/3 — parallel to ℓ, not perpendicular. C) negates the sign but never flips the fraction. D) flips the fraction but keeps the negative sign: (−2/3)(−3/2) = +1, not −1. E) a vertical line is perpendicular only to horizontal lines, and ℓ is not horizontal. Fix: compute the given line's slope first, then demand the product of slopes be exactly −1 — check the product rather than eyeballing "flipped-looking" slopes.
10. D — Correct: both squared terms are present with positive but unequal coefficients (4 and 9), which after dividing by 36 gives x²/9 + y²/4 = 1 — an ellipse. A) would require equal coefficients on x² and y². B) would require exactly one squared variable. C) would require opposite signs between the squared terms. E) would require no squared terms at all. Fix: classify conics by the squared terms alone: equal coefficients → circle, unequal positive → ellipse, opposite signs → hyperbola, one square → parabola.
11. E — Correct: the center is the midpoint of the diameter, ((−1 + 5)/2, (3 + (−5))/2) = (2, −1); the diameter's length is √(6² + 8²) = 10, so r = 5 and r² = 25: (x − 2)² + (y + 1)² = 25. A) squares the full diameter (10² = 100) instead of the radius. B) writes the center's coordinates with the signs copied directly into the parentheses instead of reversed. C) uses the diameter 10 itself as r². D) computes the "center" from half-differences ((5 − (−1))/2, (−5 − 3)/2) = (3, −4) instead of the midpoint. Fix: diameter endpoints → center = midpoint, radius = half the distance between them; then write standard form remembering the sign flips and the square on r.
12. A — Correct: (x − h)² + (y − k)² = r² is the standard form — every point (x, y) at distance r from (h, k). B) reverses the center's signs; (x + h)² is zero at x = −h, not h. C) forgets to square the radius. D) has a subtraction between the squared terms — that is a hyperbola's shape. E) drops the squares entirely, leaving a line's equation. Fix: standard form is the distance formula squared: (difference in x)² + (difference in y)² = r² — both squares and the squared radius are non-negotiable.