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Unit 3 of 5

Unit 3: Analytic Geometry

Study guide for CLEP CLEP Precalculus — Unit 3: Analytic Geometry. Practice questions, key concepts, and exam tips.

57

Practice Questions

18

Flashcards

4

Key Topics

Key Concepts to Study

Conic sections: circles, ellipses, parabolas, hyperbolas
Parametric equations and polar coordinates
Vectors in two dimensions
Transformations and symmetry

Sample Practice Questions

Try these 5 questions from this unit. Sign up for full access to all 57.

Q1EASY

Find the midpoint of the line segment joining the points (2,3) and (6,7).

A) (3,4)
B) (4,5)
C) (5,6)
D) (2,7)
E) (6,3)
Show Answer

Answer: B — "(4,5)" is correct because the midpoint formula is ((x1+x2)/2, (y1+y2)/2).

Q2MEDIUM

What is the equation of a line perpendicular to y = 2x + 1 and passing through (2, 3)?

A) y - 3 = -1/2(x - 2)
B) y - 3 = 2(x - 2)
C) y - 3 = 1/2(x - 2)
D) y - 3 = -2(x - 2)
E) y - 3 = -3(x - 2)
Show Answer

Answer: A — "y - 3 = -1/2(x - 2)" is correct because it has the correct slope.

Q3MEDIUM

What is the equation of the line that is perpendicular to the line y = 2x + 3 and passes through the point (2, 3)?

A) y = -1/2x + 4
B) y = -2x + 1
C) y = 1/2x + 2
D) y = -1/2x + 1
E) y = 2x - 1
Show Answer

Answer: A — y = -1/2x + 4 is correct because the slope of the perpendicular line is -1/2 and using point-slope form y - 3 = -1/2(x - 2) yields y = -1/2x + 4.

Q4MEDIUM

Convert the polar equation $r = 5cos( heta)$ to rectangular form. What is the resulting equation?

A) $x^2 + y^2 = 5x$
B) $x^2 + y^2 = 10x$
C) $x^2 - y^2 = 5x$
D) $x^2 + y^2 = 25$
E) $x^2 - y^2 = 25$
Show Answer

Answer: A — To convert from polar to rectangular, we use $x = rcos( heta)$ and $y = rsin( heta)$. Given $r = 5cos( heta)$, we substitute $r$ in $x = rcos( heta)$ to get $x = 5cos^2( heta)$. Since $r^2 = x^2 + y^2$ and $r = 5cos( heta)$, we have $r^2 = 5rcos( heta)$, which becomes $x^2 + y^2 = 5x$ when we substitute $rcos( heta) = x$. Thus, the correct rectangular form is $x^2 + y^2 = 5x$.

Q5MEDIUM

What are the foci of the ellipse $x^{2}$/9 + $y^{2}$/16 = 1?

A) (0, +/- 2*$\sqrt{7}$)
B) (0, +/- $\sqrt{7}$)
C) (+/- $\sqrt{7}$, 0)
D) (0, +/- 2)
E) (+/- 2, 0)
Show Answer

Answer: B — "(0, +/- $\sqrt{7}$)" is correct because $c^{2}$ = $a^{2}$ - $b^{2}$, c = $\sqrt{7}$.

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Study Tips for Unit 3: Analytic Geometry

  • Focus on understanding concepts, not memorizing facts — CLEP tests application
  • Practice with timed questions to build exam-day speed
  • Review explanations for wrong answers — they reveal common misconceptions
  • Use flashcards for key terms, practice questions for deeper understanding

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