AutoCAD Quiz: Coordinate Entry
10 questions · exam conditions
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Coordinate EntryQuestion 1 of 10

During a polyline operation, the current point is 72,−1872,-18. The next vertex must be located at the absolute coordinate 15,2415,24. Dynamic Input is off.

Which relative Cartesian entry places the next vertex at the required coordinate?

@15,24@15,24
@−57,42@-57,42
−57,42-57,42
@57,−42@57,-42
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AutoCAD Quiz

AutoCAD Quiz: Coordinate Entry

Practice Coordinate Entry in AutoCAD with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Coordinate Entry, giving you a quick way to practice the rules, question types, and explanations that matter most for AutoCAD.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

During a polyline operation, the current point is 72,−1872,-18. The next vertex must be located at the absolute coordinate 15,2415,24. Dynamic Input is off.

Which relative Cartesian entry places the next vertex at the required coordinate?

  1. @15,24@15,24
  2. @−57,42@-57,42 (correct answer)
  3. −57,42-57,42
  4. @57,−42@57,-42
Explanation: When working with coordinate entry in AutoCAD, you need to distinguish between absolute and relative coordinates. Absolute coordinates reference the origin (0,0)(0,0), while relative coordinates reference the current point. The @ symbol signals a relative entry, meaning "offset from where I am now." Here, your current point is (72,−18)(72,-18) and your target is (15,24)(15,24). To find the relative offset, subtract the current point from the destination: ΔX=15−72=−57\Delta X = 15 - 72 = -57 ΔY=24−(−18)=24+18=42\Delta Y = 24 - (-18) = 24 + 18 = 42 So the correct relative entry is @−57,42@{-57},42, which is answer B. The negative X means you move left, and the positive Y means you move up — which makes geometric sense when shifting from a point far to the right down to one at the upper-left. A (@15,24@15,24) is the trap for students who confuse relative and absolute coordinates — those are simply the destination's absolute values, not the offset. C (−57,42-57,42) has the correct offset values but omits the @ symbol, so AutoCAD would interpret it as an absolute coordinate, placing the vertex at (−57,42)(-57, 42) from the origin — completely wrong. D (@57,−42@57,-42) reverses the signs on both components, which would move you in the exact opposite direction from the target. A reliable strategy: always compute ΔX\Delta X and ΔY\Delta Y explicitly by doing destination minus current position, then prefix with @. Sign errors are the most common mistake on coordinate-entry questions.

Question 2

In a 3D drawing, a line starts at the absolute coordinate 12,−8,512,-8,5. Two successive points are entered using @−4,10,−7@-4,10,-7 and then @6,−3,9@6,-3,9.

What is the absolute coordinate of the final point?

  1. 14,−1,714,-1,7 (correct answer)
  2. 14,1,714,1,7
  3. 6,7,26,7,2
  4. 18,−11,1418,-11,14
Explanation: When working with 3D coordinates in AutoCAD, you need to distinguish between absolute coordinates (measured from the origin) and relative coordinates (measured from the previous point). The @ symbol signals a relative coordinate — each entry offsets from wherever you currently are. Starting at absolute point 12,−8,512, -8, 5, the first relative move @−4,10,−7@-4, 10, -7 adds component-by-component: 12+(−4)=812+(-4)=8, −8+10=2-8+10=2, 5+(−7)=−25+(-7)=-2, landing at 8,2,−28, 2, -2. The second relative move @6,−3,9@6, -3, 9 then offsets from that new position: 8+6=148+6=14, 2+(−3)=−12+(-3)=-1, −2+9=7-2+9=7. The final absolute coordinate is 14,−1,714, -1, 7, confirming A is correct. B 14,1,714, 1, 7 is a sign error — the Y-component 2+(−3)2+(-3) yields −1-1, not +1+1. This trap catches students who drop the negative sign during addition. C 6,7,26, 7, 2 results from incorrectly applying both relative moves directly to the original starting point instead of chaining them step-by-step — a misunderstanding of how successive relative entries accumulate. D 18,−11,1418, -11, 14 comes from treating the relative offsets as absolute coordinates and adding all three sets of values together at once, confusing the coordinate systems entirely. Your study tip: always process relative coordinates sequentially — each @ entry updates your "current position," and the next @ moves from there. Draw a quick chain on scratch paper: Start → Move 1 → Move 2. Never jump back to the original point for later relative moves.

Question 3

While drawing an orthogonal rectangular boundary, the first corner has already been entered as the absolute point 250,125250,125. The next corners must be 310,125310,125 and 310,170310,170, in that order.

Which pair of relative Cartesian entries correctly creates both remaining corners?

  1. @0,60@0,60 followed by @45,0@45,0
  2. @310,125@310,125 followed by @0,45@0,45
  3. @60,0@60,0 followed by @310,170@310,170
  4. @60,0@60,0 followed by @0,45@0,45 (correct answer)
Explanation: When working with relative Cartesian coordinates in AutoCAD, the key idea is that each @x,y entry measures displacement from the previous point, not from the drawing origin. Every time you enter a new point, your "zero" resets to wherever you just clicked. Starting at 250,125250,125, the second corner needs to be 310,125310,125. The displacement is 310−250=60310-250=60 in X and 125−125=0125-125=0 in Y, giving @60,0@60,0. From that new position at 310,125310,125, the third corner must reach 310,170310,170. The displacement is 310−310=0310-310=0 in X and 170−125=45170-125=45 in Y, giving @0,45@0,45. That sequence — @60,0@60,0 followed by @0,45@0,45 — is exactly option D, confirming it as correct. Option A reverses both values: @0,60@0,60 moves 60 units vertically first (landing at 250,185250,185, not 310,125310,125), and @45,0@45,0 then shifts horizontally — neither corner matches. Option B uses absolute coordinates disguised as relative ones; entering @310,125@310,125 would displace an additional 310 units in X and 125 in Y from the current position, overshooting far beyond the intended corner. Option C gets the first step right with @60,0@60,0, but then abandons relative notation entirely — @310,170@310,170 again adds 310 and 170 as raw offsets rather than landing on the target point. A reliable strategy: always subtract the previous point's coordinates from the target point's coordinates to find your @Δx,Δy@\Delta x, \Delta y values. Write out the subtraction explicitly before typing — it only takes a second and prevents the most common errors on coordinate-entry questions.

Question 4

A UCS is rotated 30°30° counterclockwise from the WCS about the common origin. A line begins at the WCS coordinate 100,50100,50. At the next-point prompt, the user enters @40,0@40,0.

What is the approximate WCS coordinate of the new endpoint?

  1. 140.000,50.000140.000,50.000
  2. 134.641,70.000134.641,70.000 (correct answer)
  3. 134.641,30.000134.641,30.000
  4. 34.641,20.00034.641,20.000
Explanation: When working with User Coordinate Systems (UCS) in AutoCAD, relative coordinates entered by the user are always interpreted in the current UCS, not the WCS. Your job is to convert those UCS-relative values back into WCS coordinates. Here, the UCS is rotated 30°30° counterclockwise from the WCS. The user enters @40,0@40,0, meaning a displacement of 40 units along the UCS X-axis. To find the WCS equivalent of this displacement, apply the 2D rotation transformation: ΔXWCS=40cos⁡(30°)=40×0.866=34.641\Delta X_{WCS} = 40\cos(30°) = 40 \times 0.866 = 34.641 ΔYWCS=40sin⁡(30°)=40×0.500=20.000\Delta Y_{WCS} = 40\sin(30°) = 40 \times 0.500 = 20.000 The starting WCS point is 100,50100, 50, so the new endpoint is: X=100+34.641=134.641,Y=50+20.000=70.000X = 100 + 34.641 = 134.641, \quad Y = 50 + 20.000 = 70.000 That confirms B) 134.641,70.000134.641, 70.000 as the correct answer. A) 140.000,50.000140.000, 50.000 ignores the UCS rotation entirely, treating @40,0@40,0 as a pure WCS displacement — a common mistake when students forget the coordinate transformation step. C) 134.641,30.000134.641, 30.000 correctly computes the X displacement but subtracts the Y component instead of adding it, as if the rotation were clockwise. D) 34.641,20.00034.641, 20.000 correctly calculates the displacement vector but forgets to add it to the starting point 100,50100, 50. Your study tip: always think of UCS entry as a two-step process — rotate the input vector by the UCS angle, then add it to the WCS starting point. Forgetting either step is how all three wrong answers are generated.

Question 5

The current point of a line is −10,25-10,25. Angles use the default convention: 0°0° is along positive X and positive angles are counterclockwise. The next point is entered as @30<210°@30<210°.

Approximately where is the next point located?

  1. 15.981,10.00015.981,10.000
  2. −35.981,40.000-35.981,40.000
  3. −35.981,10.000-35.981,10.000 (correct answer)
  4. 5.000,−0.9815.000,-0.981
Explanation: When you see relative polar coordinates in AutoCAD (the @distance<angle format), you're being asked to find a new point by traveling a specific distance in a specific direction from the current point. The key steps are: resolve the polar offset into X and Y components, then add those to the current coordinates. Starting at (−10,25)(-10, 25), the entry @30<210° means travel 30 units at 210°. Since 210° is in the third quadrant (180° + 30°), both X and Y components will be negative. Using trigonometry: ΔX=30cos⁡(210°)=30×(−0.866)≈−25.981\Delta X = 30\cos(210°) = 30 \times (-0.866) \approx -25.981 ΔY=30sin⁡(210°)=30×(−0.500)=−15.000\Delta Y = 30\sin(210°) = 30 \times (-0.500) = -15.000 Adding to the current point: Xnew=−10+(−25.981)≈−35.981X_{new} = -10 + (-25.981) \approx -35.981 Ynew=25+(−15)=10.000Y_{new} = 25 + (-15) = 10.000 This confirms C) (−35.981,10.000)(-35.981, 10.000) is correct. Choice A (15.981,10.00015.981, 10.000) uses a positive X component instead of negative — a sign error that ignores which quadrant 210° falls in. Choice B (−35.981,40.000-35.981, 40.000) gets the X value right but adds the Y offset instead of subtracting, treating sin⁡(210°)\sin(210°) as positive. Choice D (5.000,−0.9815.000, -0.981) appears to misapply the angle entirely, possibly confusing it with a different angle or swapping sine and cosine. A reliable strategy: always sketch the angle quickly. Angles between 180° and 270° are in the third quadrant — both components are negative. Don't let a large angle number trick you into forgetting the sign.

Question 6

In the current UCS, ANGBASE is set to 90°90° and ANGDIR is set to clockwise. A line begins at 0,00,0, and its next point is entered as @20<30°@20<30°.

Which approximate endpoint results from the polar entry?

  1. 10.000,17.32110.000,17.321 (correct answer)
  2. 17.321,10.00017.321,10.000
  3. −10.000,17.321-10.000,17.321
  4. 10.000,−17.32110.000,-17.321
Explanation: When AutoCAD's polar coordinate system is customized, you must account for two system variables before doing any trigonometry: ANGBASE (where 0° is measured from) and ANGDIR (which rotational direction is positive). By default, 0° points East and angles increase counterclockwise — but this question changes both rules. Here, ANGBASE = 90° means the zero-degree reference points North (straight up, the +Y axis), and ANGDIR = clockwise means angles increase in the clockwise direction. So when you enter @20<30°@20<30°, the angle of 30° is measured 30° clockwise from North. That places the line pointing into the upper-right quadrant, 30° from the +Y axis (equivalently, 60° from the +X axis in standard math terms). Converting to Cartesian components: X=20×sin⁡(30°)=20×0.5=10.000X = 20 \times \sin(30°) = 20 \times 0.5 = 10.000 Y=20×cos⁡(30°)=20×0.866=17.321Y = 20 \times \cos(30°) = 20 \times 0.866 = 17.321 The endpoint is (10.000, 17.321)(10.000,\ 17.321), confirming A is correct. B (17.321,10.000)(17.321, 10.000) is the trap for students who correctly identify the two component magnitudes but swap X and Y — forgetting that sine applies to X (the offset from North) and cosine applies to Y (the component along North). C (−10.000,17.321)(-10.000, 17.321) would result from rotating counterclockwise instead of clockwise. D (10.000,−17.321)(10.000, -17.321) ignores the ANGBASE shift entirely and places the line below the X-axis. Whenever you see ANGBASE and ANGDIR modified, redraw a quick mental compass: establish the new zero-degree direction first, then apply the rotation direction before computing components.

Question 7

A line currently ends at 20,1020,10. At the next-point prompt, the user enters 50<150°50<150° without an at sign. Dynamic Input is off, and the current UCS has its origin at 0,00,0 with default angle settings.

What point does AutoCAD use for this entry?

  1. −23.301,35.000-23.301,35.000
  2. 43.301,25.00043.301,25.000
  3. −43.301,25.000-43.301,25.000 (correct answer)
  4. −25.000,43.301-25.000,43.301
Explanation: When you see a polar coordinate entry like 50<150°50<150° without an at sign (@), AutoCAD treats it as an absolute polar coordinate — meaning the distance and angle are measured from the origin (0,0)(0,0), not from the last point. This is the core concept being tested here. To find the resulting point, convert 50<150°50<150° from polar to Cartesian using: x=50cos⁡(150°)=50×(−0.8660)=−43.301x = 50\cos(150°) = 50 \times (-0.8660) = -43.301 y=50sin⁡(150°)=50×(0.5000)=25.000y = 50\sin(150°) = 50 \times (0.5000) = 25.000 So AutoCAD plots the absolute point (−43.301,25.000)(-43.301, 25.000), which is answer C — confirmed as correct. Now consider why the other choices are traps. Answer A (−23.301,35.000)(−23.301, 35.000) looks like someone added the polar result to the last point's coordinates incorrectly — a partial relative calculation gone wrong. Answer B (43.301,25.000)(43.301, 25.000) is the result of using 150°150° but taking the cosine as positive — forgetting that cosine is negative in the second quadrant. Answer D (−25.000,43.301)(−25.000, 43.301) swaps the sine and cosine values, which is what you'd get if you confused the x and y components. The key study tip here: always check for the @ symbol. With @, the entry is relative to the last point; without @, it's absolute from the origin. Since Dynamic Input is off, no automatic relative mode applies. On the AutoCAD exam, questions that give you a "current endpoint" are often testing whether you recognize the @ trap — don't let the last point distract you when @ is missing.

Question 8

A segment must be exactly 100100 units long at an angle of 35°30′35°30' from the positive X-axis. The drawing accepts decimal-degree angle input, and the current point is already established.

Which relative polar entry specifies the required segment without rounding the stated angle?

  1. @100<35.30°@100<35.30°
  2. @100<35.03°@100<35.03°
  3. @100<35.50°@100<35.50° (correct answer)
  4. @100<35.83°@100<35.83°
Explanation: When working with polar coordinates in AutoCAD, you'll frequently need to convert angle measurements between degrees-minutes notation and decimal degrees. This conversion is the core skill being tested here. The stated angle is 35°30′35°30'. To convert minutes to a decimal fraction of a degree, divide the minutes by 60: 30′÷60=0.50°30' \div 60 = 0.50°. Adding that to the whole degrees gives 35°+0.50°=35.50°35° + 0.50° = 35.50°. So the correct relative polar entry is @100<35.50°@100<35.50°, which is choice C. Now let's examine why the wrong answers are tempting traps. Choice A, @100<35.30°@100<35.30°, is the most common mistake — it simply drops the degree-minutes symbol and reads 30′30' as if it were already a decimal (0.30°0.30°). This is incorrect because minutes are not tenths of a degree; there are 60 minutes in a degree, not 100. Choice B, @100<35.03°@100<35.03°, likely arises from dividing 30 by 1000 or some other erroneous scaling — it has no valid conversion basis for 30′30'. Choice D, @100<35.83°@100<35.83°, comes from incorrectly dividing 30 by 36 instead of 60, producing a plausible-looking but wrong decimal. A reliable memory trick: treat degree-minutes like hours and minutes on a clock. Just as 30 minutes is half an hour (0.50.5), 30′30' is half a degree (0.5°0.5°). Always divide minutes by 60 — never just append them as digits — before entering an angle in AutoCAD's decimal-degree format.

Question 9

In the LINE command, Dynamic Input is off and the current UCS uses the default Cartesian axes and angle direction. The first point is entered as 125,80125,80. The next two points are entered as @−35,20@-35,20 and @50<270°@50<270°.

What absolute Cartesian coordinate is established by the final entry?

  1. 90,5090,50 (correct answer)
  2. 140,50140,50
  3. 90,15090,150
  4. 40,10040,100
Explanation: When working with AutoCAD coordinate entry, you need to distinguish between three input types: absolute coordinates (X,YX,Y), relative Cartesian (@ΔX,ΔY@\Delta X,\Delta Y), and relative polar (@distance<angle@distance<angle). The @@ symbol always means "relative to the last point entered." Start by establishing your points step by step. The first point is entered as 125,80125,80 — this is absolute, so your starting position is (125,80)(125, 80). The second entry, @−35,20@-35,20, shifts the position by −35-35 in X and +20+20 in Y, giving you 125+(−35), 80+20=(90,100)125 + (-35),\ 80 + 20 = (90, 100). Now for the final entry: @50<270°@50<270° is relative polar, meaning travel 50 units at 270°. In AutoCAD's default setup, 0° points right (east) and angles increase counterclockwise, so 270° points straight down (south). This means ΔX=50cos⁡(270°)=0\Delta X = 50\cos(270°) = 0 and ΔY=50sin⁡(270°)=−50\Delta Y = 50\sin(270°) = -50. Adding to (90,100)(90, 100): 90+0, 100+(−50)=(90,50)90 + 0,\ 100 + (-50) = (90, 50), which confirms A is correct. Choice B, (140,50)(140, 50), results from adding 50 to X instead of applying the polar angle correctly — treating 270° as an eastward move. Choice C, (90,150)(90, 150), comes from adding +50+50 to Y rather than subtracting, confusing 270° with 90°. Choice D, (40,100)(40, 100), likely reflects a sign error in the first relative entry, misapplying the −35-35 offset. As a study tip, always sketch the displacement direction for polar entries — 90° is up, 270° is down, 180° is left, and 0° is right. This quick mental compass prevents the most common angle-direction mistakes on the AutoCAD exam.

Question 10

A line currently ends at 40,3040,30 and must connect to the absolute point −20,−15-20,-15. The current UCS uses default angle settings, and the next point will be entered with a relative polar coordinate.

Which entry most accurately reaches the required point?

  1. @75<36.87°@75<36.87°
  2. @75<143.13°@75<143.13°
  3. 75<216.87°75<216.87°
  4. @75<216.87°@75<216.87° (correct answer)
Explanation: When working with relative polar coordinates in AutoCAD, two things must be correct simultaneously: the distance and the angle. The @ symbol makes a coordinate relative to the last point, and the angle is measured counterclockwise from the positive X-axis by default. Start by finding the displacement from (40,30)(40, 30) to (−20,−15)(-20, -15): Δx=−20−40=−60\Delta x = -20 - 40 = -60 and Δy=−15−30=−45\Delta y = -15 - 30 = -45. The distance is (−60)2+(−45)2=3600+2025=5625=75\sqrt{(-60)^2 + (-45)^2} = \sqrt{3600 + 2025} = \sqrt{5625} = 75. So the magnitude of 75 is confirmed. Next, find the angle. The reference angle is arctan⁡(4560)=36.87°\arctan\left(\frac{45}{60}\right) = 36.87°, but since both components are negative, the vector points into the third quadrant (down and to the left). The true angle is 180°+36.87°=216.87°180° + 36.87° = 216.87°. The correct entry is therefore @75<216.87°@75<216.87°, confirming D. Choice A (@75<36.87°@75<36.87°) uses the reference angle only, pointing into the first quadrant — the exact opposite direction. Choice B (@75<143.13°@75<143.13°) places the angle in the second quadrant, which would move up and to the left, not down. Choice C (75<216.87°75<216.87°) has the right angle but is missing the @ symbol, making it an absolute polar coordinate measured from the origin — a completely different point. Always remember: @ = relative, no @ = absolute. When your displacement is negative in both X and Y, add 180°180° to your reference angle to land in the correct third-quadrant direction.