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The Coordinate Plane: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The location label

The graph shows points $A$ and $C$ . A location label uses the x-coordinate of $A$ and the y-coordinate of $C$ . Write the ordered pair on the label.

Points $A$ and $C$ on the coordinate plane.

Text description of this figure

A square coordinate grid. The horizontal x-axis and the vertical y-axis each run from negative five to five, with tick marks, number labels and gridlines at every whole number and equal unit lengths on both axes, and the origin labeled 0. Two points are plotted and labeled with their letters only: point A, four units left of the y-axis and three units above the x-axis, at negative four, three; and point C, two units right of the y-axis and two units below the x-axis, at two, negative two. No coordinate pairs, guide lines or other points are shown.
1. Hint 1
  An ordered pair records horizontal position before vertical position.
2. Hint 2
  Read straight from $A$ to the x-axis and from $C$ to the y-axis.
Answer

$(-4, -2)$ .

Full solution

Point $A$ lines up with $-4$ on the x-axis.
Point $C$ lines up with $-2$ on the y-axis.
Put the horizontal reading first.
$(-4, -2)$

The named location would lie directly below $A$ and directly left of $C$ , which agrees with the two readings.

Answer

$(-4, -2)$ .

Key idea

A location can combine the horizontal reading of one point with the vertical reading of another.
Problem 1: Mark complete
Problem 2 A mark on the segment

The graph shows segment $MN$ . Mark and label point $T$ where this segment meets the y-axis, and write its ordered pair.

Segment $MN$ on the coordinate plane.

Text description of this figure

A square coordinate grid. The horizontal x-axis and the vertical y-axis each run from negative four to four, with tick marks, number labels and gridlines at every whole number and equal unit lengths on both axes, and the origin labeled 0. Point M is plotted at negative three, negative one, and point N at three, negative one, each labeled with its letter only. A solid horizontal segment joins M to N and passes across the y-axis. The place where it crosses the y-axis is not marked or labeled, and no other points are shown.
1. Hint 1
  Points on the y-axis have no horizontal movement from the origin.
2. Hint 2
  Follow the drawn segment to the vertical axis and read its height.
Answer

$T$ at $(0, -1)$ .

Full solution

The segment stays at height $-1$ as it runs from left to right.
At the y-axis, the x-coordinate is $0$ .
Mark the crossing and label it $T$ .
$T=(0,-1)$

The mark lies between $M$ and $N$ and on the y-axis, as required.

Answer

$T$ at $(0, -1)$ .

Key idea

The point where a horizontal segment crosses the y-axis has x-coordinate zero and the segment's height as its y-coordinate.
Problem 2: Mark complete
Problem 3 A point $4.5$ units from $E$

A point has the same x-coordinate as $E=(-2,1)$ and is $4.5$ units from $E$ . Find every possible ordered pair for the point.
1. Hint 1
  Points sharing an x-coordinate lie on one vertical line.
2. Hint 2
  The required point can lie above or below $E$ ; change the y-coordinate by the stated distance in each direction.
Answer

$(-2, 5.5)$ and $(-2, -3.5)$ .

Full solution

The x-coordinate stays $-2$ .
A point $4.5$ units above $E$ has y-coordinate
$1+4.5=5.5$
A point $4.5$ units below $E$ has y-coordinate
$1-4.5=-3.5$
Thus the possible points are $(-2,5.5)$ and $(-2,-3.5)$ .

Check the vertical distances.
$|5.5-1|=4.5$ $|-3.5-1|=4.5$
Above and below are the two directions on this vertical line, so these are all the possibilities.

Answer

$(-2, 5.5)$ and $(-2, -3.5)$ .

Key idea

A given positive vertical distance from a point allows one location above it and one below it.
Problem 3: Mark complete
Problem 4 The storage map

Each named zone in the map is exactly one quadrant. Give each zone's quadrant number and the sign pattern of its ordered pairs. Which zone, if any, contains a marker placed at $(-3,0)$ ? Explain.

The storage map on the coordinate plane.

Text description of this figure

A square coordinate grid used as a map. The horizontal x-axis and the vertical y-axis each run from negative four to four, with tick marks, number labels and gridlines at every whole number and equal unit lengths on both axes, and the origin labeled 0. The axes divide the grid into four regions, and one zone name is written in each, with no point markers: Sand centered at two, two, above and right of the origin; Lake centered at negative two, two, above and left; Birch centered at negative two, negative two, below and left; and Ridge centered at two, negative two, below and right. Nothing else is labeled.
1. Hint 1
  Quadrant numbering starts in the upper right and follows the counterclockwise direction.
2. Hint 2
  For each zone, decide whether its points lie right or left of the y-axis and above or below the x-axis.
3. Hint 3
  Write the horizontal sign before the vertical sign.
4. Hint 4
  For the marker, notice which of its coordinates is zero and what that says about where it lies.
Answer

Sand: I, $(+,+)$ ; Lake: II, $(-,+)$ ; Birch: III, $(-,-)$ ; Ridge: IV, $(+,-)$ . The marker at $(-3,0)$ : on the x-axis, in no zone.

Full solution

Quadrant I is the upper-right region, and the numbering runs counterclockwise from there.
Sand is the upper-right zone, so its points have positive x-coordinates and positive y-coordinates.
It is Quadrant I, with pattern $(+,+)$ .

Moving counterclockwise reaches Lake in the upper left.
It is Quadrant II, with pattern $(-,+)$ .
Next is Birch in the lower left, Quadrant III, with pattern $(-,-)$ .
Ridge in the lower right is Quadrant IV, with pattern $(+,-)$ .

The marker at $(-3,0)$ has y-coordinate $0$ , so it lies on the x-axis, on the boundary between Lake and Birch.
The axes are the boundaries of the quadrants and belong to none of them, so the marker is in no zone.

Answer

Sand: I, $(+,+)$ ; Lake: II, $(-,+)$ ; Birch: III, $(-,-)$ ; Ridge: IV, $(+,-)$ . The marker at $(-3,0)$ : on the x-axis, in no zone.

Key idea

Quadrants are numbered counterclockwise from the upper right, each sign pattern follows from the point's side of each axis, and a point on an axis lies in no quadrant.
Problem 4: Mark complete
Problem 5 The robot route

The graph shows a robot's route from $S$ through $A$ and $B$ to $T$ . It travels along the three drawn segments without turning back. Find the total distance traveled and the ordered pair for its final location.

The robot's route from $S$ through $A$ and $B$ to $T$ .

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative six to three and the vertical y-axis from negative five to six, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Four points are labeled with their letters only: S at negative five, four; A at one, four; B at one, negative three; and T at negative two, negative three. A solid segment runs from S to A, a second from A to B, and a third from B to T, and each segment carries a small arrow pointing in the direction of travel: rightward from S to A, downward from A to B, and leftward from B to T. There is no segment from S to T, and no lengths or totals are shown.
1. Hint 1
  Find the length of each part of the route, then combine the lengths.
2. Hint 2
  A horizontal part uses the gap between x-coordinates; a vertical part uses the gap between y-coordinates.
3. Hint 3
  Read the coordinates of $T$ in horizontal, then vertical order.
Answer

$16$ units; final location $(-2,-3)$ .

Full solution

Read the route points as $S=(-5,4)$ , $A=(1,4)$ , $B=(1,-3)$ , and $T=(-2,-3)$ .
The first and last segments are horizontal, and the middle segment is vertical.

Find each length using the coordinate that changes.
$SA=|1-(-5)|$ $SA=6$ $AB=|-3-4|$ $AB=7$ $BT=|-2-1|$ $BT=3$

Add all three traveled lengths.
$6+7+3=16$
The robot travels $16$ units and ends at $(-2,-3)$ .
Counting the spaces on the route gives $6$ right, $7$ down, and $3$ left, which checks the total.

Answer

$16$ units; final location $(-2,-3)$ .

Key idea

The length of a route made of horizontal and vertical segments is the sum of their individual lengths.
Problem 5: Mark complete
Problem 6 Fence posts and paving tiles

The graph shows the four corners of a rectangular garden. Each coordinate unit represents $1$ meter. A fence post stands at each corner, and the posts are spaced $1$ meter apart all the way around the garden's boundary. Square paving tiles, $1$ meter on each side, cover the garden with no gaps or overlaps. How many fence posts are there, and how many tiles?

The four corners of the rectangular garden.

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative three to six and the vertical y-axis from negative four to three, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Four corners are plotted and labeled with their letters only: A at negative two, negative three; B at five, negative three; C at five, two; and D at negative two, two. They are joined in the order A, B, C, D and back to A, outlining a rectangle whose sides run along gridlines. Below the grid a note reads: One coordinate unit represents one meter. No side lengths, perimeter, area, posts or tiles are shown.
1. Hint 1
  The posts are spread along the garden's perimeter, and each tile fills one square meter of its area.
2. Hint 2
  Read one horizontal side and one vertical side from the corner coordinates.
3. Hint 3
  Walk once around the boundary from a corner post: each $1$ -meter step ends at a post, and the last step ends back at the post you started from.
Answer

$24$ fence posts and $35$ tiles.

Full solution

The left and right corners have x-coordinates $-2$ and $5$ .
The lower and upper corners have y-coordinates $-3$ and $2$ .
Thus the side lengths in meters are
$l=|5-(-2)|$ $l=7$ $w=|2-(-3)|$ $w=5$

The garden's perimeter in meters is
$P=2(7+5)$ $P=24$
Start at a corner post and walk once around the boundary in $1$ -meter steps.
The first $23$ steps each end at a new post, and the $24$ th step ends back at the starting post.
So the garden has $1+23=24$ posts, one for each meter of boundary.

Check by sides.
Each horizontal side holds $8$ posts and each vertical side holds $6$ , but each of the $4$ corner posts belongs to two sides and is counted twice.
$2(8)+2(6)-4=24$

Each tile covers $1$ square meter, so the number of tiles equals the area in square meters.
$A=7\times5$ $A=35$
The garden needs $35$ tiles.

Answer

$24$ fence posts and $35$ tiles.

Key idea

Posts spaced one unit apart around a closed boundary number the same as its perimeter, and unit square tiles covering a rectangle with whole-number sides number the same as its area.
Problem 6: Mark complete
Problem 7 The display panel

Segment $MN$ in the graph is the lower edge of a rectangular panel with sides along the gridlines. The panel lies above this segment and has area $54$ square units. Plot its upper corners, labeling the one directly above $M$ as $P$ and the one directly above $N$ as $Q$ , and give their ordered pairs.

Segment $MN$ , the lower edge of the panel.

Text description of this figure

A coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from negative three to seven and the vertical y-axis from negative four to nine, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Point M is plotted at negative one, negative two, and point N at five, negative two, each labeled with its letter only, and a solid horizontal segment joins them. The rest of the grid, including everything above the segment, is empty: no other corners, edges, lengths or shading are drawn.
1. Hint 1
  The given edge determines the width, and the area determines the height.
2. Hint 2
  Divide the area by the width to find the vertical distance from the lower edge to the upper edge.
3. Hint 3
  The upper corners keep the x-coordinates of the corners directly below them.
Answer

$P=(-1,7)$ and $Q=(5,7)$ .

Full solution

The graph gives $M=(-1,-2)$ and $N=(5,-2)$ .
The width is
$w=|5-(-1)|$ $w=6$

Since area equals width times height, the height is
$h=54\div6$ $h=9$
Add this height to the lower edge's y-coordinate.
$-2+9=7$

The point above $M$ retains x-coordinate $-1$ , so plot $P=(-1,7)$ .
The point above $N$ retains x-coordinate $5$ , so plot $Q=(5,7)$ .
The height is $|7-(-2)|=9$ , and $6\times9=54$ checks the required area.

Answer

$P=(-1,7)$ and $Q=(5,7)$ .

Key idea

A rectangle's area and one plotted edge determine how far away its parallel edge must be.
Problem 7: Mark complete
Problem 8 Lea's reversed pairs

Lea says that if reversing the order of a point's coordinates leaves it in the same quadrant, it must leave it in the same place. Is her claim correct? Explain your decision.

Then describe every point whose location does not change when its coordinates are reversed.
1. Hint 1
  A quadrant contains many different locations.
2. Hint 2
  Choose unequal coordinates with the same sign, then compare their positions before and after reversing the order.
3. Hint 3
  Reversing $(a,b)$ gives $(b,a)$ . Ask what must be true of $a$ and $b$ for these two pairs to name the same point.
Answer

No; for example, $(1,4)$ and $(4,1)$ are both in Quadrant I but are different points. A point keeps its location exactly when its two coordinates are equal, such as $(5,5)$ , $(-2,-2)$ or the origin $(0,0)$ .

Full solution

Take the point $(1,4)$ .
Reversing its coordinates gives $(4,1)$ .
Both coordinates in each pair are positive, so both points are in Quadrant I.

The first point is $1$ unit right and $4$ units up from the origin.
The second is $4$ units right and $1$ unit up.
Their horizontal positions differ, so
$(1,4)\ne(4,1)$
They lie in the same quadrant but not at the same place, disproving Lea's claim.

Now reverse any point $(a,b)$ to get $(b,a)$ .
Two ordered pairs name the same point exactly when their first coordinates agree and their second coordinates agree, and here both conditions say $a=b$ .
So a point keeps its location exactly when its two coordinates are equal, such as $(5,5)$ , $(-2,-2)$ or the origin $(0,0)$ .
Every other point moves when its coordinates are reversed, even if it stays in the same quadrant.

Answer

No; for example, $(1,4)$ and $(4,1)$ are both in Quadrant I but are different points. A point keeps its location exactly when its two coordinates are equal, such as $(5,5)$ , $(-2,-2)$ or the origin $(0,0)$ .

Key idea

A reversed ordered pair can land in the same quadrant and still name a different point; it names the same point only when the two coordinates are equal.
Problem 8: Mark complete
Problem 9 Mina's map update

Two points share the same y-coordinate. A map update adds $3$ to both x-coordinates and subtracts $2$ from both y-coordinates. Mina says that the horizontal distance between the points stays the same for every possible starting pair. Is she correct? Explain.
1. Hint 1
  Horizontal distance depends on the gap between the x-coordinates.
2. Hint 2
  Check whether the new y-coordinates still match, then compare the old and new x-coordinate differences.
3. Hint 3
  Write the starting x-coordinates as $a$ and $b$ and simplify $(b+3)-(a+3)$ .
Answer

Yes, for every possible starting pair.

Full solution

Call the starting x-coordinates $a$ and $b$ and the shared y-coordinate $c$ .
The updated points both have y-coordinate $c-2$ , so the distance between them is still horizontal.

The original horizontal distance is $|b-a|$ .
The new distance is
$d=|(b+3)-(a+3)|$ $d=|b-a|$
The added threes cancel, so the distances agree for every choice of $a$ , $b$ , and $c$ .
Mina is correct.

Answer

Yes, for every possible starting pair.

Key idea

Changing both points' x-coordinates by the same amount preserves their horizontal gap when their y-coordinates remain equal.
Problem 9: Mark complete
Problem 10 The two frames

The graph shows rectangular frame $U$ . A printer draws frame $V$ by reversing the two numbers in each corner's ordered pair and joining the new corners in the same order. One coordinate unit has the same length in both frames. Nolan says the two frames need equal amounts of trim around their boundaries and equal amounts of fabric to fill their insides. Is he correct? Justify your answer with the boundary lengths and inside areas.

Frame $U$ on the coordinate plane.

Text description of this figure

A coordinate grid headed Frame U, with equal unit lengths on both axes. The horizontal x-axis runs from negative two to eight and the vertical y-axis from negative three to two, with tick marks, number labels and gridlines at every whole number, and the origin labeled 0. Four corners are plotted and labeled with their letters only: A at negative one, negative two; B at seven, negative two; C at seven, zero; and D at negative one, zero. They are joined in the order A, B, C, D and back to A to outline a rectangle, whose top side from D to C lies along the x-axis. No second frame, side lengths, perimeter or area are shown.
1. Hint 1
  Reversing an ordered pair changes the roles of its horizontal and vertical readings.
2. Hint 2
  Read the four corners of $U$ and write each pair in reverse order to find the corners of $V$ .
3. Hint 3
  Find one horizontal and one vertical side for each frame, then compare the two boundary lengths and the two areas.
Answer

Yes; each boundary is $20$ units, and each inside area is $16$ square units.

Full solution

The corners of $U$ , in order, are $(-1,-2)$ , $(7,-2)$ , $(7,0)$ , and $(-1,0)$ .
Reversing each ordered pair gives the corners of $V$ : $(-2,-1)$ , $(-2,7)$ , $(0,7)$ , and $(0,-1)$ .

Frame $U$ has horizontal side length $|7-(-1)|=8$ and vertical side length
$|0-(-2)|=2.$
Frame $V$ has horizontal side length $|0-(-2)|=2$ and vertical side length
$|7-(-1)|=8.$
The side lengths have exchanged roles.

The boundary lengths are
$P_U=2(8+2)$ $P_U=20$ $P_V=2(2+8)$ $P_V=20$
The inside areas are
$A_U=8\times2$ $A_U=16$ $A_V=2\times8$ $A_V=16$
Each frame needs $20$ units of trim and $16$ square units of fabric, so Nolan is correct.

Answer

Yes; each boundary is $20$ units, and each inside area is $16$ square units.

Key idea

Reversing every corner's ordered pair of a rectangle with sides along the gridlines exchanges its horizontal and vertical side lengths, so its perimeter and area stay the same.
Problem 10: Mark complete

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