Precalculus · Unit 1 · Lesson 1 · Section 1.1

Rectangular coordinates

Give every point a place. Use coordinates to describe position, measure distance, and find the halfway point.

Video lesson9 worked examples4 printable PDFs

01 / THE GOAL

By the end, you can…

Plot ordered pairs and identify quadrants and axes.
Use distance and midpoint formulas, including exact radical answers.
Verify right triangles and interpret coordinates in real situations.

Before you start: Signed numbers, absolute value, simplifying square roots, and the Pythagorean Theorem.

02 / WATCH & LEARN

A little guidance goes a long way.

Keep your notes nearby. Pause whenever you need to try a step yourself.

The rectangular coordinate system and plottingProfessor Leonard · Lesson video

Professor LeonardGuest lesson · Shawn Todd’s video will replace this when available.

Watch on YouTube ↗

03 / THE MAIN IDEA

Let’s make it click.

An ordered pair gives two directions.

In (x, y), move horizontally first and vertically second. Positive x moves right; negative x moves left. Positive y moves up; negative y moves down. The origin is (0, 0).

\mathrm{I}:(+,+)\quad\mathrm{II}:(-,+)\quad\mathrm{III}:(-,-)\quad\mathrm{IV}:(+,-)

A point on either axis is not in a quadrant. Points (0, y) lie on the y-axis; points (x, 0) lie on the x-axis.

Distance comes from a right triangle.

The horizontal and vertical changes form the legs of a right triangle. Use the Pythagorean Theorem to find the straight-line distance. Square each change before adding; do not add the changes first.

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Simplify the radical completely. For this practice sheet, give a two-decimal approximation when the radical cannot be simplified further.

A midpoint is an average in each direction.

Average the x-coordinates together, then average the y-coordinates together. Keep the answer as an ordered pair. In an application, interpret both coordinates and include units.

M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

04 / SEE THE STEPS

One step. Then the next.

These examples follow the problem types on your practice sheet. Use Next step to reveal the reasoning, or open the complete solution to read at your own pace.

GROUP A · PRACTICE 1–6

Plot a point and identify its quadrant

Plot (3, −5), then name its quadrant.

(x,y)=(3,-5)

Start at the origin. The first coordinate controls horizontal movement.

Step 1 of 4

Read the complete solution

$(x,y)=(3,-5)$
Start at the origin. The first coordinate controls horizontal movement.
$x=3\quad\text{(right 3)}$
Move 3 units right along the x-axis.
$y=-5\quad\text{(down 5)}$
Move 5 units down. Mark the point at (3, −5).
$(+,-)\ \Longrightarrow\ \boxed{\text{Quadrant IV}}$
Positive x and negative y place the point in the lower-right quadrant.

GROUP A · PRACTICE 4

Recognize an axis point

Where is (0, 7)?

(0,7)

The x-coordinate is zero, so there is no horizontal movement.

Step 1 of 2

Read the complete solution

$(0,7)$
The x-coordinate is zero, so there is no horizontal movement.
$x=0\ \Longrightarrow\ \boxed{\text{on the }y\text{-axis}}$
Move up 7 units. The point is on an axis, not in a quadrant.

GROUP A · PRACTICE 7

Let the meaning of the coordinates guide you

For h(t) = 10 + 12t + 4.9t², where do the points lie when t is elapsed time?

(t,h)\quad\text{with}\quad t\ge0

Elapsed time is nonnegative, so only the right half of the coordinate plane is relevant.

Step 1 of 3

Read the complete solution

$(t,h)\quad\text{with}\quad t\ge0$
Elapsed time is nonnegative, so only the right half of the coordinate plane is relevant.
$h(0)=10,\qquad h(t)>0\text{ for }t\ge0$
For the formula printed on the sheet, every term is nonnegative when t ≥ 0, and the constant term is 10.
$\boxed{\text{Quadrant I for }t>0}$
At t = 0, the point (0, 10) is on the positive vertical axis. For positive time, both coordinates are positive.

GROUP B · PRACTICE 8, 10, 11

Find a distance

Find the distance between (1, 2) and (4, 6).

\Delta x=4-1=3,\quad\Delta y=6-2=4

Find the horizontal and vertical changes. These are the legs of a right triangle.

Step 1 of 3

Read the complete solution

$\Delta x=4-1=3,\quad\Delta y=6-2=4$
Find the horizontal and vertical changes. These are the legs of a right triangle.
$d=\sqrt{(4-1)^2+(6-2)^2}$
Substitute into the Distance Formula.
$d=\sqrt{3^2+4^2}=\sqrt{25}=\boxed{5}$
Square, add, and take the square root. Distance is nonnegative.

GROUP B · PRACTICE 9 + NOTES WARM-UP

Simplify a radical—or approximate at the end

Compare √72 with the distance from (−3, 1) to (2, −1).

\sqrt{72}=\sqrt{36\cdot2}=\boxed{6\sqrt2}

Look for a perfect-square factor. This radical simplifies, so keep the simplified exact form.

Step 1 of 3

Read the complete solution

$\sqrt{72}=\sqrt{36\cdot2}=\boxed{6\sqrt2}$
Look for a perfect-square factor. This radical simplifies, so keep the simplified exact form.
$d=\sqrt{(2-(-3))^2+(-1-1)^2}$
For the two points, subtract carefully. A double negative becomes addition.
$d=\sqrt{25+4}=\sqrt{29}\approx\boxed{5.39}$
29 has no perfect-square factor greater than 1. Keep √29 in your work and round the final result to two decimals, as the practice directs.

GROUP C · PRACTICE 12–15

Find the halfway point

Find the midpoint of (−4, 6) and (2, −2).

M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

Average the horizontal coordinates and vertical coordinates separately.

Step 1 of 3

Read the complete solution

$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
Average the horizontal coordinates and vertical coordinates separately.
$M=\left(\frac{-4+2}{2},\frac{6+(-2)}{2}\right)$
Keep each coordinate in its own fraction.
$M=\left(\frac{-2}{2},\frac{4}{2}\right)=\boxed{(-1,2)}$
The midpoint is an ordered pair, not a distance.

GROUP D · PRACTICE 16

Verify a right triangle

Use A(1, 1), B(4, 1), and C(4, 5) to verify a right triangle.

AB=|4-1|=3,\quad BC=|5-1|=4

AB is horizontal and BC is vertical. Find their lengths using absolute differences.

Step 1 of 3

Read the complete solution

$AB=|4-1|=3,\quad BC=|5-1|=4$
AB is horizontal and BC is vertical. Find their lengths using absolute differences.
$AC=\sqrt{(4-1)^2+(5-1)^2}=5$
Use the Distance Formula for the slanted side. It is the longest side.
$3^2+4^2=9+16=25=5^2$
The side lengths satisfy a² + b² = c². By the converse of the Pythagorean Theorem, the triangle is right, with its right angle at B.

GROUP D · PRACTICE 17

Estimate a value using a midpoint

Sales were $8.2 million in 2019 and $11.6 million in 2023. Estimate 2021 sales assuming a linear trend.

(2019,8.2),\quad(2023,11.6)

Treat the year as x and sales in millions of dollars as y.

Step 1 of 3

Read the complete solution

$(2019,8.2),\quad(2023,11.6)$
Treat the year as x and sales in millions of dollars as y.
$M=\left(\frac{2019+2023}{2},\frac{8.2+11.6}{2}\right)$
2021 is halfway between 2019 and 2023, so average the sales values.
$M=(2021,9.9)$
The estimate is $9.9 million in 2021. This midpoint estimate relies on the stated linear-trend assumption.

GROUP D · PRACTICE 18

Turn map units into real distance

Entrances are at (2, 3) and (14, 8). Each map unit represents 100 meters.

\Delta x=14-2=12,\quad\Delta y=8-3=5

First calculate distance in map units.

Step 1 of 3

Read the complete solution

$\Delta x=14-2=12,\quad\Delta y=8-3=5$
First calculate distance in map units.
$d=\sqrt{12^2+5^2}=\sqrt{169}=13$
The straight-line distance is 13 map units, not 13 meters.
$13\text{ units}\cdot\frac{100\text{ m}}{1\text{ unit}}=\boxed{1300\text{ m}}$
Multiply by the scale factor. State the result in meters.