This section is all about straight lines on a grid. Good news: there are only a few ideas here, and the ISEE asks about them in very predictable ways. Find the slope, write the line, find where it crosses the axes, check if two lines are parallel or perpendicular, and measure between two points. Learn the handful of formulas below, learn the traps, and these become free points.
The practice is mixed into the lesson: each question sits right after the idea it needs, and the Mixed Practice at the bottom takes that hint away. Flip the switch above a question to type your answer instead: a number, a fraction like 4/3, a root like 5√2, a point like (1, -5), an equation like y = 2x - 3, or a quadrant like IV. When the answer has to be one of the choices, that version stays multiple choice, just like the real ISEE.
Everything in this section lives on one grid, so let's name its parts first.
The coordinate plane is the grid made by two number lines crossing at a right angle: the horizontal $x$-axis and the vertical $y$-axis. They cross at the origin, $(0, 0)$.
Every point gets an address, $(x, y)$: first how far left or right of the origin, then how far up or down. The two axes chop the plane into four regions.
The four quadrants (I–IV) divide the plane by the axes and indicate the signs of $(x, y)$.
These exact same four quadrants come back in trigonometry, where the signs of $x$ and $y$ decide whether sine and cosine are positive or negative. See the Angles page when you get there.
They're numbered counterclockwise, starting from the top right. Instead of memorizing the picture, memorize the signs.
So a quadrant question is really a sign question. Here's one with a catch.
Where are the points (a) $(-3, 7)$ and (b) $(0, -4)$?
(a) Negative $x$, positive $y$: left and up, so Quadrant II.
(b) Trick question. With $x = 0$, the point sits right on the $y$-axis, so it isn't in any quadrant.
That second one is the trap to remember.
A point with a $0$ in it is on an axis, never inside a quadrant. The ISEE loves to sneak one in as a wrong answer.
Sort a few points by their signs and watch where they land.
Your turn: read the signs, name the quadrant.
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The ISEE also likes to hide the numbers: it tells you which quadrant $(a, b)$ is in and asks about a rearranged point, like $(b, -a)$. Same trick, just track the signs. In Quadrant II, $a$ is negative and $b$ is positive, so $(b, -a)$ is $(+, +)$: Quadrant I.
Now with no picture at all.
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Slope is the single most important number about a line. It tells you how steep the line is and which way it tilts.
Slope ($m$) is how steep a line is: how much $y$ changes for every step of $1$ in $x$.
To compute it, pick two points on the line and compare how far it rises to how far it runs. Rise over run.
Watch it happen: the run, then the rise, then the division. Then the line tips through every other kind of slope.
So lines that rise have positive slope, lines that fall have negative slope, flat lines have slope $0$, and vertical lines have undefined slope, because their run is $0$.
A vertical line has slope $\frac{\text{rise}}{0}$. I finally get to divide by zero, and the ISEE just calls it "undefined." Rude.
Flat and vertical lines also have the simplest equations of all.
What are the equations of the horizontal line and the vertical line through $(-2, 5)$?
Every point on the horizontal line has height $5$, so it's $y = 5$ (slope $0$). Every point on the vertical line is at $x = -2$, so it's $x = -2$ (slope undefined). Watch out: the line called $x = -2$ is the vertical one, because $x$ is the thing that never changes along it.
Try one. Ask yourself which letter stays the same along the line.
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Back to slanted lines, with actual numbers this time.
Calculate the slope of the line through $A(-2, 3)$ and $B(4, 7)$.
$$m = \frac{7 - 3}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}$$
Two things go wrong here all the time.
Subtract in the same order on top and bottom. If you start with $B$'s $y$ on top, start with $B$'s $x$ on the bottom. Mixing the order flips the sign. And watch the double negative: $4 - (-2)$ is $6$, not $2$.
Now you try it, from points or from a graph.
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Now get a feel for it. The rise and the run each have a sign: right and up are positive, left and down are negative.
One trap with graphs: a steep line looks "big," but if it falls, its slope is negative. Bigger slope means more uphill, not more tilted.
Slope also works as walking directions. Slope $\frac{2}{3}$ means: every $3$ steps right, go $2$ up. So from one point you know, you can find more points on the line without ever writing its equation.
A line has slope $\frac{3}{4}$ and passes through $(-2, 1)$. Is the point $(6, 7)$ on the line?
From $x = -2$ to $x = 6$ is a run of $8$: two batches of $4$ steps. Each batch rises $3$, so the line rises $6$ in all, from $1$ up to $7$. Yes, $(6, 7)$ is on the line.
Here's that walk on the grid, one batch at a time.
The same walk works backward, too: if you know the slope and one point, a missing coordinate is just a few steps away.
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The same line can be written three different ways. The ISEE switches between them on purpose, so you need to recognize all three.
Slope-intercept form writes a line as $y = mx + b$, where $m$ is slope and $b$ is the $y$-intercept.
This is the friendliest form: the slope and the starting height are just sitting there. The next one is better when you know a point instead of the intercept.
Point-slope form writes a line through $(x_1, y_1)$ with slope $m$ as $y - y_1 = m(x - x_1)$.
They're the same line, just started from different places. Watch both get built on one line.
Point-slope form looks like someone took $y = mx + b$, shook it, and subtracted things until it stopped moving.
And the third form hides the slope on purpose.
A line in standard form is written $Ax + By = C$, where $A$ and $B$ aren't both $0$.
You could solve for $y$ every time, but there's a shortcut.
The line $Ax + By = C$ has slope $\displaystyle m = -\frac{A}{B}$ (when $B \neq 0$).
Here's the shortcut in action, with a check the long way.
Find the slope of $6x - 2y = 3$.
$A = 6$ and $B = -2$, so $m = -\frac{6}{-2} = 3$. (Check by solving for $y$: $-2y = -6x + 3$, so $y = 3x - \frac{3}{2}$.)
Where does that $-\frac{A}{B}$ come from? Watch the equation get $y$ alone, one move at a time.
Every standard-form line works the same way: move the $x$-term, divide by $B$, and the slope is $-\frac{A}{B}$. Your turn.
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Now put it together. Back to the points from before.
Find the equation of the line through $A(-2, 3)$ and $B(4, 7)$ in $y = mx + b$ form.
We already found $m = \frac{2}{3}$. Plug in point $A$ to find $b$: $3 = \frac{2}{3}(-2) + b$, so $3 = -\frac{4}{3} + b$, so $b = \frac{13}{3}$.
So the line is $y = \frac{2}{3}x + \frac{13}{3}$. Ugly numbers are fine. The ISEE usually picks nicer ones.
Slope first, then $b$. That order never changes, whatever you're handed.
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One more move: a point is on a line exactly when its coordinates make the equation true. So if a problem gives you a point with a missing coordinate, plug in what you know and solve.
The point $(3, k)$ lies on the line $y = -2x + 4$. Find $k$.
Plug in $x = 3$: $k = -2(3) + 4 = -2$.
When the missing coordinate is the $x$ instead, it's a two-step equation, but the idea is the same.
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On the ISEE the line is often just drawn for you. Then the graph hands you both numbers, as long as you read it carefully.
What is the equation of the line shown?
The line crosses the $y$-axis at the marked point $(0, -1)$, so $b = -1$. From there, walk to the other marked point, $(3, 1)$: right $3$, up $2$. So $m = \frac{2}{3}$, and the line is $y = \frac{2}{3}x - 1$.
The one thing to be careful about is which points you trust.
Only trust points where the line passes exactly through a grid corner, like the marked ones. Guessing a point that's "about" on the line is how you end up with a slope that's slightly off, and it'll be one of the choices.
Read this one the same way: $b$ first, then the step.
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Here's every slope-intercept line at your fingertips. The sign of $b$ decides which side of the origin the line crosses the $y$-axis on, and the sign of $m$ decides which way it tilts. Together they pick the three quadrants it visits.
So you can tell which quadrants a line crosses without drawing it.
Which quadrants does the line $y = -2x + 3$ pass through?
$b = 3$ is positive, so the line crosses the $y$-axis above the origin, in between Quadrants I and II. $m = -2$ is negative, so it falls from left to right: it comes down out of Quadrant II, crosses into Quadrant I, and keeps falling into Quadrant IV. So Quadrants I, II, and IV, and it never touches Quadrant III.
First the forward direction: equation in, quadrants out.
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And backward: from where the line goes to the signs of $m$ and $b$.
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Intercepts are where a line crosses the axes. They're the easiest points to find on any line.
Intercepts are where a graph meets the axes: the $x$-intercept at $y = 0$, the $y$-intercept at $x = 0$.
Finding an intercept is just solving a linear equation: plug in the $0$, then undo the operations one at a time.
Let's find both intercepts of the line from before.
Find the $y$-intercept and the $x$-intercept of $y = \frac{2}{3}x + \frac{13}{3}$.
$y$-intercept: set $x = 0$, and you get $y = \frac{13}{3}$. In this form it's just $b$.
$x$-intercept: set $y = 0$. Then $0 = \frac{2}{3}x + \frac{13}{3}$, so $\frac{2}{3}x = -\frac{13}{3}$ and $x = -\frac{13}{2}$.
Notice which variable got set to $0$ each time.
The names are backwards from what you'd guess. To find the $x$-intercept, you set $y$ to $0$. If you plug in $x = 0$ when they asked for the $x$-intercept, you'll get the $y$-intercept, and it'll be one of the wrong answers.
Some lines are written so the intercepts are basically handed to you.
The line $\frac{x}{2} + \frac{y}{3} = 1$ meets the axes at $(x_0, 0)$ and $(0, y_0)$. Which is bigger, $x_0$ or $y_0$?
Set $y = 0$: $\frac{x}{2} = 1$, so $x_0 = 2$. Set $x = 0$: $\frac{y}{3} = 1$, so $y_0 = 3$. So $x_0 < y_0$. In this form, each intercept is the number under its variable.
For a line in standard form there's a quick way to do this in your head: cover up one term, and what's left solves itself.
Now go the other way: read the intercepts off a few graphs and match each one to its equation.
Your turn. Decide which letter gets the $0$ before you do anything else.
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Two lines can relate to each other in two special ways, and both come down to slope.
Parallel lines are two different lines that never meet.
Same tilt, different starting point, so they run side by side forever. The other relationship is all about the right angle, and it needs one new word first.
The negative reciprocal of a nonzero number flips it upside down and changes its sign.
It takes both moves, every time.
Find the negative reciprocals of $\frac{2}{3}$ and $-4$.
Flip $\frac{2}{3}$ to get $\frac{3}{2}$, then change the sign: $-\frac{3}{2}$. Write $-4$ as $-\frac{4}{1}$, flip it to $-\frac{1}{4}$, then change the sign: $\frac{1}{4}$.
Now the right angle:
Perpendicular lines meet at a right angle.
Both relationships turn into a test on the slopes.
(parallel: two different lines, neither one vertical; perpendicular: neither line horizontal or vertical. Two different vertical lines are parallel, and a horizontal line is always perpendicular to a vertical one.)
Why the flip? Turn a line a quarter turn and its rise and run trade places, with one of them changing sign. That trade is the negative reciprocal.
Parallel lines: same slope, done. Perpendicular lines: flip it AND change the sign. Why does one of them get two chores?
Here's the flip doing real work.
A line $\ell$ passes through $(0, -2)$ and is perpendicular to $y = \frac{1}{3}x + 5$. What is the equation of line $\ell$?
Flip $\frac{1}{3}$ to get $3$, then change the sign: $m = -3$. The point $(0, -2)$ is on the $y$-axis, so $b = -2$. The line is $y = -3x - 2$.
The wrong answers here are predictable.
Two half-done answers are always waiting for you: just flipping ($3$) and just negating ($-\frac{1}{3}$). A negative reciprocal needs both.
The ISEE also likes to write the answer in standard form, so you can't just match slopes by eye.
Which line is parallel to $y = -2x + 7$: (a) $2x + y = 10$ or (b) $x + 2y = 10$?
We need slope $-2$. For (a), $m = -\frac{2}{1} = -2$. For (b), $m = -\frac{1}{2}$. So (a) is parallel.
Turn a line yourself and watch the two slopes keep their deal.
First just the slope.
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Then the whole line through a point, the way the perpendicular example did it.
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Last two formulas. Both take two points, $A(x_1, y_1)$ and $B(x_2, y_2)$.
The midpoint is the average of the endpoints:
$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
The midpoint answers "where is the halfway point?" The other formula answers "how far apart are they?"
The distance formula gives the length of $\overline{AB}$:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
The distance formula is the Pythagorean theorem in disguise. The horizontal and vertical gaps are the legs of a right triangle, and the distance is the hypotenuse. Full story in Section 6.
Here's the disguise coming off: the gaps grow into the legs of a right triangle, and the segment is its long side.
Now both formulas with numbers.
Find the midpoint of $A(-2, 3)$ and $B(4, 7)$, and the distance from $(0, 0)$ to $(-6, 8)$.
Midpoint: $\left(\frac{-2 + 4}{2}, \frac{3 + 7}{2}\right) = (1, 5)$.
Distance: $\sqrt{(-6)^2 + 8^2}$ $= \sqrt{36 + 64}$ $= \sqrt{100} = 10$.
Before you trust your eyes, guess a few midpoints and check them.
Average the $x$'s, average the $y$'s. Your turn.
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For distance, the trap is the one the animation showed.
Don't add the legs. From $(0, 0)$ to $(-6, 8)$ is not $6 + 8 = 14$. That's the walking distance around the corner, and it's always one of the choices. Square, add, then square root.
Drag the points and build some distances of your own.
Now a distance question, sometimes on a town map.
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Sometimes the ISEE gives you the midpoint and asks for an endpoint. The midpoint is halfway, so just take the same step again.
The midpoint of $\overline{AB}$ is $M(2, -1)$, and $A = (-1, 3)$. Find $B$.
From $A$ to $M$, $x$ goes up by $3$ and $y$ goes down by $4$. Do the same step again from $M$: $B = (2 + 3, -1 - 4) = (5, -5)$.
Check: $\left(\frac{-1 + 5}{2}, \frac{3 + (-5)}{2}\right) = (2, -1)$. ✓
Here's that step, taken twice.
The usual wrong turn is reaching for the formula out of habit.
Plugging $A$ and $M$ into the midpoint formula gives the point halfway between them, which stops short of $M$ instead of going past it. It's always one of the choices.
Your turn: same step again, past the midpoint.
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One last term, which mixes midpoints with perpendicular lines (and shows up in the Spicy Bonus at the end).
The perpendicular bisector of a segment is the line through its midpoint that meets it at a right angle.
Keep it in your back pocket. Everything it needs is already on this page.
A line's equation is a rule: put in an $x$, get out a $y$. Function notation just gives the rule a name, so $f(x) = 3x + 1$ means "the rule $f$ triples a number and adds one," and $f(4) = 13$.
Functions get a whole page of their own in precalculus: Function Notation.
The ISEE loves to invent its own symbols for rules, just to see if you panic.
A defined operation is a made-up symbol with a rule attached.
To use one, plug the number into the rule. That's all.
If $n^* = 4n + 3$, what is the value of $8^*$?
Replace $n$ with $8$: $8^* = 4(8) + 3 = 35$. The star isn't an exponent or anything scary; it's just the rule's name.
Whatever the symbol, it's the same move. Try one.
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Rules can also be stacked, one inside another. Always work from the inside out.
If $f(x) = 2x + 1$ and $g(x) = x^2$, what is $f(g(3))$?
Inside first: $g(3) = 9$. Then $f(9) = 2(9) + 1 = 19$.
Doing $f$ first gives $g(f(3)) = g(7) = 49$, a completely different number. The same goes for a made-up rule used twice: with $n^* = 4n + 3$, the value of $(2^*)^*$ starts with $2^* = 11$, and then $11^* = 47$.
Watch the same $3$ go through both orders.
Same input, same two rules, two answers. Find the inside rule first.
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Comparing two functions is the same idea: plug in, then compare.
If $f(x) = \sqrt{x}$ and $g(x) = x^2$, which is true?
$g(0.9) = 0.81$ and $f(0.9) \approx 0.95$, while $f(1.1) \approx 1.05$ and $g(1.1) = 1.21$.
So $g(0.9) < f(0.9) < f(1.1) < g(1.1)$: the second one.
There's a rule of thumb that saves you the arithmetic.
Between $0$ and $1$, squaring makes a number smaller and a square root makes it bigger. Above $1$, it's the other way around. No calculator needed.
Watch the three of them swap places as $x$ crosses $1$.
First decide which side of $1$ you're on. Then the order is automatic.
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Here's a rule that refuses some inputs: you can't take the square root of a negative number (not on the ISEE, anyway).
Now let's move shapes around the grid. There are three moves.
A translation slides every point of a figure the same distance in the same direction.
A slide is the easiest move to do with numbers: add to the coordinates.
The point $(2, -1)$ is translated $3$ units left and $4$ units up. Where does it land?
Left is negative and up is positive: $(2 - 3, -1 + 4) = (-1, 3)$.
The place a point lands, like $(-1, 3)$ here, has a name.
The image of a point or figure is where it lands after a move. Its corners get a prime mark: corner $A$ lands on $A'$ (say "A prime").
Or flip it instead of sliding it:
A reflection flips a figure over a line, like a mirror image.
The mirror keeps one coordinate and flips the other.
Reflect $A(3, 2)$ over the $x$-axis. Where is $A'$?
The $x$-axis is the mirror, so $A$ stays $3$ across and flips from $2$ above it to $2$ below it: $A' = (3, -2)$.
Or spin it:
A rotation turns a figure around a point (on the ISEE, usually the origin) by some angle.
Every move has a coordinate rule:
Here are the four mirrors at work on one point, $(4, 1)$.
Moves can also be chained. Just follow one corner through each rule in turn.
A figure has a corner at $(2, 5)$. Reflect it over the $y$-axis, rotate it $180^\circ$ about the origin, then reflect it over the $x$-axis. Does it end up exactly where it started?
Yes. Follow one corner: $(2, 5) \to (-2, 5) \to (2, -5) \to (2, 5)$. Back where it started! The rules agree for every point: $(x, y) \to (-x, y) \to (x, -y) \to (x, y)$.
That wasn't luck, and it's worth knowing why.
Reflecting over the $x$-axis and the $y$-axis does the same thing as a $180^\circ$ rotation. So "reflect, rotate $180^\circ$, reflect" cancels itself out. Track one point through the moves and you'll never get lost.
Now play with the moves yourself.
Now predict before you move: where will corner $A$ end up?
Your turn, without the grid moving for you.
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And the question the ISEE likes best, turned around: the figure has already moved. Which moves did it?
Test each choice on one corner. The right route lands it on $A'$.
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One more idea, and it's slope wearing a costume.
The average rate of change of $f$ from $x_1$ to $x_2$ is $\dfrac{f(x_2) - f(x_1)}{x_2 - x_1}$.
It's a rate, just like miles per hour: change in output per change in input. Slope is the same thing for a line.
That's just slope between the two points $(x_1, f(x_1))$ and $(x_2, f(x_2))$. It works even when the graph isn't a line.
Find the average rate of change of $f(x) = x^2$ from $x = 1$ to $x = 3$.
$f(3) = 9$ and $f(1) = 1$, so the rate is $\frac{9 - 1}{3 - 1} = \frac{8}{2} = 4$.
Here's that same calculation on the graph, one step at a time.
Then drag the two points along the curve. The line through them is the "average," and its slope is the rate.
Outputs first, then the subtraction on top, then the division. Your turn.
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Every question so far came right after its idea, which is a hint the real test never gives. Down here the hint is gone. First, the quantitative comparison: two columns, four fixed choices, and any idea from this section fair game. Watch out for "cannot be determined": a sign or a quadrant often isn't enough to settle a comparison.
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Next, the mixed set. Every question type on this page except the quantitative comparison goes into one hat, and each click draws a new one at random. Flip the switch to type a number answer; points, equations, and quadrants stay multiple choice down here.
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Alright, if you've made it this far, it's time for something with a little heat. This one chains together almost everything on this page: a midpoint, a slope, a negative reciprocal, and an equation. (It asks for a perpendicular bisector, from the Midpoint & Distance section.) Take it one step at a time.
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