Digital SAT Math · Geometry & Trigonometry

Coordinate geometry

The coordinate plane turns geometry into algebra. Distance is Pythagoras on the differences, midpoint is an average, perpendicular slope is the negative reciprocal, and a circle is one equation that hands you the centre and radius if you read the signs. One toolkit — four decisions.

On the test

Recognition cues: in the xy-plane; two points written $(x_1,y_1)$ and $(x_2,y_2)$; midpoint, distance, diameter; perpendicular or negative reciprocal; equation with two squared binomials equal to a constant.

Where this skill ends and its neighbours begin

Matching a drawn line by slope sign and intercept lives on graphs of linear equations — picture-first Algebra. Function language $f(x)$, tables, input-vs-output lives on linear functions. Pure right-triangle side lengths without coordinates will sit on the Pythagorean-theorem page; arc length, sectors, inscribed angles will sit on circles. This page is the coordinate toolkit: points → length, halfway, slope relation, or circle equation.

Pattern recognition

Sort by what the question is actually asking for:

1. Length — distance formula (or leave under a square root / report the square if asked).
2. Halfway — midpoint averages; diameter centre is a midpoint.
3. Direction relation — parallel (same slope) vs perpendicular (product $-1$).
4. Circle features — read $(h,k)$ and $r$ from standard form, or complete the square when forced.
5. Mixed — diameter endpoints → midpoint centre → distance radius → equation.

Method

1. Sketch or list the given points. Write $\Delta x$ and $\Delta y$ with a consistent order (second minus first).
2. Distance when length is asked. $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Squares kill the sign of each difference.
3. Midpoint when halfway is asked. $M=\Bigl(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\Bigr)$. For a missing endpoint, double the midpoint and subtract the known end.
4. Perpendicular slope = negative reciprocal. If $m=\dfrac{a}{b}$, then $m_\perp=-\dfrac{b}{a}$. Reciprocal alone (drop the minus) and sign flip alone (keep the fraction) are both wrong and both appear in choices.
5. Circle: read standard form first. $(x-h)^2+(y-k)^2=r^2$ → centre $(h,k)$, radius $r=\sqrt{\text{RHS}}$. Completing the square only when the equation is expanded.

Four tools, one grid

Practice

Answer before you open the explanation. Four items depend on a figure — read the grid or the circle on axes before the choices. Two items are student-produced response. Every wrong choice below is a specific, named error: midpoint-where-distance, reciprocal-not-negative-reciprocal, $h,k$ unflipped, or $r^2$ as $r$.

Common mistakes

1. Using the midpoint when distance was asked — averaging the coordinates yields a point, not a length. If the answer choices include an ordered pair and a number, re-read the verb.
2. Taking the reciprocal without the negative for a perpendicular slope. Parallel keeps the slope; perpendicular multiplies to $-1$.
3. Reading $h$ and $k$ with unflipped signs — $(x+4)^2$ means $h=-4$; $(y-1)^2$ means $k=1$. Write centre as $(h,k)$ only after rewriting each binomial as $(x-h)$ and $(y-k)$.
4. Reporting $r^2$ as the radius — the right-hand side of standard form is already squared; take the square root (and keep it positive).
5. Forgetting to square before adding in the distance formula — treating $\lvert\Delta x\rvert+\lvert\Delta y\rvert$ as the length.
6. Leaving the distance as the sum of squares — computing $36+64=100$ and stopping without $\sqrt{\phantom{0}}$.
7. Dividing one coordinate average but not the other when finding a midpoint, or summing without dividing by 2.
8. Mismatched subtraction order on only one difference when building slope (sign error) — reverse both differences or reverse neither.
9. Completing the square but forgetting to add the constant to both sides — the radius comes out wrong even when the centre looks right.

FAQ

Distance or midpoint — how do I tell in two seconds? Look at the noun the question wants. Distance, length, how far → root of sum of squares. Midpoint, halfway, centre of the segment/diameter → averages. If both tools appear in the choices, you already know which trap they built.

Why negative reciprocal and not just opposite sign? Opposite sign alone ($-\frac{2}{3}$ when $m=\frac{2}{3}$) does not make the product $-1$. You need both the flip and the minus: $m\cdot m_\perp=-1$.

Do I always complete the square for a circle? Only when the equation is expanded ($x^2+y^2+Dx+Ey+F=0$). If it is already $(x-h)^2+(y-k)^2=r^2$, read centre and radius directly.

Can a grid-in answer here be negative or a radical? Yes for coordinates and for signed slopes. For a length, the distance is non-negative; enter exact values such as 10 or, when the stem asks for the square of the distance, the integer under the root.

Is the diameter’s midpoint always the centre? Yes — every diameter is a chord through the centre, so its midpoint is the centre. Radius is then the distance from that centre to either endpoint.

Editorial team · last reviewed 2026-08-12