38. Coordinate Geometry · Mastering AMC 10/12

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Line in Coordinate Plane

In this section you will be able to turn lines into equations: slope, both forms of a line, the distance formula and the distance from a point to a line.

Put the picture on a grid and geometry becomes algebra. Every line is an equation, and every length is a square root.

Two points pin down a line. Its slope is rise over run, and it is also $\tan\theta$ , where $\theta$ is the angle the line makes with the $x$ -axis.

Lengths come from the Pythagorean theorem in disguise: the distance formula. The shortest way from a point to a line is straight along the perpendicular, and one formula gives that length directly.

Look: when a problem gives a rectangle or a right angle, put a corner at the origin and the sides on the axes. The numbers get small and the equations get easy.

Exploreplay first, formulas later

m = \dfrac{\text{rise}}{\text{run}} = \dfrac{3}{4}

\theta \approx 36.9^\circ,\ \tan\theta = \tfrac{3}{4}

y = \tfrac{3}{4}x + \tfrac{1}{4}

3x - 4y + 1 = 0

\text{x-intercept: } -\tfrac{1}{3},\quad \text{y-intercept: } \tfrac{1}{4}

Try this: Drag A and B. Read the rise and the run. Does rise ÷ run match the slope m?
Try this: Turn on Show θ and tilt the line to 45°. Is the slope tan 45° = 1?
Try this: Press Distance. Which right triangle is hiding under the distance formula?
Try this: Press Point to line and drag P. Does PF match |ax₀ + by₀ + c| / √(a² + b²)?

Name it

Definition 38.1.1 · Equation of a Line

The equation of a line is

ax + by + c = 0.

Remark 38.1.7

Be careful not to get the equation of the line confused with $ax + by = c$ . In $ax + by = c$ the constant sits on the other side, so its sign flips when you move it over.

Definition 38.1.2 · Slope-Intercept Form

An equation of a line in slope-intercept form is

y = mx + b,

where $m$ is the slope and $b$ is the $y$ -intercept.

Theorem 38.1.3 · Slope of a line through 2 points

The slope of a line passing through 2 points $(x_1, y_1)$ and $(x_2, y_2)$ is

\frac{y_2 - y_1}{x_2 - x_1}.

Theorem 38.1.4 · Slope of a line through angle

The slope of a line with an angle of $\theta$ above the $x$ -axis is

\tan\theta.

Theorem 38.1.5 · Distance between 2 points

The distance between 2 points $(x_1, y_1)$ and $(x_2, y_2)$ is

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

Theorem 38.1.6 · Point to line formula

The distance between a point $(x_0, y_0)$ and a line $ax + by + c = 0$ is

\frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}}.

Remark 38.1.9

This formula is a bit confusing, so here is an easy way to remember the numerator: it's just the equation of the line with the point's coordinates plugged in as the $x$ and $y$ values.

Watch itpredict, then reveal

Example 38.3

AMC 10

In rectangle $ABCD$ , $AB = 6$ and $BC = 3$ . Point $E$ between $B$ and $C$ , and point $F$ between $E$ and $C$ are such that $BE = EF = FC$ . Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$ , respectively. The ratio $BP : PQ : QD$ can be written as $r : s : t$ where the greatest common factor of $r$ , $s$ , and $t$ is $1$ . What is $r + s + t$ ?

1
Put $D$ at the origin: $D = (0, 0)$ , $C = (6, 0)$ , $B = (6, 3)$ , $A = (0, 3)$ . Side $BC$ is cut into thirds. Where are $E$ and $F$ ?

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Try it

Problem 38.5.9(additional)

AMC 10

The line $12x + 5y = 60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

Contest cornerreal contest problems

Problem 38.5.2

AMC 12

A laser is placed at the point $(3, 5)$ . The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$ -axis, then hit and bounce off the $x$ -axis, then hit the point $(7, 5)$ . What is the total distance the beam will travel along this path?

Problem 38.5.3

AMC 10/12

The point $P(a, b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1, 5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6, 3)$ . What is $b - a$ ?

Problem 38.5.4

AMC 10

The line segment from $A(1, 2)$ to $B(3, 3)$ can be transformed to the line segment from $A'(3, 1)$ to $B'(4, 3)$ , sending $A$ to $A'$ and $B$ to $B'$ , by a rotation centered at the point $P(s, t)$ . What is $|s - t|$ ?

Problem 38.5.11(additional)

AMC 10

The lines with equations $ax - 2y = c$ and $2x + by = -c$ are perpendicular and intersect at $(1, -5)$ . What is $c$ ?

Problem 38.5.15(additional)

AMC 12

Two lines with slopes $\frac{1}{2}$ and $2$ intersect at $(2, 2)$ . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?

Checkpoint for section 38.1

Get 2 of 3 right to master this section. One try each, no hints.

38.20%

Circle in Coordinate Plane

In this section you will be able to write the equation of a circle from its center and radius, and read the center and radius back off an equation.

A circle is every point at distance r from a center. Square the distance formula and you have its equation.

The point $(x, y)$ is on the circle with center $(a, b)$ and radius $r$ exactly when $(x - a)^2 + (y - b)^2 = r^2$ .

Going backwards takes one move: complete the square in $x$ and in $y$ . Then the center and the radius can be read straight off.

Tangent circles are easiest with coordinates too. Put the centers where the distances are simple, and subtract one circle’s equation from another to kill the squared terms.

Exploreplay first, formulas later

Try this: Drag the center O. Which numbers in the equation change?
Try this: Press + on the radius. Which part of the equation changes, and how?
Try this: Look at the expanded form x² + y² + Dx + Ey + F = 0. Can you read the center back off D and E?

Name it

Theorem 38.2.1 · Equation of a Circle

A circle with center $(a, b)$ and radius $r$ has equation

(x - a)^2 + (y - b)^2 = r^2.

Watch itpredict, then reveal

Example 38.2

AMC 10

A triangle with sides of $5$ , $12$ , and $13$ has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?

1
$5^2 + 12^2 = 13^2$ , so this is a right triangle. Put the right angle at the origin: $C = (0, 0)$ , $B = (12, 0)$ , $A = (0, 5)$ . Where is the circumcenter $O$ of a right triangle?

Predict first:
Click a choice or type its letter (A–E).

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Example 38.5

PUMaC

Let $ABCD$ be a square with side length $8$ . Let $M$ be the midpoint of $BC$ and let $\omega$ be the circle passing through $M$ , $A$ , and $D$ . Let $O$ be the center of $\omega$ , $X$ be the intersection point (besides $A$ ) of $\omega$ with $AB$ , and $Y$ be the intersection point of $OX$ and $AM$ . Find the length $OY$ .

1
Set $A = (0, 0)$ , $B = (8, 0)$ , $C = (8, 8)$ , $D = (0, 8)$ . Then $M = (8, 4)$ . The center $O$ is equally far from $A$ and $D$ . Which line must it lie on?

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Contest cornerreal contest problems

Problem 38.5.6

AMC 10

The area of the region bounded by the graph of

x^2 + y^2 = 3|x - y| + 3|x + y|

is $m + n\pi$ , where $m$ and $n$ are integers. What is $m + n$ ?

Problem 38.5.7

AMC 10

A circle of radius $r$ is surrounded by three circles, whose radii are $1$ , $2$ , and $3$ , all externally tangent to the inner circle and externally tangent to each other, as shown in the diagram below. What is $r$ ?

Problem 38.5.13(additional)

AMC 10

Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$ . Point $C$ , not equal to $A$ or $B$ , lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

Problem 38.5.14(additional)

AMC 12

This problem also appears in the book as 34.5.33.

Right triangle $ABC$ has side lengths $BC = 6$ , $AC = 8$ , and $AB = 10$ . A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$ . A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$ . What is $OP$ ?

Checkpoint for section 38.2

Get 2 of 3 right to master this section. One try each, no hints.

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Shoelace Theorem to Find Area

In this section you will be able to find the area of any polygon from its vertices with the shoelace formula.

List the corners, criss-cross multiply, halve the difference. That’s any polygon’s area in one table.

The shoelace formula needs only the coordinates of the vertices, in order around the polygon. Clockwise or counterclockwise doesn’t matter, since you take the absolute value at the end.

Look: you choose where the axes go. Put a right angle at the origin and many products in the table become $0$ .

Exploreplay first, formulas later

#	x	y	↘ x·y′	↙ x′·y
A	1	1	1	7
B	7	1	42	7
C	7	6	21	24
D	4	3	24	3
E	1	6	1	6
A	1	1	Σ 89	Σ 47

Area = ½ × |89 − 47| = 21

signed: 21 (counter-clockwise)

Try this: Press Reverse order. What happens to the signed area? And to the area?
Try this: Drag a vertex inward to make the polygon concave. Does the formula still give the area?
Try this: Drag a vertex across a side so two edges cross. Why does the number stop being the area?

Name it

Theorem 38.3.1 · Shoelace Theorem using coordinates

Suppose the polygon $P$ has vertices $(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$ , listed in clockwise order. Then the area $A$ of $P$ is

A = \frac{1}{2}\left|(a_1 b_2 + a_2 b_3 + \cdots + a_n b_1) - (b_1 a_2 + b_2 a_3 + \cdots + b_n a_1)\right|.

You can also go in counterclockwise order, as long as you take the absolute value of the answer.

The Shoelace Theorem gets its name from the picture you get when you list the coordinates in a column, repeat the first pair at the bottom, and mark the pairs of coordinates to be multiplied: the criss-cross lines look like the laces of a shoe. (This is Theorem 30.1.12 again.)

Remark 38.3.2 · Intuitive Way of Thinking about Shoelace Theorem

Steps to the Shoelace Theorem:

Line up all of your polygon's coordinates in a vertical column.
Repeat your first coordinate at the bottom of the column.
Let $A$ be the sum of the products of all the rightward (down-right) diagonal pairs.
Let $B$ be the sum of the products of all the leftward (down-left) diagonal pairs.
Find

\frac{1}{2}|A - B|.

Watch itpredict, then reveal

Example 38.1

AMC 10

Let points $A = (0, 0)$ , $B = (1, 2)$ , $C = (3, 3)$ , and $D = (4, 0)$ . Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$ . This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$ , where these fractions are in lowest terms. What is $p + q + r + s$ ?

#	x	y	↘ x·y′	↙ x′·y
A	0	0	0	0
B	1	2	3	6
C	3	3	0	12
D	4	0	0	0
A	0	0	Σ 3	Σ 18

Area = ½ × |3 − 18| = ?

signed: −7.5 (clockwise)

1
Start with the whole area. Use the shoelace formula on $A, B, C, D$ . What is $[ABCD]$ ?

Predict first:
Your prediction

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Example 38.4

AMC 10

In the right triangle $\triangle ACE$ , we have $AC = 12$ , $CE = 16$ , and $EA = 20$ . Points $B$ , $D$ , and $F$ are located on $\overline{AC}$ , $\overline{CE}$ , and $\overline{EA}$ , respectively, so that $AB = 3$ , $CD = 4$ , and $EF = 5$ . What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$ ?

#	x	y	↘ x·y′	↙ x′·y
D	4	0	36	0
B	0	9	0	108
F	12	3	0	12
D	4	0	Σ 36	Σ 120

Area = ½ × |36 − 120| = ?

signed: −42 (clockwise)

1
$EA = 20$ is the longest side, so the right angle is at $C$ . Put $C$ at the origin, $A$ on the $y$ -axis and $E$ on the $x$ -axis. What are their coordinates?

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Contest cornerreal contest problems

Problem 38.5.10(additional)

AMC 12

Triangle $ABC$ has vertices $A = (3, 0)$ , $B = (0, 3)$ , and $C$ , where $C$ is on the line $x + y = 7$ . What is the area of $\triangle ABC$ ?

Problem 38.5.16(additional)

AMC 12

A triangle has vertices $(0, 0)$ , $(1, 1)$ , and $(6m, 0)$ , and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$ ?

Problem 38.5.21(additional)

AIME

Consider the parallelogram with vertices $(10, 45)$ , $(10, 114)$ , $(28, 153)$ , and $(28, 84)$ . A line through the origin cuts this figure into two congruent polygons. The slope of the line is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

#	x	y	↘ x·y′	↙ x′·y
P	10	45	1140	450
Q	10	114	1530	3192
R	28	153	2352	4284
S	28	84	1260	840
P	10	45	Σ 6282	Σ 8766

Area = ½ × |6282 − 8766| = 1242

signed: −1242 (clockwise)

Problem 38.5.23(additional)

AIMEChallenge

Triangle $ABC$ lies in the cartesian plane and has an area of $70$ . The coordinates of $B$ and $C$ are $(12, 19)$ and $(23, 20)$ , respectively, and the coordinates of $A$ are $(p, q)$ . The line containing the median to side $\overline{BC}$ has slope $-5$ . Find the largest possible value of $p + q$ .

Checkpoint for section 38.3

Get 2 of 3 right to master this section. One try each, no hints.

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Pick's Theorem (Optional)

In this section you will be able to find the area of a lattice polygon by counting dots, and count dots from the area.

On a lattice, area is just counting dots: the dots inside, plus half the dots on the boundary, minus one.

Pick’s theorem works for any polygon whose corners are lattice points, convex or not. It also runs backwards. Know the area (from the shoelace formula) and the boundary count, and you know how many lattice points are inside.

Counting boundary points is quick: a side from $(x_1, y_1)$ to $(x_2, y_2)$ has $\gcd(|x_2 - x_1|, |y_2 - y_1|)$ lattice steps.

Exploreplay first, formulas later

Try this: Drag a vertex so one more lattice point is inside. Does the area go up by 1?
Try this: Now add a point on the boundary instead. Why does the area only go up by ½?
Try this: Press Shoelace. Do Pick's count and the shoelace formula agree?

Name it

Theorem 38.4.1 · Pick's Theorem

If a polygon has vertices with integer coordinates (lattice points), then the area of the polygon is

i + \frac{b}{2} - 1,

where $i$ is the number of lattice points inside the polygon and $b$ is the number of lattice points on the boundary of the polygon. (This is Theorem 30.1.14 again.)

Contest cornerreal contest problems

Problem 38.5.19(additional)

AIME

Find the area of the region enclosed by the graph of $|x - 60| + |y| = \left|\frac{x}{4}\right|$ .

Problem 38.5.20(additional)

AIME

Find the number of ordered pairs $(x, y)$ of positive integers that satisfy $x \le 2y \le 60$ and $y \le 2x \le 60$ .

Checkpoint for section 38.4

Get 2 of 3 right to master this section. One try each, no hints.

38.5

Practice Problems

In this section you will be able to mix every coordinate tool of the chapter on contest problems.

The rest of the practice set, from a quick 3-D warm-up to two AIME problems. Ask first: where should the axes go?

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 38.5.1

AMC 10

The interior of a quadrilateral is bounded by the graphs of $(x + ay)^2 = 4a^2$ and $(ax - y)^2 = a^2$ , where $a$ is a positive real number. What is the area of this region in terms of $a$ , valid for all $a > 0$ ?

Problem 38.5.5

AMC 10

What is the length of the boundary of the region in the $xy$ -plane consisting of points of the form $(2u - 3w, v + 4w)$ with $0 \le u \le 1$ , $0 \le v \le 1$ , and $0 \le w \le 1$ ?

Problem 38.5.8

AIMEChallenge

Let $ABCD$ be a rhombus whose vertices all lie on the hyperbola $\frac{x^2}{20} - \frac{y^2}{24} = 1$ and are in that order. If its diagonals intersect at the origin, find the largest number less than $BD^2$ for all rhombuses $ABCD$ .

Problem 38.5.12(additional)

AMC 12

The point $P = (1, 2, 3)$ is reflected in the $xy$ -plane, then its image $Q$ is rotated by $180^\circ$ about the $x$ -axis to produce $R$ , and finally, $R$ is translated by $5$ units in the positive- $y$ direction to produce $S$ . What are the coordinates of $S$ ?

Problem 38.5.17(additional)

AMC 12

Let $A$ , $B$ and $C$ be three distinct points on the graph of $y = x^2$ such that line $AB$ is parallel to the $x$ -axis and $\triangle ABC$ is a right triangle with area $2008$ . What is the sum of the digits of the $y$ -coordinate of $C$ ?

Problem 38.5.18(additional)

AMC 10

In rectangle $ABCD$ , we have $A = (6, -22)$ , $B = (2006, 178)$ , $D = (8, y)$ , for some integer $y$ . What is the area of rectangle $ABCD$ ?

Problem 38.5.22(additional)

AIME

Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A = (24, -1)$ , and let $M$ be the line perpendicular to line $L$ that contains the point $B = (5, 6)$ . The original coordinate axes are erased, and line $L$ is made the $x$ -axis and line $M$ the $y$ -axis. In the new coordinate system, point $A$ is on the positive $x$ -axis, and point $B$ is on the positive $y$ -axis. The point $P$ with coordinates $(-14, 27)$ in the original system has coordinates $(\alpha, \beta)$ in the new coordinate system. Find $\alpha + \beta$ .

Problem 38.5.24(additional)

AMC 10

This problem also appears in the book as 32.2.30.

In rectangle $ABCD$ , we have $AB = 8$ , $BC = 9$ , $H$ is on $\overline{BC}$ with $BH = 6$ , $E$ is on $AD$ with $DE = 4$ , line $EC$ intersects line $AH$ at $G$ , and $F$ is on line $AD$ with $\overline{GF} \perp \overline{AF}$ . Find the length $GF$ .