Chapter 16 — The Power of Coordinates
In this chapter
- 16.1 Labelling the Plane
- 16.2 What’s it Good For?
- 16.3 Straight and Narrow (Lines)
- 16.4 Plotting a Line
- 16.5 The Distance Formula and Circles
- 16.6 Went Down to the Crossroads (Intersections)
- 16.7 Analytic Geometry
- Key Formulas Table
- Exercises & Problems
Key Formulas
| Formula | Expression |
|---|---|
| Slope | $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ |
| Slope-intercept form | $y = mx + b$ |
| Standard form | $Ax + By = C$ |
| Point-slope form | $y - y_1 = m(x - x_1)$ |
| Distance formula | $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ |
| Midpoint formula | $M = \left(\dfrac{x_1+x_2}{2},\;\dfrac{y_1+y_2}{2}\right)$ |
| Circle equation | $(x-h)^2 + (y-k)^2 = r^2$ |
| General circle | $x^2 + y^2 + Ax + By + C = 0$ |
| Parallel lines | Same slope: $m_1 = m_2$ |
| Perpendicular lines | Slopes multiply to $-1$: $m_1 \cdot m_2 = -1$ |
16.1 Labelling the Plane
Cartesian Coordinates. Each point in the plane is labelled by an ordered pair $(x, y)$:
- $x$ = distance right (negative for left) — the $x$-coordinate (abscissa)
- $y$ = distance up (negative for down) — the $y$-coordinate (ordinate)
The horizontal axis is the $x$-axis, the vertical the $y$-axis. They cross at the origin $(0,0)$.
The plane is divided into four quadrants, numbered counterclockwise starting from the upper right:
| Quadrant | $x$ | $y$ |
|---|---|---|
| I (upper right) | $+$ | $+$ |
| II (upper left) | $-$ | $+$ |
| III (lower left) | $-$ | $-$ |
| IV (lower right) | $+$ | $-$ |
Polar Coordinates. A point is labelled $(r, \theta)$ where $r$ is the distance from the origin and $\theta$ is the angle from the positive $x$-axis. The relationship to Cartesian coordinates is:
\(x = r\cos\theta, \qquad y = r\sin\theta, \qquad r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac{y}{x}\)
Exercise 16-1. Plot the Cartesian points $(4.5, 3)$, $(1, -7/2)$, and $(-6, -4)$ and the polar points $(2, 3\pi/4)$, $(3/2, 270°)$, and $(1, 0°)$.
Exercise 16-2. In which quadrant is $x$ positive and $y$ negative?
16.2 What’s it Good For?
The beauty of coordinates is that we can convert geometric pictures into equations.
- The equation $x = 1$ represents a vertical line (all points where $x$-coordinate is 1).
- The equation $y = -5$ represents a horizontal line.
- In polar coordinates, $r = 3$ represents a circle of radius 3 centered at the origin.
Rule of thumb:
- Cartesian coordinates are best for straight lines and line segments.
- Polar coordinates are often better for circles and rotations.
- Under translation, Cartesian changes simply: $(x,y) \to (x+a, y+b)$.
- Under rotation about the origin, polar changes simply: $(r,\theta) \to (r, \theta + \alpha)$.
Exercise 16-3. What curve does the equation $\theta = 47°$ represent?
Exercise 16-4. What happens to $(x, y)$ and $(r, \theta)$ under reflection across the $x$-axis? The $y$-axis?
16.3 Straight and Narrow
Intercepts
- The $y$-intercept is the point where a line crosses the $y$-axis: set $x = 0$ and solve for $y$.
- The $x$-intercept is the point where a line crosses the $x$-axis: set $y = 0$ and solve for $x$.
Example 16-1. Find the $x$- and $y$-intercepts of $2x + 7y = 14$.
Solution. Set $y = 0$: $2x = 14 \Rightarrow x = 7$, so the $x$-intercept is $(7, 0)$. Set $x = 0$: $7y = 14 \Rightarrow y = 2$, so the $y$-intercept is $(0, 2)$.
Exercise 16-5. Find the area of the region enclosed between the line $3x + 4y = 12$ and the two coordinate axes. (Hint: you don’t need to draw a picture.)
Slope-Intercept Form
Slope-Intercept Form. A non-vertical line can be written as:
\[y = mx + b\]where $m$ is the slope (“rise over run”) and $b$ is the $y$-intercept.
The slope $m$ tells us: for every 1 unit we move right, we move $m$ units up. A positive $m$ means the line goes up-right; negative means down-right. A horizontal line has $m = 0$; a vertical line has undefined slope.
Computing Slope from Two Points
Given two points $(x_1, y_1)$ and $(x_2, y_2)$:
\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}\]Example 16-2. Find the slope of the line containing $(15, 16)$ and $(-2, -18)$.
Solution. $m = \frac{-18 - 16}{-2 - 15} = \frac{-34}{-17} = 2$.
Example 16-4. Find the equation of a line passing through $(-1, -1)$ and $(3, 7)$.
Solution. Slope: $m = \frac{7-(-1)}{3-(-1)} = \frac{8}{4} = 2$. Using slope-intercept form: $y = 2x + b$. Substituting $(-1, -1)$: $-1 = -2 + b \Rightarrow b = 1$. The line is $y = 2x + 1$.
Exercise 16-8. Find the equation of a line if its $x$-intercept is at $(-4, 0)$ and its $y$-intercept is at $(0, 3)$.
Exercise 16-9. Find the equation of a line whose intercepts are $(2, 0)$ and $(0, -6)$.
Standard Form and Two-Point Form
Standard Form. $Ax + By = C$, where $A$, $B$, $C$ are integers (if possible) and $A \ge 0$.
Given two points $(p_1, q_1)$ and $(p_2, q_2)$, the line through them is:
\((q_1 - q_2)x - (p_1 - p_2)y = p_2 q_1 - p_1 q_2\)
Parallel and Perpendicular Lines
- Parallel: Two lines are parallel $\iff$ they have the same slope ($m_1 = m_2$).
- Perpendicular: Two lines are perpendicular $\iff$ their slopes multiply to $-1$ ($m_1 \cdot m_2 = -1$).
Equivalently, if one line has slope $\frac{a}{b}$, a perpendicular line has slope $-\frac{b}{a}$.
Warning. The perpendicular-slope rule $m_1 \cdot m_2 = -1$ does not apply to horizontal/vertical line pairs (one has slope 0, the other undefined).
Example 16-6. Find the slope, $x$-intercept, and $y$-intercept of $2x + 3y = 6$.
Solution. $x$-intercept: set $y = 0 \Rightarrow x = 3$; point $(3, 0)$. $y$-intercept: set $x = 0 \Rightarrow y = 2$; point $(0, 2)$. Solving for $y$: $y = -\frac{2}{3}x + 2$, so slope $m = -\frac{2}{3}$.
In general, for $Ax + By + C = 0$, the slope is $-\frac{A}{B}$.
Exercise 16-11. Find the equation for a line with slope $m$ containing the point $(p, q)$.
Exercise 16-13. Draw the lines $y = \frac{1}{2}x + 1$ and $y = -2x + 3$ (whose slopes multiply to $-1$) to confirm they are perpendicular.
16.4 Plotting a Line
To plot a line from its equation:
- Plug in various values of $x$ to find corresponding $y$ values.
- Plot the points and connect them. Two points suffice for a line.
Shortcut. If the equation is in slope-intercept form $y = mx + b$:
- Start at the $y$-intercept $(0, b)$.
- Use “rise over run”: if $m = \frac{p}{q}$, go right $q$, up $p$, and repeat.
Example 16-7. Plot the line $3x + 4y = 5$.
Solution. Plugging in $x = -1, 0, 1, 2$:
| $x$ | $y = \frac{5-3x}{4}$ |
|---|---|
| $-1$ | $2$ |
| $0$ | $5/4$ |
| $1$ | $1/2$ |
| $2$ | $-1/4$ |
Plot these points and draw the line through them.
Exercise 16-14. Plot the lines $y = -\frac{1}{2}x + 2$ and $(x - 2) = 3(y + 4)$.
16.5 The Distance Formula and Circles
The Distance Formula
The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This is a direct application of the Pythagorean Theorem: the horizontal and vertical displacements form the legs of a right triangle, and the distance is the hypotenuse.
Example 16-8. Find all $x$ such that the point $(x, 3)$ is 5 units away from $(-1, 7)$.
Solution. $5 = \sqrt{(-1-x)^2 + 16}$. Squaring: $25 = (x+1)^2 + 16$, so $(x+1)^2 = 9$, giving $x+1 = \pm 3$. Thus $x = 2$ or $x = -4$.
Circle Equation
A circle with center $(h, k)$ and radius $r$ consists of all points $(x, y)$ satisfying:
\((x - h)^2 + (y - k)^2 = r^2\)
Example 16-9. The circle $(x+1)^2 + (y-2)^2 = 9$ has center $(-1, 2)$ and radius 3. To plot: start at the center and go 3 units in all four cardinal directions to find points $(2,2)$, $(-4,2)$, $(-1,-1)$, and $(-1,5)$.
General Form and Completing the Square
The general form $x^2 + y^2 + Ax + By + C = 0$ can be converted to standard form by completing the square.
Example 16-10. Find the center and radius of $2x^2 + 2y^2 + 8x - 12y + 3 = 0$.
Solution. Divide by 2: $x^2 + y^2 + 4x - 6y + \frac{3}{2} = 0$. Group and complete the square:
\[(x^2 + 4x + 4) + (y^2 - 6y + 9) = -\frac{3}{2} + 4 + 9 = \frac{23}{2}\] \[(x + 2)^2 + (y - 3)^2 = \frac{23}{2}\]Center: $(-2, 3)$, radius: $\sqrt{\frac{23}{2}} = \frac{\sqrt{46}}{2}$.
The Midpoint Formula
The midpoint of the segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is:
\(M = \left(\frac{x_1 + x_2}{2},\;\frac{y_1 + y_2}{2}\right)\)
Exercise 16-15. In the completing-the-square process, what if the number on the right side is 0? Negative?
16.6 Went Down to the Crossroads
Finding intersections of curves = solving their equations simultaneously.
Example 16-11. Find the intersections of $x + y = -2$ and $(x+3)^2 + (y-8)^2 = 25$.
Solution. From the line: $y = -2 - x$. Substitute into the circle equation:
\[(x+3)^2 + (-10-x)^2 = 25\]Expanding: $x^2 + 6x + 9 + x^2 + 20x + 100 = 25$, so $2x^2 + 26x + 84 = 0$, i.e., $x^2 + 13x + 42 = 0$. Factoring: $(x+6)(x+7) = 0$, giving $x = -6$ or $x = -7$.
The intersection points are $(-6, 4)$ and $(-7, 5)$.
Exercise 16-16. Find the intersection of $x + y = -3$ and $-2x + 3y = 2$ by solving simultaneously, then verify graphically.
Exercise 16-17. In how many points can two distinct lines intersect? Two circles? A line and a circle? Justify using equations.
16.7 Analytic Geometry
Analytic geometry uses coordinate equations rather than purely geometric reasoning to solve problems. It is extremely powerful but can become algebraically heavy. Always try a geometric approach first!
Example 16-12. Prove that the midpoint of the segment from $(x_1, y_1)$ to $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
Proof. The two right triangles formed by the midpoint $M$, the endpoints, and horizontal/vertical lines are congruent (equal hypotenuse since $M$ is the midpoint, equal angles from parallel lines). Each horizontal leg has length $\frac{x_2-x_1}{2}$, so the $x$-coordinate of $M$ is $x_1 + \frac{x_2-x_1}{2} = \frac{x_1+x_2}{2}$. Similarly for $y$. $\square$
Example 16-13. Prove that the centroid of a triangle exists (all three medians are concurrent).
Proof. Place the triangle with vertices $A = (a, 0)$, $B = (b, 0)$, $C = (0, c)$. The midpoints opposite $A$, $B$, $C$ are $\left(\frac{b}{2}, \frac{c}{2}\right)$, $\left(\frac{a}{2}, \frac{c}{2}\right)$, $\left(\frac{a+b}{2}, 0\right)$.
Writing the equations of the three medians and solving, all three pass through the single point:
\[G = \left(\frac{a+b}{3},\;\frac{c}{3}\right)\]This is the centroid — the average of the three vertices’ coordinates! $\square$
Caution. Even simple analytic-geometry problems can become heavily algebraic. Don’t use coordinates on every geometry problem — you may miss the elegant geometric solution.
Exercise 16-18. Prove that if $G$ is the centroid of $\triangle ABC$ and $AM$ is a median, then $AG = 2GM$.
Exercise 16-19. The centroid is the average of the three coordinates of the triangle. Show this is true regardless of how the triangle is oriented in the plane.
Problems to Solve for Chapter 16
Problem 278. Find the distance between $(2, 12)$ and $(-4, 10)$.
Problem 279. A right triangle has legs of lengths 3 on the $x$-axis and 4 on the $y$-axis along the positive coordinate axes. Find the coordinates of the midpoint of the hypotenuse.
Problem 280. How many points with integer coordinates are exactly 5 units from $(0,0)$?
Problem 281. Find the equation of a line through $(5, 7)$ that bisects the area of the circle $(x+12)^2 + (y+3)^2 = 4$.
Problem 282. Find the centroid of a triangle with vertices $(10, 66)$, $(19, 72)$, and $(17, 56)$.
Problem 283. Find the distance from $(5, 7)$ to the center of the circle $4x^2 + 8x + 4y^2 - 16y - 16 = 0$.
Problem 284. If $a, b > 0$ and the triangle in the first quadrant bounded by the coordinate axes and $ax + by = 6$ has area 6, find $ab$. (AHSME 1989)
Problem 285. Parallelogram $ABCD$ has vertices $A(0,0)$, $B(2,4)$, and $D(5,1)$. If vertex $C$ is in the first quadrant, find its coordinates. (MATHCOUNTS 1992)
Problem 286. The graphs of $x + 3y = 6$ and $kx + 2y = 12$ are perpendicular. Find $k$. (MATHCOUNTS 1989)
Problem 287. What are the coordinates of the point that is two-thirds of the way from $(2,4)$ to $(-1,1)$? (MATHCOUNTS 1990)
Problem 288. Given the line $3x + 5y = 15$ and a point on this line equidistant from the coordinate axes, in which quadrants can such a point exist? (AHSME 1960)
Problem 289. Prove analytically that the diagonals of a rectangle bisect each other.
Problem 290. Find $y$ if $(3, y)$ lies on the line joining $(0, 3/2)$ and $(9/4, 0)$. (MAΘ 1987)
Problem 291. Describe analytically all lines which bisect the area of the square with vertices $(0,0)$, $(s,0)$, $(s,s)$, and $(0,s)$.
Problem 292. Find the vertices of a square centered at $(-17, 23)$ with side length 4 and diagonals parallel to the coordinate axes.
Problem 293. A circle is drawn with center at the origin and radius 3. Find the coordinates of all intersections of this circle with an origin-centered square of side length 4 whose sides are parallel to the axes.