Module 0g — Coordinate Geometry: Lines, Circles, and Parabolas
Course: MAT1CJ101 — Differential Calculus (prerequisite refresher)
Status: Prerequisite background material — not part of the official 60 taught hours, not examined.
0. Why This Module Exists
Preliminaries §2 of Thomas & Finney sets up the coordinate plane before functions and their graphs are introduced in §3–4. Distance, slope, and the equations of lines and circles aren’t just algebra drills — they resurface constantly once Module I introduces tangent lines (a tangent is a line, described by exactly this machinery) and whenever a graph needs to be read or sketched precisely. This module is a fast, self-contained refresher on that machinery.
1. The Cartesian Plane
A point in the plane is an ordered pair $(x, y)$: $x$ is the horizontal (signed) distance from the origin, $y$ the vertical one. The two axes divide the plane into four quadrants, numbered counter-clockwise from the upper right:
| Quadrant | Sign of $x$ | Sign of $y$ |
|---|---|---|
| I | $+$ | $+$ |
| II | $-$ | $+$ |
| III | $-$ | $-$ |
| IV | $+$ | $-$ |
Worked Example 1: In which quadrant does $(-3, 5)$ lie? What about $(4, -2)$?
$(-3, 5)$: $x < 0$, $y > 0$ — Quadrant II.
$(4, -2)$: $x > 0$, $y < 0$ — Quadrant IV.
2. Distance and Midpoint
| For two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, the segment joining them is the hypotenuse of a right triangle with legs $ | x_2 - x_1 | $ and $ | y_2 - y_1 | $, so Pythagoras gives: |
The midpoint is just the average of the coordinates:
\[M = \left( \frac{x_1 + x_2}{2}, \ \frac{y_1 + y_2}{2} \right)\]Worked Example 2: Find the distance and midpoint between $P_1 = (-1, 4)$ and $P_2 = (5, -2)$.
\[d = \sqrt{(5-(-1))^2 + (-2-4)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}\] \[M = \left( \frac{-1+5}{2}, \frac{4+(-2)}{2} \right) = (2, 1)\]3. Slope of a Line
The slope measures steepness: “rise over run” between any two points on the line.
\[m = \frac{y_2 - y_1}{x_2 - x_1} \qquad (x_1 \neq x_2)\]| Case | Slope |
|---|---|
| Horizontal line | $m = 0$ |
| Vertical line | undefined (division by zero) |
| Line rises left to right | $m > 0$ |
| Line falls left to right | $m < 0$ |
Parallel lines have equal slopes: $m_1 = m_2$.
Perpendicular lines have slopes that are negative reciprocals: $m_1 m_2 = -1$.
Worked Example 3: Find the slope of the line through $(2, 3)$ and $(-4, 9)$. Is it parallel or perpendicular to a line of slope $\tfrac{1}{2}$?
\[m = \frac{9-3}{-4-2} = \frac{6}{-6} = -1\]Since $-1 \neq \tfrac12$, not parallel. Since $(-1)\left(\tfrac12\right) = -\tfrac12 \neq -1$, not perpendicular either — just two unrelated lines.
4. Equations of a Line
Three equivalent forms, useful in different situations:
| Form | Equation | Best used when |
|---|---|---|
| Point-slope | $y - y_1 = m(x - x_1)$ | you know a point and the slope |
| Slope-intercept | $y = mx + b$ | you want to read off slope $m$ and $y$-intercept $b$ directly |
| Standard (general) | $Ax + By = C$ | comparing/combining lines, or when the line is vertical |
Worked Example 4: Find the equation of the line through $(1, -2)$ with slope $3$, in all three forms.
Point-slope: $y - (-2) = 3(x - 1) \implies y + 2 = 3(x-1)$
Slope-intercept: expand — $y + 2 = 3x - 3 \implies y = 3x - 5$
Standard form: $3x - y = 5$
Worked Example 5: Find the equation of the line through $(0, 4)$ that is perpendicular to $y = -2x + 1$.
The given line has slope $-2$, so the perpendicular slope is $m = \tfrac{1}{2}$ (negative reciprocal). Using point-slope through $(0,4)$:
\[y - 4 = \tfrac12(x - 0) \implies y = \tfrac12 x + 4\]5. Circles
A circle is the set of all points at a fixed distance (the radius $r$) from a fixed point (the center $(h,k)$). Applying the distance formula to a general point $(x,y)$ on the circle gives the standard equation:
\[(x - h)^2 + (y - k)^2 = r^2\]Expanding this standard form always produces a general form $x^2 + y^2 + Dx + Ey + F = 0$. To go the other way — from general form back to center/radius — complete the square in $x$ and in $y$ separately.
Worked Example 6: Find the center and radius of the circle $x^2 + y^2 - 6x + 4y - 3 = 0$.
Group and complete the square:
\((x^2 - 6x) + (y^2 + 4y) = 3\) \((x^2 - 6x + 9) + (y^2 + 4y + 4) = 3 + 9 + 4\) \((x-3)^2 + (y+2)^2 = 16\)
Center $(3, -2)$, radius $\sqrt{16} = 4$.
6. Parabolas
The graph of $y = ax^2 + bx + c$ ($a \neq 0$) is a parabola. Its axis of symmetry is the vertical line
\[x = -\frac{b}{2a}\]| and its vertex is the point where the axis meets the parabola — found by substituting $x = -\tfrac{b}{2a}$ back into the equation to get the vertex’s $y$-coordinate. The parabola opens upward if $a > 0$ and opens downward if $a < 0$; the larger $ | a | $ is, the narrower the parabola. |
Worked Example 8: Find the vertex, axis, and $x$-intercepts of $y = x^2 - 2x - 3$.
Axis: $x = -\dfrac{-2}{2(1)} = 1$. Vertex $y$-value: $1^2 - 2(1) - 3 = -4$, so the vertex is $(1, -4)$.
$x$-intercepts (set $y=0$): $x^2 - 2x - 3 = (x-3)(x+1) = 0 \implies x = 3, -1$.
Since $a = 1 > 0$, the parabola opens upward.
7. Line–Circle Intersection
Substituting a line’s equation into a circle’s equation gives a quadratic in one variable. Its discriminant tells you how the line and circle meet:
| Discriminant | Meaning |
|---|---|
| $> 0$ | line crosses the circle at two points (secant) |
| $= 0$ | line touches the circle at exactly one point (tangent) |
| $< 0$ | line misses the circle entirely |
Worked Example 7: Does the line $y = x + 1$ intersect the circle $x^2 + y^2 = 4$? If so, where?
Substitute: $x^2 + (x+1)^2 = 4 \implies x^2 + x^2 + 2x + 1 = 4 \implies 2x^2 + 2x - 3 = 0$.
Discriminant: $2^2 - 4(2)(-3) = 4 + 24 = 28 > 0$ — two intersection points. Solving with the quadratic formula: $x = \dfrac{-2 \pm \sqrt{28}}{4} = \dfrac{-1 \pm \sqrt{7}}{2}$, then $y = x+1$ for each.
8. Summary Table
| Concept | Formula |
|---|---|
| Distance | $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ |
| Midpoint | $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$ |
| Slope | $m = \frac{y_2-y_1}{x_2-x_1}$ |
| Parallel lines | $m_1 = m_2$ |
| Perpendicular lines | $m_1 m_2 = -1$ |
| Point-slope form | $y-y_1 = m(x-x_1)$ |
| Slope-intercept form | $y = mx+b$ |
| Standard form | $Ax+By=C$ |
| Circle (standard) | $(x-h)^2+(y-k)^2=r^2$ |
| Circle (general) | $x^2+y^2+Dx+Ey+F=0$ — complete the square to find $(h,k), r$ |
| Parabola axis | $x = -\dfrac{b}{2a}$ for $y=ax^2+bx+c$ |
| Parabola opening | upward if $a>0$, downward if $a<0$ |
9. Practice Problems
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Plot $(2, -5)$ and $(-3, -1)$; state which quadrant each lies in.
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Find the distance between $(-2, 3)$ and $(4, -5)$.
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Find the midpoint of the segment joining $(7, -1)$ and $(-3, 5)$.
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A circle has diameter endpoints $(1, 2)$ and $(7, 8)$. Find its center and radius. (Hint: the center is the midpoint of the diameter.)
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Find the slope of the line through $(3, -2)$ and $(3, 6)$. What kind of line is this?
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Find the slope of the line through $(-4, 1)$ and $(2, -4)$.
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Determine whether the lines through $(0,0),(2,4)$ and through $(1,1),(3,5)$ are parallel, perpendicular, or neither.
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Find the point-slope equation of the line through $(-1, 3)$ with slope $-2$, then convert it to slope-intercept form.
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Find the equation (slope-intercept form) of the line through $(2, 5)$ and $(6, -3)$.
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Find the equation of the line through $(4, -1)$ that is parallel to $y = 3x - 2$.
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Find the equation of the line through $(0, -3)$ that is perpendicular to $2x + y = 5$. (Hint: rewrite the given line in slope-intercept form first to read off its slope.)
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Write the standard equation of the circle with center $(-2, 5)$ and radius $6$.
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Find the center and radius of $x^2 + y^2 + 8x - 2y + 8 = 0$ by completing the square.
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Find the center and radius of $x^2 + y^2 - 10x = 0$. (Hint: there is no $y$ term to complete the square on — treat its coefficient as $0$.)
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(Harder) Determine whether the line $y = -x + 5$ intersects the circle $(x-2)^2 + (y-1)^2 = 4$, and if so, find the point(s) of intersection.
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(Harder) Find the value(s) of $k$ for which the line $y = 2x + k$ is tangent to the circle $x^2 + y^2 = 5$. (Hint: substitute, then require the resulting quadratic’s discriminant to equal zero.)
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Find the vertex, axis of symmetry, and $x$-intercepts of $y = x^2 - 4x + 3$.
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Find the vertex, axis of symmetry, and $y$-intercept of $y = -x^2 - 2x + 8$. Does the parabola open upward or downward?
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(Harder) A parabola has vertex $(2, -1)$ and passes through the point $(0, 3)$. Find its equation in the form $y = a(x-2)^2 - 1$. (Hint: substitute the known point to solve for $a$.)