Graphing Lines: Chapter Review

Vocabulary and notation

Graph of an equation

The set of all points $(x, y)$ making the equation true. For a linear equation it is a straight line, so "the graph" and "the solution set" name the same thing.

$x$-intercept

The point where a line crosses the $x$-axis. Every point on that axis has $y = 0$, so it looks like $(a, 0)$.

$y$-intercept

The point where a line crosses the $y$-axis. Every point on that axis has $x = 0$, so it looks like $(0, b)$; the number $b$ is its height.

Rise and run

Between two points of a line, the rise is the vertical change and the run is the horizontal change. Both are signed: downward is a negative rise, leftward a negative run.

Slope $m$

A line's steepness and direction in one number, rise divided by run. Every pair of points on the line reports the same value.

Slope triangle

The right triangle between two points of a line, its horizontal leg the run and its vertical leg the rise.

Negative reciprocal

The result of flipping a fraction over and changing its sign: the negative reciprocal of $\tfrac{2}{5}$ is $-\tfrac{5}{2}$. Only a nonzero number has one.

Parallel lines

Two distinct lines in the plane that never meet.

Perpendicular lines

Two lines crossing at a right angle.

Formulas and theorems

• Slope, as rise over run

  $$m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}$$

  Use when $x_2 \neq x_1$. Subtract in the SAME order on top and bottom. Which point you call first does not matter.

• What the slope tells you

  $m > 0$: the line rises left to right. $m < 0$: it falls. $m = 0$: horizontal. Run $0$, so no $m$ at all: vertical.

  Use when Every line except a vertical one has a real-number slope.

• Stepping to another point from the slope

  From a point on the line, moving across by the run and up or down by the rise lands on another point of the line.

  Use when Read $m$ as a fraction first; a whole number $m$ is $\tfrac{m}{1}$. Stepping backwards, left by the run and opposite the rise, also stays on the line.

  e.g. Slope $-\tfrac{2}{3}$ from $(3, 4)$: right $3$, down $2$, reaching $(6, 2)$.

• Point-slope form

  $$y - y_1 = m(x - x_1)$$

  Use when Needs one point on the line and a defined slope, so it cannot write a vertical line. Both halves are subtractions, so a negative coordinate turns into an addition.

  e.g. Through $(-4, 5)$ with $m = -\tfrac{3}{4}$: $y - 5 = -\tfrac{3}{4}(x + 4)$.

• Slope-intercept form

  $$y = mx + b$$

  Use when Valid only with $y$ alone and coefficient $1$. Every non-vertical line has one; a vertical line has none, because it has no $m$ to write.

  e.g. In $y = 5 - 2x$ the slope is $-2$ and the $y$-intercept is $(0, 5)$, not the other way round.

• Standard form

  $$Ax + By = C$$

  Use when $A$, $B$, $C$ integers, $A$ and $B$ not both zero, $A \ge 0$ by convention, fractions cleared. It is the same form written $ax + by = c$ when a system is being solved; only the capitals here carry the integer and sign conventions. Unlike point-slope and slope-intercept form, this one can write a vertical line, as $B = 0$.

  e.g. $y - 1 = -\tfrac{2}{3}(x - 4)$ clears to $2x + 3y = 11$.

• Horizontal line

  $$y = c$$

  Use when Every point sits at height $c$ and $x$ runs free. Slope $0$. Its $y$-intercept is $(0, c)$; no $x$-intercept unless $c = 0$, when the line is the $x$-axis.

• Vertical line

  $$x = c$$

  Use when Every point has $x$-coordinate $c$ and $y$ runs free. Slope undefined, so neither point-slope nor slope-intercept form applies; write it directly. Its $x$-intercept is $(c, 0)$; no $y$-intercept unless $c = 0$, when the line is the $y$-axis.

• Finding the two intercepts

  Set $y = 0$ and solve for $x$ to get the $x$-intercept; set $x = 0$ and solve for $y$ to get the $y$-intercept.

  Use when Works from any form, no rearranging needed. A line through the origin has both intercepts at $(0, 0)$, and a horizontal or vertical line off the axes has only one of the two, so each needs a second point found another way.

  e.g. $4x - 5y = 20$: $y = 0$ gives $(5, 0)$, and $x = 0$ gives $(0, -4)$.

• Parallel test

  $$m_1 = m_2$$

  Use when Both lines non-vertical, and their $y$-intercepts must DIFFER: equal slopes with equal intercepts is one line written twice. Any two distinct vertical lines are parallel too, settled by sight since neither has a slope.

  e.g. $y = 4x - 1$ and $y = 4x + 6$: same slope $4$, different intercepts, so parallel.

• Perpendicular test

  $$m_1 \cdot m_2 = -1, \qquad m_2 = -\frac{1}{m_1}$$

  Use when Both lines must have defined slopes, so neither is vertical. Solving for the partner slope needs $m_1 \neq 0$ as well: a horizontal line's perpendicular is vertical, which this rule cannot produce.

  e.g. Slopes $\tfrac{3}{2}$ and $-\tfrac{2}{3}$ multiply to $-1$, so those lines meet at a right angle.

Problem types, step by step

Exam traps

• Trap Subtracting in opposite orders on the two halves of the slope formula, as in $\dfrac{y_2 - y_1}{x_1 - x_2}$.

  Fix That flips the sign of the answer. Swapping BOTH is harmless, since $\dfrac{-a}{-b} = \dfrac{a}{b}$; swapping one is not.

• Trap Calling a vertical line's slope $0$ or "infinite", or a horizontal line's slope undefined.

  Fix Horizontal is rise $0$ over a nonzero run, so $m = 0$, an ordinary number. Vertical has run $0$, and division by zero has no result, so there is no slope at all.

• Trap Forcing $m_1 \cdot m_2 = -1$ onto a horizontal line and a vertical line, or hunting for the negative reciprocal of $0$.

  Fix That pair really is perpendicular, but you confirm it by sight: a vertical line has no slope to multiply, and $-\tfrac{1}{0}$ does not exist. The product rule needs two ordinary slopes.

• Trap Changing only the sign, or only the fraction, to build a perpendicular slope.

  Fix A negative reciprocal does both. The perpendicular partner of $3$ is $-\tfrac{1}{3}$: $-3$ gives a product of $-9$, and $\tfrac{1}{3}$ gives $+1$.

• Trap Reading a slope off an equation that is not solved for $y$, so $6x + 2y = 1$ looks like slope $6$.

  Fix Only the coefficient of $x$ in $y = mx + b$ is the slope. Solve first: $y = -3x + \tfrac{1}{2}$, so $m = -3$.

• Trap Calling two equal-slope equations the same line, or calling one line written twice a parallel pair.

  Fix Equal slopes with different $y$-intercepts are parallel, two separate lines. Only equal slopes AND equal intercepts describe one line.

• Trap Ranking steepness by the signed slope, so a slope of $1$ looks steeper than $-4$.

  Fix Steepness is the absolute value. Since $|-4| = 4 > 1$, the slope $-4$ line is far steeper; it just falls instead of rising.

• Trap Deciding a two-point answer is wrong because the other point gave a different-looking equation.

  Fix Either point yields the same line: $y - 3 = \tfrac{1}{2}(x - 2)$ and $y - 1 = \tfrac{1}{2}(x + 2)$ both simplify to $x - 2y = -4$.

• Trap Dropping the given point straight into $y = mx + b$ as $b$ when building a parallel or perpendicular line.

  Fix The point supplies $b$ only when its $x$-coordinate is $0$. Otherwise run it through point-slope form: slope $3$ through $(2, -1)$ gives $y = 3x - 7$, never $y = 3x - 1$.