Chapter Review · a rapid pre-test review (speedrun)

Graphing Lines: Chapter Review

A rapid review before the test: the chapter's vocabulary and notation, every formula with the conditions to use it, the standard problem types step by step, and the traps that cost points.

Vocabulary and notation

Graph of an equation
The set of all points (x,y)(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.
xx-intercept
The point where a line crosses the xx-axis. Every point on that axis has y=0y = 0, so it looks like (a,0)(a, 0).
yy-intercept
The point where a line crosses the yy-axis. Every point on that axis has x=0x = 0, so it looks like (0,b)(0, b); the number bb 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 mm
A line's steepness and direction in one number, rise divided by run. Every pair of points on a nonvertical 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 25\tfrac{2}{5} is −52-\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=riserun=y2−y1x2−x1m = \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1}

    Use when x2≠x1x_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>0m > 0: the line rises left to right. m<0m < 0: it falls. m=0m = 0: horizontal. Run 00, so no mm 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 mm as a fraction first; a whole number mm is m1\tfrac{m}{1}. Stepping backwards, left by the run and opposite the rise, also stays on the line.

    e.g. Slope −23-\tfrac{2}{3} from (3,4)(3, 4): right 33, down 22, reaching (6,2)(6, 2).

  • Point-slope form

    y−y1=m(x−x1)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)(-4, 5) with m=−34m = -\tfrac{3}{4}: y−5=−34(x+4)y - 5 = -\tfrac{3}{4}(x + 4).

  • Slope-intercept form

    y=mx+by = mx + b

    Use when Valid only with yy alone and coefficient 11. Every nonvertical line has one; a vertical line has none, because it has no mm to write.

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

  • Standard form

    Ax+By=CAx + By = C

    Use when AA, BB, CC integers, AA and BB not both zero, A≥0A \ge 0 by convention, fractions cleared, and the three numbers sharing no common factor other than 11. Unlike point-slope and slope-intercept form, this one can write a vertical line, as B=0B = 0.

    e.g. y−1=−23(x−4)y - 1 = -\tfrac{2}{3}(x - 4) clears to 2x+3y=112x + 3y = 11.

  • Horizontal line

    y=cy = c

    Use when Every point sits at height cc and xx runs free. Slope 00. Its yy-intercept is (0,c)(0, c); no xx-intercept unless c=0c = 0, when the line is the xx-axis.

  • Vertical line

    x=cx = c

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

  • Finding the two intercepts

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

    Use when Works from any form, no rearranging needed. A line through the origin has both intercepts at (0,0)(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=204x - 5y = 20: y=0y = 0 gives (5,0)(5, 0), and x=0x = 0 gives (0,−4)(0, -4).

  • Parallel test

    m1=m2m_1 = m_2

    Use when Both lines nonvertical, and their yy-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−1y = 4x - 1 and y=4x+6y = 4x + 6: same slope 44, different intercepts, so parallel.

  • Perpendicular test

    m1⋅m2=−1,m2=−1m1m_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 m1≠0m_1 \neq 0 as well: a horizontal line's perpendicular is vertical, which this rule cannot produce.

    e.g. Slopes 32\tfrac{3}{2} and −23-\tfrac{2}{3} multiply to −1-1, so those lines meet at a right angle.

Problem types, step by step

Graph a linear equation from a table of values

  1. If the equation has no yy-term, it reduces to x=cx = c, a vertical line: plot it directly and stop, since a vertical line cannot be solved for yy. Otherwise, solve the equation for yy if it is not already.
  2. Choose three or four xx-values, including 00 and a negative one, and compute each yy.
  3. Plot the pairs; a point missing the line the others make is an arithmetic slip, so recheck it.
  4. Draw the line through them, extended past the outer points with arrowheads.

e.g. x+y=4x + y = 4 becomes y=4−xy = 4 - x, giving (−1,5)(-1, 5), (0,4)(0, 4), (1,3)(1, 3), (2,2)(2, 2).

Graph a line from its two intercepts

  1. Set y=0y = 0 and solve for xx; set x=0x = 0 and solve for yy.
  2. Plot both points and draw the line through them.
  3. If the two coincide at the origin, or one is an awkward fraction, pick any convenient xx instead and compute its yy for the second point.

Find the slope of a line

  1. Check for a vertical line first: two points with the same xx-coordinate, or a graph running straight up and down, has no slope at all. Stop there.
  2. From two points, label them (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) and compute m=y2−y1x2−x1m = \dfrac{y_2 - y_1}{x_2 - x_1}, same order top and bottom.
  3. From a graph, pick two points on grid corners, count the run rightward, then the rise (downward is negative), and write rise over run.
  4. Simplify and check the sign against the picture.

e.g. Through (−1,5)(-1, 5) and (2,−1)(2, -1): m=−1−52−(−1)=−63=−2m = \dfrac{-1 - 5}{2 - (-1)} = \dfrac{-6}{3} = -2.

Read the slope and intercept, then graph, from y=mx+by = mx + b

  1. Get yy alone with coefficient 11, reordering the terms into mx+bmx + b if they arrive shuffled.
  2. Read mm as the coefficient of xx and bb as the constant, each with its own sign; the yy-intercept is (0,b)(0, b).
  3. Plot (0,b)(0, b) on the yy-axis.
  4. Read mm as rise over run, step to a second point, and draw the line through both.

e.g. y=12x+1y = \tfrac{1}{2}x + 1: plot (0,1)(0, 1), run 22 and rise 11 to (2,2)(2, 2).

Rewrite an equation in a requested form

  1. To slope-intercept form: isolate the yy-term, then divide EVERY term by the coefficient of yy.
  2. To standard form: distribute the slope, multiply through to clear fractions, then collect xx and yy left and the constant right.
  3. For standard form, multiply the whole equation by −1-1 if AA came out negative.
  4. Check by substituting a point known to be on the line.

e.g. 2x+3y=122x + 3y = 12 gives 3y=−2x+123y = -2x + 12, then y=−23x+4y = -\tfrac{2}{3}x + 4.

Write the equation of a line

  1. If given two points that share an xx-coordinate, skip everything else and write x=cx = c; if they share a yy-coordinate, write y=cy = c.
  2. Otherwise, given two points, compute the slope first; given a point and a slope, you already have both.
  3. Substitute into y−y1=m(x−x1)y - y_1 = m(x - x_1), letting a negative coordinate turn the subtraction into an addition.
  4. Simplify to whatever form the question names, or leave point-slope form if it names none.
  5. Check by substituting the original point back in; for a two-point problem check both, since a wrong slope can still satisfy one.

e.g. Through (−2,1)(-2, 1) and (2,3)(2, 3): m=12m = \tfrac{1}{2}, so y−3=12(x−2)y - 3 = \tfrac{1}{2}(x - 2).

Classify a pair of lines as parallel, perpendicular, or neither

  1. If either line is vertical, decide by sight before computing any slope: a distinct vertical is parallel to a vertical, perpendicular to horizontal, and neither to any other slanted line.
  2. Otherwise, put each line into slope-intercept form, or compute each slope from two given points.
  3. Equal slopes, different yy-intercepts: parallel. Equal slopes, same intercept: one line.
  4. Slopes whose product is −1-1: perpendicular.
  5. Neither pattern: the lines cross at an angle that is not a right angle.

e.g. 3x+4y=83x + 4y = 8 and 4x−3y=124x - 3y = 12 have slopes −34-\tfrac{3}{4} and 43\tfrac{4}{3}, product −1-1, so perpendicular.

Write a line through a point, parallel or perpendicular to a given line

  1. If the given line is vertical or horizontal, skip the slope arithmetic entirely: a vertical line's parallel partner is vertical and its perpendicular partner is horizontal, and a horizontal line's parallel partner is horizontal and its perpendicular partner is vertical. Write that equation directly through the point.
  2. Otherwise, find the given line's slope, solving for yy first if it arrived in standard form.
  3. Take the same slope for parallel, or the negative reciprocal (flip AND change the sign) for perpendicular.
  4. Substitute that slope and the given point into point-slope form.
  5. Simplify to the requested form and confirm the point satisfies the result.

e.g. Through (4,1)(4, 1) perpendicular to y=2x−5y = 2x - 5: m=−12m = -\tfrac{1}{2}, so y=−12x+3y = -\tfrac{1}{2}x + 3.

Decide whether a point is on a line, or find a missing coordinate

  1. Substitute the coordinates you have into the equation.
  2. For an on-or-off question, compare the sides: equal means on the line, unequal means off it.
  3. For a missing coordinate, solve the resulting one-variable equation for the letter left over.

e.g. On 3x+2y=123x + 2y = 12: (2,4)(2, 4) gives 14≠1214 \neq 12, so it is off the line; at x=1x = 1, 2y=92y = 9, so the point is (1,92)\left(1, \tfrac{9}{2}\right).

Exam traps

  • Trap Subtracting in opposite orders on the two halves of the slope formula, as in y2−y1x1−x2\dfrac{y_2 - y_1}{x_1 - x_2}.

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

  • Trap Calling a vertical line's slope 00 or "infinite", or a horizontal line's slope undefined.

    Fix Horizontal is rise 00 over a nonzero run, so m=0m = 0, an ordinary number. Vertical has run 00, and division by zero has no result, so there is no slope at all.

  • Trap Forcing m1⋅m2=−1m_1 \cdot m_2 = -1 onto a horizontal line and a vertical line, or hunting for the negative reciprocal of 00.

    Fix That pair really is perpendicular, but you confirm it by sight: a vertical line has no slope to multiply, and −10-\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 33 is −13-\tfrac{1}{3}: −3-3 gives a product of −9-9, and 13\tfrac{1}{3} gives +1+1.

  • Trap Reading a slope off an equation that is not solved for yy, so 6x+2y=16x + 2y = 1 looks like slope 66.

    Fix Only the coefficient of xx in y=mx+by = mx + b is the slope. Solve first: y=−3x+12y = -3x + \tfrac{1}{2}, so m=−3m = -3.

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

    Fix Equal slopes with different yy-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 11 looks steeper than −4-4.

    Fix Steepness is the absolute value. Since ∣−4∣=4>1|-4| = 4 > 1, the slope −4-4 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=12(x−2)y - 3 = \tfrac{1}{2}(x - 2) and y−1=12(x+2)y - 1 = \tfrac{1}{2}(x + 2) both simplify to x−2y=−4x - 2y = -4.

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

    Fix The point supplies bb only when its xx-coordinate is 00. Otherwise run it through point-slope form: slope 33 through (2,−1)(2, -1) gives y=3x−7y = 3x - 7, never y=3x−1y = 3x - 1.

Chapter Test Questions from across the chapter