Parallel and perpendicular lines

1. The given line y = 3x - 5 is in slope-intercept form, so its slope is m = 3.
2. Parallel lines have the same slope and a different y-intercept.
3. Check each choice's slope:
   • y = 3x + 2: slope 3 ✓, intercept 2 \ne -5 ✓
   • y = -3x - 5: slope -3 ✗
   • y = -\tfrac{1}{3}x + 2: slope -\tfrac{1}{3} ✗ (this one is perpendicular)
   • y = 2x - 5: slope 2 ✗

Answer: y = 3x + 2

Check: Both lines rise 3 for every 1 to the right, and they cross the y-axis at different points (-5 and 2), so they never meet ✓

Tip: Look only at the number in front of x. The intercept does not make lines parallel — it just needs to be different so the lines are not the same line.

1. The given line has slope m = -2 (the number in front of x).
2. Write -2 as a fraction: -\tfrac{2}{1}.
3. Flip: -\tfrac{1}{2}. Switch the sign: +\tfrac{1}{2}.

Answer: \tfrac{1}{2}

Check: -2 \cdot \tfrac{1}{2} = -1 ✓

Tip: The negative reciprocal of a negative number is positive. A falling line's perpendicular always rises (and vice versa) — so a perpendicular slope always has the opposite sign.

1. Two different lines in a plane never intersect only if they are parallel.
2. Parallel lines have the same slope and different y-intercepts.
3. The slope of y = -4x + 1 is -4.
4. Check the choices: only y = -4x - 6 has slope -4 (and a different intercept, -6).
   • y = 4x + 1 has slope 4.
   • y = \tfrac{1}{4}x + 1 has slope \tfrac{1}{4} (it is perpendicular).
   • y = -\tfrac{1}{4}x + 3 has slope -\tfrac{1}{4}.

Answer: y = -4x - 6

Check: Try to solve -4x + 1 = -4x - 6: the x-terms cancel and leave 1 = -6, which is impossible — so there is no intersection point ✓.

Tip: "Never intersect," "never cross" and "run side by side" are GED code words for parallel: same slope, different intercept.

1. The line y = 3 contains every point whose y-coordinate is 3, such as (0, 3), (2, 3), (-5, 3). It is a horizontal line with slope 0.
2. A line perpendicular to a horizontal line must be vertical.
3. Vertical lines have equations of the form x = c.
4. Among the choices, only x = -2 is vertical.

Answer: x = -2

Check: The vertical line x = -2 crosses y = 3 at the point (-2, 3), forming a right angle like the corner of a sheet of graph paper ✓.

Tip: The "flip and switch" rule does not work for horizontal and vertical lines (slope 0 has no reciprocal). Just remember: horizontal ⊥ vertical, and y = number is horizontal, x = number is vertical.

1. The given slope is -\tfrac{1}{2}. A parallel line has the same slope: y = -\tfrac{1}{2}x + b.
2. Substitute (4, 5): 5 = -\tfrac{1}{2}(4) + b.
3. Multiply: 5 = -2 + b.
4. Add 2: b = 7.
5. Equation: y = -\tfrac{1}{2}x + 7.

Answer: y = -\tfrac{1}{2}x + 7

Check: -\tfrac{1}{2}(4) + 7 = -2 + 7 = 5 ✓ Same slope as the given line, different intercept (7 \ne 3) ✓

Tip: Keep the slope, change the intercept. If your new line has the same intercept as the old one, you have just rewritten the old line.

1. Given slope: \tfrac{3}{4}. Perpendicular slope: flip and switch → -\tfrac{4}{3}.
2. Substitute (3, -1) into y = -\tfrac{4}{3}x + b: -1 = -\tfrac{4}{3}(3) + b = -4 + b
3. Add 4: b = 3.
4. Equation: y = -\tfrac{4}{3}x + 3.

Answer: y = -\tfrac{4}{3}x + 3

Check: Slopes: \tfrac{3}{4} \cdot \left(-\tfrac{4}{3}\right) = -1 ✓ Point: -\tfrac{4}{3}(3) + 3 = -4 + 3 = -1 ✓

Tip: -\tfrac{4}{3} \times 3 = -4 (negative × positive = negative). Then undo -4 by adding 4.

1. Marked points on line l: (-4, -1) and (2, 3).
2. Slope of l: \dfrac{3 - (-1)}{2 - (-4)} = \dfrac{4}{6} = \dfrac{2}{3}.
3. Perpendicular slope: flip and switch → -\tfrac{3}{2}.
4. The new line crosses the y-axis at (0, -2), so b = -2.
5. Equation: y = -\tfrac{3}{2}x - 2.

Answer: y = -\tfrac{3}{2}x - 2

Check: \tfrac{2}{3} \cdot \left(-\tfrac{3}{2}\right) = -1 ✓ and at x = 0, y = -2 ✓

Tip: Line l rises, so any perpendicular line must fall (negative slope). That alone removes two choices.

1. Parallel lines have the same slope, so m = \tfrac{2}{5}.
2. Substitute (-5, 4) into y = \tfrac{2}{5}x + b: 4 = \tfrac{2}{5}(-5) + b
3. Multiply: \tfrac{2}{5} \times (-5) = -2, so 4 = -2 + b.
4. Add 2: b = 6.

Answer: y = \tfrac{2}{5}x + 6

Check: \tfrac{2}{5}(-5) + 6 = -2 + 6 = 4 ✓, and the slope matches the original line ✓.

Tip: For a parallel line, copy the slope but never the intercept — the new line goes through a different point, so its b must be recalculated.

1. The given slope is -3 = -\tfrac{3}{1}.
2. Perpendicular slope: flip to -\tfrac{1}{3}, switch the sign to +\tfrac{1}{3}.
3. Substitute (6, 1) into y = \tfrac{1}{3}x + b: 1 = \tfrac{1}{3}(6) + b = 2 + b.
4. Subtract 2: b = -1.

Answer: y = \tfrac{1}{3}x - 1

Check: Slopes: -3 \times \tfrac{1}{3} = -1 ✓ (perpendicular). Point: \tfrac{1}{3}(6) - 1 = 1 ✓.

Tip: A whole-number slope like -3 is the fraction -\tfrac{3}{1}. Writing the 1 makes the flip easy to see.

1. Solve 3x - 4y = 12 for y:
   • Subtract 3x: -4y = -3x + 12.
   • Divide every term by -4: y = \tfrac{3}{4}x - 3.
2. The slope is \tfrac{3}{4}.
3. A parallel line has slope \tfrac{3}{4}. Only y = \tfrac{3}{4}x + 5 has that slope.

Answer: y = \tfrac{3}{4}x + 5

Check: Shortcut for Ax + By = C: slope = -\tfrac{A}{B} = -\tfrac{3}{-4} = \tfrac{3}{4} ✓.

Tip: In standard form the slope is not the number in front of x. Solve for y (or use -\tfrac{A}{B}), and divide by a negative carefully.

1. The slopes are 2 and -2.
2. Parallel needs equal slopes: 2 \ne -2, so the lines are not parallel.
3. Perpendicular needs slopes whose product is -1 (negative reciprocals): 2 \times (-2) = -4 \ne -1, so they are not perpendicular.
4. The perpendicular slope to 2 would be -\tfrac{1}{2}, not -2.
5. So the lines are neither parallel nor perpendicular. They cross, but not at a right angle.

Answer: Sam is wrong; the slopes multiply to -4, so the lines are neither parallel nor perpendicular.

Check: 2 \times (-\tfrac{1}{2}) = -1 ✓ — that pair would be perpendicular, which confirms -2 is not.

Tip: Opposite slopes only change the sign. Perpendicular slopes change the sign and flip the fraction.

1. Slope of the bike path: m = \dfrac{9 - 3}{4 - 0} = \dfrac{6}{4} = 1.5.
2. The parallel walking path has the same slope: y = 1.5x + b.
3. Substitute the fountain (2, -1): -1 = 1.5(2) + b = 3 + b.
4. Subtract 3: b = -4.

Answer: y = 1.5x - 4

Check: 1.5(2) - 4 = 3 - 4 = -1 ✓ (passes through the fountain). The bike path is y = 1.5x + 3; same slope, different intercept, so the paths never meet ✓.

Tip: Real-world "parallel" questions are the same as textbook ones: same slope, new point, new intercept.

1. x = 4 contains all points with x-coordinate 4, such as (4, 0) and (4, 7). It is a vertical line.
2. A line perpendicular to a vertical line is horizontal, with an equation y = c.
3. A horizontal line through (-2, 5) has every y-value equal to 5: y = 5.

Answer: y = 5

Check: y = 5 contains (-2, 5) ✓ and crosses x = 4 at (4, 5) at a right angle ✓.

Tip: A horizontal line uses the point's y-coordinate (y = 5); a vertical line uses the point's x-coordinate (x = -2). Decide the direction first, then pick the coordinate.

1. Slopes of the sides:
   • AB: \dfrac{3 - 1}{5 - 1} = \dfrac{2}{4} = \dfrac{1}{2}
   • BC: \dfrac{5 - 3}{4 - 5} = \dfrac{2}{-1} = -2
   • CD: \dfrac{3 - 5}{0 - 4} = \dfrac{-2}{-4} = \dfrac{1}{2}
   • DA: \dfrac{1 - 3}{1 - 0} = \dfrac{-2}{1} = -2
2. Opposite sides: AB \parallel CD (both \tfrac{1}{2}) and BC \parallel DA (both -2). So it is a parallelogram.
3. Neighboring sides: \tfrac{1}{2} \cdot (-2) = -1, so every corner is a right angle. It is a rectangle.
4. Is it a square? Side lengths: AB = \sqrt{4^2 + 2^2} = \sqrt{20}; BC = \sqrt{1^2 + 2^2} = \sqrt{5}. Not equal, so not a square.

Answer: ABCD is a rectangle that is not a square.

Check: DA: from (0, 3) down 2 and right 1 to (1, 1) ✓ slope -2; AB: right 4, up 2 ✓ slope \tfrac{1}{2}.

Tip: Equal slopes → parallel sides; slopes multiplying to -1 → right angles. To decide "square," you still need the side lengths.

1. Slope of \overline{AB}: \dfrac{3 - 1}{2 - (-2)} = \dfrac{2}{4} = \dfrac{1}{2}.
2. Slope of \overline{BC}: \dfrac{-1 - 3}{4 - 2} = \dfrac{-4}{2} = -2.
3. Slope of \overline{AC}: \dfrac{-1 - 1}{4 - (-2)} = \dfrac{-2}{6} = -\dfrac{1}{3}.
4. Test the products: \tfrac{1}{2} \times (-2) = -1 ✓, so \overline{AB} \perp \overline{BC}. They share vertex B.
5. The other products: \tfrac{1}{2} \times (-\tfrac{1}{3}) = -\tfrac{1}{6} and -2 \times (-\tfrac{1}{3}) = \tfrac{2}{3} — no other right angles.

Answer: ABC is a right triangle with the right angle at B.

Check: Pythagorean theorem: AB^2 = 4^2 + 2^2 = 20, BC^2 = 2^2 + 4^2 = 20, AC^2 = 6^2 + 2^2 = 40, and 20 + 20 = 40 ✓.

Tip: The right angle is at the vertex shared by the two perpendicular sides. \overline{AB} and \overline{BC} both contain B.

1. Step 1 is wrong. Nia changed the sign but did not flip the fraction. The perpendicular slope to \tfrac{4}{3} is -\tfrac{3}{4}. Test: \tfrac{4}{3} \times (-\tfrac{4}{3}) = -\tfrac{16}{9} \ne -1.
2. Steps 2–4 correctly use her wrong slope.
3. Correct work: -1 = -\tfrac{3}{4}(8) + b = -6 + b, so b = 5 and y = -\tfrac{3}{4}x + 5.

Answer: Step 1

Check: \tfrac{4}{3} \times (-\tfrac{3}{4}) = -1 ✓ and -\tfrac{3}{4}(8) + 5 = -6 + 5 = -1 ✓.

Tip: Flip and switch. Testing that the slopes multiply to -1 takes seconds and catches both half-done versions.

1. Slope of line a: \dfrac{4 - 2}{3 - (-1)} = \dfrac{2}{4} = \dfrac{1}{2}.
2. Slope of line b: \dfrac{1 - 5}{2 - 0} = \dfrac{-4}{2} = -2.
3. Compare: the slopes are not equal, so the lines are not parallel.
4. Product: \tfrac{1}{2} \times (-2) = -1, so the lines are perpendicular.

Answer: The lines are perpendicular.

Check: -2 is the negative reciprocal of \tfrac{1}{2}: flip \tfrac{1}{2} \to 2, switch sign \to -2 ✓.

Tip: When lines are given by points, you need two slope calculations before you can compare. Keep the subtraction order the same in each one.

1. Marked points: (-3, 3) and (3, -1). Slope: \dfrac{-1 - 3}{3 - (-3)} = \dfrac{-4}{6} = -\dfrac{2}{3}.
2. The parallel line has slope -\tfrac{2}{3} and passes through P(3, 4): 4 = -\tfrac{2}{3}(3) + b = -2 + b
3. Add 2: b = 6.

Answer: y = -\tfrac{2}{3}x + 6

Check: -\tfrac{2}{3}(3) + 6 = -2 + 6 = 4 ✓. The shown line is y = -\tfrac{2}{3}x + 1; same slope, different intercept ✓.

Tip: On graph items, count the slope carefully: from (-3, 3) to (3, -1) is down 4, right 6. Then reduce: -\tfrac{4}{6} = -\tfrac{2}{3}.

1. Solve the first equation for y: 2y = -ax + 8, so y = -\tfrac{a}{2}x + 4. Its slope is -\tfrac{a}{2}.
2. The second line has slope \tfrac{2}{3}. A perpendicular slope must be -\tfrac{3}{2}.
3. Set equal: -\tfrac{a}{2} = -\tfrac{3}{2}.
4. Multiply both sides by -2: a = 3.

Answer: a = 3

Check: With a = 3: 3x + 2y = 8 → y = -\tfrac{3}{2}x + 4. Then -\tfrac{3}{2} \cdot \tfrac{2}{3} = -1 ✓

Tip: When a letter sits in an equation, treat it like a number: solve for y, read the slope (an expression with a), set it equal to the slope you need, and solve.

1. Slope through (1, 2) and (5, -4): \dfrac{-4 - 2}{5 - 1} = \dfrac{-6}{4} = -\dfrac{3}{2}.
2. Solve 3x + ay = 6 for y: ay = -3x + 6 \Rightarrow y = -\dfrac{3}{a}x + \dfrac{6}{a}. Its slope is -\dfrac{3}{a}.
3. Parallel lines have equal slopes: -\dfrac{3}{a} = -\dfrac{3}{2}
4. The numerators match, so a = 2.

Answer: a = 2

Check: 3x + 2y = 6 \Rightarrow y = -\tfrac{3}{2}x + 3. Slope -\tfrac{3}{2} ✓. Also, (1, 2) gives 3 + 4 = 7 \ne 6, so it is a different (parallel) line ✓.

Tip: For Ax + By = C the slope is -\tfrac{A}{B}. Write the slope with the unknown in it, then set up the parallel (equal) or perpendicular (product -1) condition.

1. Line j: both points have x = -2, so j is the vertical line x = -2. (The slope formula gives \tfrac{-7}{0}, which is undefined.)
2. Line k: both points have y = 3, so k is the horizontal line y = 3 (slope \tfrac{0}{9} = 0).
3. Line n: both points have x = 1, so n is the vertical line x = 1.
4. A vertical and a horizontal line are perpendicular: j \perp k.
5. Two different vertical lines are parallel: j \parallel n.

Answer: j is perpendicular to k, and j is parallel to n.

Check: k is horizontal and n is vertical, so k \perp n as well; j and n never meet because x can't be both -2 and 1 ✓.

Tip: Same x-values → vertical line (x = c). Same y-values → horizontal line (y = c). The product rule -1 doesn't apply, but horizontal ⊥ vertical always.

1. Step 1 is correct: subtract 4x from both sides.
2. Step 2 is correct: divide every term by -6: \tfrac{-4}{-6} = \tfrac{2}{3} and \tfrac{3}{-6} = -\tfrac{1}{2}.
3. Step 3 is wrong: Omar divided only the 8 by 2. Every term must be divided: y = -\tfrac{3}{2}x + 4
4. Correct conclusion: \tfrac{2}{3} \times (-\tfrac{3}{2}) = -1, so the lines are perpendicular.

Answer: Step 3 is the first error; the lines are actually perpendicular.

Check: Shortcut -\tfrac{A}{B}: first line -\tfrac{4}{-6} = \tfrac{2}{3}; second line -\tfrac{3}{2}. Product -1 ✓.

Tip: Dividing "both sides by 2" means dividing every term on both sides. Missing one term changes the slope and the conclusion.