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ISEE Upper Level Mathematics Achievement Quiz

ISEE Upper Level Mathematics Achievement Quiz: Finding Slope

Practice Finding Slope in ISEE Upper Level Mathematics Achievement with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 16

0 of 16 answered

Line segment PQ has slope 34\frac{3}{4}43​. If P has coordinates (7, 2) and Q has coordinates (a, 11), what is the value of a?

Select an answer to continue

What this quiz covers

This quiz focuses on Finding Slope, giving you a quick way to practice the rules, question types, and explanations that matter most for ISEE Upper Level Mathematics Achievement.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Line segment PQ has slope 34\frac{3}{4}43​. If P has coordinates (7, 2) and Q has coordinates (a, 11), what is the value of a?

  1. 191919 (correct answer)
  2. 151515
  3. 121212
  4. 999

Explanation: When you encounter a question about finding coordinates using slope, you're working with the fundamental slope formula: slope=y2−y1x2−x1\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}slope=x2​−x1​y2​−y1​​. Given that line segment PQ has slope 34\frac{3}{4}43​, point P is at (7, 2), and point Q is at (a, 11), you can substitute these values into the slope formula: 34=11−2a−7\frac{3}{4} = \frac{11 - 2}{a - 7}43​=a−711−2​ Simplifying the numerator: 34=9a−7\frac{3}{4} = \frac{9}{a - 7}43​=a−79​ Cross-multiply to solve for a: 3(a−7)=4(9)3(a - 7) = 4(9)3(a−7)=4(9) This gives you: 3a−21=363a - 21 = 363a−21=36 Adding 21 to both sides: 3a=573a = 573a=57 Therefore: a=19a = 19a=19 Let's check why the other answers are incorrect: Choice B (15): If a = 15, the slope would be 915−7=98\frac{9}{15-7} = \frac{9}{8}15−79​=89​, which doesn't equal 34\frac{3}{4}43​. Choice C (12): If a = 12, the slope would be 912−7=95\frac{9}{12-7} = \frac{9}{5}12−79​=59​, which is too large. Choice D (9): If a = 9, the slope would be 99−7=92\frac{9}{9-7} = \frac{9}{2}9−79​=29​, which is much too large. The correct answer is A) 19. Strategy tip: Always substitute your answer back into the original slope formula to verify. This catches calculation errors and ensures your coordinate produces the given slope. Remember that slope is "rise over run" – the change in y-coordinates over the change in x-coordinates.

Question 2

A triangle has vertices at A(2, 6), B(8, 3), and C(5, 9). What is the slope of the altitude from vertex C to side AB?

  1. 12\frac{1}{2}21​
  2. −12-\frac{1}{2}−21​
  3. 222 (correct answer)
  4. −2-2−2

Explanation: When you encounter altitude problems in coordinate geometry, remember that an altitude is perpendicular to the side it meets, which means their slopes are negative reciprocals of each other. To find the slope of the altitude from C to side AB, you first need the slope of side AB itself. Using the slope formula with points A(2, 6) and B(8, 3): slope of AB=3−68−2=−36=−12\text{slope of AB} = \frac{3-6}{8-2} = \frac{-3}{6} = -\frac{1}{2}slope of AB=8−23−6​=6−3​=−21​ Since the altitude is perpendicular to AB, its slope is the negative reciprocal of −12-\frac{1}{2}−21​. To find the negative reciprocal, flip the fraction and change the sign: the negative reciprocal of −12-\frac{1}{2}−21​ is 222. Looking at the wrong answers: Choice A gives 12\frac{1}{2}21​, which is just the reciprocal of AB's slope but not the negative reciprocal. Choice B gives −12-\frac{1}{2}−21​, which is actually the slope of side AB itself, not its altitude. Choice D gives −2-2−2, which would be the negative reciprocal if AB had slope 12\frac{1}{2}21​ instead of −12-\frac{1}{2}−21​. The correct answer is C: 222. Study tip: For perpendicular lines, always remember the negative reciprocal relationship. If one line has slope mmm, the perpendicular line has slope −1m-\frac{1}{m}−m1​. Practice this concept because it appears frequently in problems involving altitudes, perpendicular bisectors, and rectangular shapes.

Question 3

In the coordinate plane, line ℓ passes through points (2, -3) and (6, 5). Line m passes through points (-1, 4) and (3, 12). What is the relationship between the slopes of these two lines?

  1. The slopes are equal, so the lines are parallel (correct answer)
  2. The slopes are negative reciprocals, so the lines are perpendicular
  3. The slopes have the same absolute value but opposite signs
  4. One slope is twice the other slope

Explanation: When you encounter questions about line relationships, you need to find and compare the slopes of both lines using the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. For line ℓ passing through (2, -3) and (6, 5): mℓ=5−(−3)6−2=84=2m_ℓ = \frac{5 - (-3)}{6 - 2} = \frac{8}{4} = 2mℓ​=6−25−(−3)​=48​=2 For line m passing through (-1, 4) and (3, 12): mm=12−43−(−1)=84=2m_m = \frac{12 - 4}{3 - (-1)} = \frac{8}{4} = 2mm​=3−(−1)12−4​=48​=2 Since both slopes equal 2, the lines are parallel. This confirms answer choice A is correct. Let's examine why the other options are wrong. Choice B incorrectly identifies the relationship as perpendicular. For lines to be perpendicular, their slopes must be negative reciprocals (like 2 and -1/2), but both slopes here are positive 2. Choice C suggests the slopes have the same absolute value but opposite signs. While they do have the same absolute value (2), they have the same sign (both positive), not opposite signs. Choice D claims one slope is twice the other, but since both slopes equal 2, neither is twice the other. Remember this key pattern: when slopes are equal, lines are parallel; when slopes are negative reciprocals, lines are perpendicular. Always calculate both slopes carefully and compare them systematically. Watch out for calculation errors, especially with negative coordinates, as they can lead you to incorrect relationship conclusions.

Question 4

Point P(4, 9) lies on a line with slope 32\frac{3}{2}23​. If point Q also lies on this line and has coordinates (10, k), what is the value of k?

  1. 151515
  2. 181818 (correct answer)
  3. 212121
  4. 242424

Explanation: When you encounter a problem about points on a line with a given slope, you're working with the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. This formula relates any two points on the same line. Since both points P(4, 9) and Q(10, k) lie on a line with slope 32\frac{3}{2}23​, you can substitute these coordinates into the slope formula. Let P be your first point and Q be your second point: 32=k−910−4\frac{3}{2} = \frac{k - 9}{10 - 4}23​=10−4k−9​ Simplifying the denominator: 32=k−96\frac{3}{2} = \frac{k - 9}{6}23​=6k−9​ Cross-multiply to solve for k: 3×6=2(k−9)3 \times 6 = 2(k - 9)3×6=2(k−9) This gives you: 18=2k−1818 = 2k - 1818=2k−18 Adding 18 to both sides: 36=2k36 = 2k36=2k Therefore: k=18k = 18k=18 The answer is B. Looking at the wrong answers: A) 15 likely comes from forgetting to add back the 18 after cross-multiplying, stopping at k−9=9k - 9 = 9k−9=9. C) 21 might result from incorrectly calculating 9+32×69 + \frac{3}{2} \times 69+23​×6 as if you're adding the slope times the horizontal distance directly to the y-coordinate. D) 24 could come from computational errors in the cross-multiplication step. Remember this key strategy: when you have two points on a line and know the slope, the slope formula becomes your bridge to finding missing coordinates. Always set up the equation carefully, keeping track of which point is (x1,y1)(x_1, y_1)(x1​,y1​) and which is (x2,y2)(x_2, y_2)(x2​,y2​).

Question 5

Line segment AB has endpoints A(-5, 2) and B(3, 6). Point C divides AB in the ratio 3:1. What is the slope of line segment AC?

  1. 12\frac{1}{2}21​ (correct answer)
  2. 34\frac{3}{4}43​
  3. 111
  4. 222

Explanation: When you encounter a problem involving points that divide line segments in a given ratio, you need to use the section formula to find the coordinates of the dividing point, then calculate slope using those coordinates. Since point C divides AB in the ratio 3:1, you can find C's coordinates using the section formula: C=(m⋅x2+n⋅x1m+n,m⋅y2+n⋅y1m+n)C = \left(\frac{m \cdot x_2 + n \cdot x_1}{m + n}, \frac{m \cdot y_2 + n \cdot y_1}{m + n}\right)C=(m+nm⋅x2​+n⋅x1​​,m+nm⋅y2​+n⋅y1​​) where the ratio is m:n. Here, A(-5, 2), B(3, 6), and the ratio is 3:1. For the x-coordinate: 3(3)+1(−5)3+1=9−54=1\frac{3(3) + 1(-5)}{3 + 1} = \frac{9 - 5}{4} = 13+13(3)+1(−5)​=49−5​=1 For the y-coordinate: 3(6)+1(2)3+1=18+24=5\frac{3(6) + 1(2)}{3 + 1} = \frac{18 + 2}{4} = 53+13(6)+1(2)​=418+2​=5 So C is at (1, 5). Now you can find the slope of AC using A(-5, 2) and C(1, 5): Slope = y2−y1x2−x1=5−21−(−5)=36=12\frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 2}{1 - (-5)} = \frac{3}{6} = \frac{1}{2}x2​−x1​y2​−y1​​=1−(−5)5−2​=63​=21​ Choice A is correct. Choice B (34\frac{3}{4}43​) might result from miscalculating the coordinates or confusing the ratio. Choice C (1) could come from incorrectly finding C at the midpoint instead of using the 3:1 ratio. Choice D (2) is the slope of the entire segment AB, which you'd get if you ignored point C altogether and calculated from A to B directly. Remember: when a point divides a segment in a specific ratio, always use the section formula first to find the exact coordinates before calculating any geometric properties like slope or distance.

Question 6

A rhombus has vertices at R(0, 0), S(5, 3), T(8, 8), and U(3, 5). What is the slope of diagonal RT?

  1. 111 (correct answer)
  2. 35\frac{3}{5}53​
  3. 53\frac{5}{3}35​
  4. 85\frac{8}{5}58​

Explanation: When you encounter coordinate geometry problems involving diagonals of polygons, your goal is to find the slope using the standard slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. To find the slope of diagonal RT, you need the coordinates of points R and T. From the given information, R is at (0, 0) and T is at (8, 8). Applying the slope formula: m=8−08−0=88=1m = \frac{8 - 0}{8 - 0} = \frac{8}{8} = 1m=8−08−0​=88​=1 The slope of diagonal RT is 1, making A correct. Let's examine why the other answers are incorrect. Choice B (35\frac{3}{5}53​) is the slope of side RS, calculated as 3−05−0=35\frac{3-0}{5-0} = \frac{3}{5}5−03−0​=53​. This is a common trap where students confuse a diagonal with a side of the rhombus. Choice C (53\frac{5}{3}35​) is simply the reciprocal of choice B, which might catch students who flip the slope formula. Choice D (85\frac{8}{5}58​) appears to mix coordinates incorrectly, perhaps using the x-coordinate of T with the x-coordinate of S. Remember that diagonal problems require you to identify the correct endpoints first. Don't let the mention of "rhombus" distract you—the shape's properties aren't needed here, just basic coordinate geometry. Always double-check that you're using the right points for the line segment in question, as test makers often include slopes of other segments as distractors.

Question 7

The slope of line ℓ₁ is 23\frac{2}{3}32​ and the slope of line ℓ₂ is −32-\frac{3}{2}−23​. Line ℓ₃ is perpendicular to line ℓ₁. What is the slope of a line parallel to ℓ₃?

  1. 23\frac{2}{3}32​
  2. −23-\frac{2}{3}−32​
  3. 32\frac{3}{2}23​
  4. −32-\frac{3}{2}−23​ (correct answer)

Explanation: When you encounter questions about perpendicular and parallel lines, focus on the key relationships between their slopes: perpendicular lines have slopes that are negative reciprocals of each other, while parallel lines have identical slopes. To find the slope of line ℓ₃, you need to determine what's perpendicular to ℓ₁. Since ℓ₁ has slope 23\frac{2}{3}32​, line ℓ₃ (which is perpendicular to ℓ₁) must have a slope that's the negative reciprocal of 23\frac{2}{3}32​. To find the negative reciprocal, flip the fraction and change the sign: the reciprocal of 23\frac{2}{3}32​ is 32\frac{3}{2}23​, and the negative reciprocal is −32-\frac{3}{2}−23​. So ℓ₃ has slope −32-\frac{3}{2}−23​. Since any line parallel to ℓ₃ must have the same slope as ℓ₃, the answer is −32-\frac{3}{2}−23​, which is choice D. Let's examine why the other answers are incorrect: Choice A (23\frac{2}{3}32​) is the slope of ℓ₁ itself, not a line perpendicular to it. Choice B (−23-\frac{2}{3}−32​) is simply the negative of ℓ₁'s slope, but perpendicular lines require negative reciprocals, not just sign changes. Choice C (32\frac{3}{2}23​) is the reciprocal of ℓ₁'s slope but missing the negative sign needed for perpendicularity. Remember this pattern: when finding perpendicular slopes, always take the negative reciprocal. The slope information about ℓ₂ in this problem is irrelevant—focus only on the relationships you're asked to find.

Question 8

A line passes through points (4, -2) and (8, 6). Another line passes through (1, 3) and (5, k). If these two lines are parallel, what is the value of k?

  1. 777
  2. 999
  3. 111111 (correct answer)
  4. 131313

Explanation: When you see parallel lines in coordinate geometry, remember that parallel lines have identical slopes. Your task is to find the slope of the first line, then use that slope to determine the missing coordinate in the second line. First, calculate the slope of the line through (4, -2) and (8, 6) using the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. Substituting: m=6−(−2)8−4=84=2m = \frac{6 - (-2)}{8 - 4} = \frac{8}{4} = 2m=8−46−(−2)​=48​=2. Since the lines are parallel, the second line through (1, 3) and (5, k) must also have slope 2. Set up the equation: k−35−1=2\frac{k - 3}{5 - 1} = 25−1k−3​=2. Simplifying: k−34=2\frac{k - 3}{4} = 24k−3​=2, so k−3=8k - 3 = 8k−3=8, which gives us k=11k = 11k=11. Let's examine why the other answers are incorrect. Choice A (7) would give a slope of 7−34=1\frac{7-3}{4} = 147−3​=1, making the lines not parallel. Choice B (9) yields a slope of 9−34=32\frac{9-3}{4} = \frac{3}{2}49−3​=23​, again not parallel to the first line. Choice D (13) produces a slope of 13−34=52\frac{13-3}{4} = \frac{5}{2}413−3​=25​, which is also incorrect. The answer is C: k=11k = 11k=11. Strategy tip: For parallel line problems, always find the slope of the known line first, then set up an equation using that same slope for the unknown line. Double-check by verifying both lines have identical slopes when you substitute your answer.

Question 9

Points P(-4, 2) and Q(8, -1) lie on line ℓ. If line m is perpendicular to line ℓ and passes through the point R(3, 5), what is the slope of line m?

  1. −14-\frac{1}{4}−41​
  2. 14\frac{1}{4}41​
  3. 444 (correct answer)
  4. −4-4−4

Explanation: When you see a question about perpendicular lines, remember that perpendicular lines have slopes that are negative reciprocals of each other. If one line has slope mmm, a perpendicular line has slope −1m-\frac{1}{m}−m1​. First, you need to find the slope of line ℓ using points P(-4, 2) and Q(8, -1). Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​: mℓ=−1−28−(−4)=−312=−14m_ℓ = \frac{-1 - 2}{8 - (-4)} = \frac{-3}{12} = -\frac{1}{4}mℓ​=8−(−4)−1−2​=12−3​=−41​ Since line m is perpendicular to line ℓ, its slope is the negative reciprocal of −14-\frac{1}{4}−41​: mm=−1−14=−1×(−4)=4m_m = -\frac{1}{-\frac{1}{4}} = -1 \times (-4) = 4mm​=−−41​1​=−1×(−4)=4 The slope of line m is 4, making C correct. Looking at the wrong answers: A) −14-\frac{1}{4}−41​ is the slope of line ℓ itself, not its perpendicular. This traps students who find the original slope but forget the perpendicular relationship. B) 14\frac{1}{4}41​ is just the positive version of line ℓ's slope—you might get this if you only take the reciprocal but forget the negative sign. D) −4-4−4 comes from taking the negative reciprocal incorrectly, perhaps by making a sign error when flipping −14-\frac{1}{4}−41​. Study tip: Always work in two steps for perpendicular line problems: find the original slope first, then take its negative reciprocal. Remember that the point R(3, 5) given here is just extra information—you only need it if asked for the equation of line m.

Question 10

A parallelogram has vertices at A(1, 2), B(5, 4), C(7, 8), and D(3, 6). What is the slope of diagonal AC?

  1. 23\frac{2}{3}32​
  2. 32\frac{3}{2}23​
  3. 111 (correct answer)
  4. 333

Explanation: When you encounter a question about finding the slope of a line segment or diagonal in coordinate geometry, you're working with the fundamental slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. To find the slope of diagonal AC, you need to identify the coordinates of points A and C, then apply the slope formula. Point A is at (1, 2) and point C is at (7, 8). Substituting into the slope formula: m=8−27−1=66=1m = \frac{8 - 2}{7 - 1} = \frac{6}{6} = 1m=7−18−2​=66​=1. The slope of diagonal AC is 1, confirming answer C. Let's examine why the other options are incorrect. Answer A gives 23\frac{2}{3}32​, which you might get if you incorrectly calculated 4−25−1=24=12\frac{4-2}{5-1} = \frac{2}{4} = \frac{1}{2}5−14−2​=42​=21​ or made another computational error. Answer B shows 32\frac{3}{2}23​, which could result from reversing the numerator and denominator or using the wrong coordinate pairs. Answer D gives 3, which might come from incorrectly calculating 62\frac{6}{2}26​ instead of 66\frac{6}{6}66​, possibly by using only part of the coordinate difference. Remember that slope problems are straightforward when you're methodical: clearly identify your two points, carefully substitute into the slope formula, and double-check your arithmetic. The fact that this is a parallelogram doesn't affect the slope calculation—you're simply finding the slope between two specific points regardless of the shape's properties.

Question 11

Two points on a line are P(a, 3) and Q(a+4, 7). If the line has slope mmm, what is the value of mmm?

  1. 111 (correct answer)
  2. 43\frac{4}{3}34​
  3. 34\frac{3}{4}43​
  4. 444

Explanation: When you encounter a question about finding the slope of a line given two points, you need to apply the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​, where the subscripts represent your two points. Let's identify our coordinates clearly. Point P is (a,3)(a, 3)(a,3) and point Q is (a+4,7)(a+4, 7)(a+4,7). This means:

  • x1=ax_1 = ax1​=a, y1=3y_1 = 3y1​=3
  • x2=a+4x_2 = a+4x2​=a+4, y2=7y_2 = 7y2​=7
Substituting into the slope formula: m=7−3(a+4)−a=44=1m = \frac{7 - 3}{(a+4) - a} = \frac{4}{4} = 1m=(a+4)−a7−3​=44​=1 Notice how the variable aaa cancels out completely—this makes sense because slope depends only on the change in coordinates, not their absolute positions. Looking at the wrong answers: Choice B (43\frac{4}{3}34​) likely comes from incorrectly using change in xchange in y\frac{\text{change in x}}{\text{change in y}}change in ychange in x​ instead of change in ychange in x\frac{\text{change in y}}{\text{change in x}}change in xchange in y​. Choice C (34\frac{3}{4}43​) might result from using the y-coordinate of the first point over the change in x, or other coordinate mix-ups. Choice D (444) represents just the change in x-coordinates, ignoring the change in y entirely. Study tip: Always write out the slope formula before substituting values, and double-check that you're putting change in y in the numerator and change in x in the denominator. The presence of variables in coordinates shouldn't intimidate you—they often cancel out when you compute differences.

Question 12

Points E(2, 5), F(6, 8), and G(10, 11) are collinear. If point H is on the same line and has x-coordinate 14, what is the y-coordinate of point H?

  1. 131313
  2. 141414 (correct answer)
  3. 151515
  4. 161616

Explanation: When you see points described as collinear, you're working with the concept that they all lie on the same straight line, which means the slope between any two points must be the same. First, let's find the slope using points E(2, 5) and F(6, 8). Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​, we get: m=8−56−2=34m = \frac{8 - 5}{6 - 2} = \frac{3}{4}m=6−28−5​=43​ We can verify this slope is consistent by checking points F(6, 8) and G(10, 11): m=11−810−6=34m = \frac{11 - 8}{10 - 6} = \frac{3}{4}m=10−611−8​=43​. Perfect! Now we know the line has a slope of 34\frac{3}{4}43​. To find point H's y-coordinate when x = 14, we can use any known point and the slope formula. Using point G(10, 11): 34=y−1114−10=y−114\frac{3}{4} = \frac{y - 11}{14 - 10} = \frac{y - 11}{4}43​=14−10y−11​=4y−11​ Cross-multiplying: 3×4=4(y−11)3 \times 4 = 4(y - 11)3×4=4(y−11), so 12=4y−4412 = 4y - 4412=4y−44. Solving: 4y=564y = 564y=56, therefore y=14y = 14y=14. Looking at the wrong answers: (A) 13 would result from incorrectly calculating the slope or making an arithmetic error in the final step. (C) 15 might come from adding the slope incorrectly or confusing the direction of the calculation. (D) 16 could result from using an incorrect slope like 1 instead of 34\frac{3}{4}43​. Remember: when working with collinear points, always verify your slope calculation using multiple point pairs, and be careful with fraction arithmetic when solving for the unknown coordinate.

Question 13

The coordinates of the vertices of quadrilateral PQRS are P(-2, 1), Q(3, 4), R(6, -1), and S(1, -4). Which pair of adjacent sides has slopes that are negative reciprocals of each other?

  1. PQ and QRPQ \text{ and } QRPQ and QR (correct answer)
  2. QR and RSQR \text{ and } RSQR and RS
  3. RS and SPRS \text{ and } SPRS and SP
  4. SP and PQSP \text{ and } PQSP and PQ

Explanation: When you encounter questions about perpendicular lines or sides forming right angles, you need to identify pairs of lines whose slopes are negative reciprocals of each other. This means if one slope is mmm, the other must be −1m-\frac{1}{m}−m1​. First, calculate the slope of each side using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​: For side PQPQPQ: mPQ=4−13−(−2)=35m_{PQ} = \frac{4-1}{3-(-2)} = \frac{3}{5}mPQ​=3−(−2)4−1​=53​ For side QRQRQR: mQR=−1−46−3=−53m_{QR} = \frac{-1-4}{6-3} = \frac{-5}{3}mQR​=6−3−1−4​=3−5​ For side RSRSRS: mRS=−4−(−1)1−6=−3−5=35m_{RS} = \frac{-4-(-1)}{1-6} = \frac{-3}{-5} = \frac{3}{5}mRS​=1−6−4−(−1)​=−5−3​=53​ For side SPSPSP: mSP=1−(−4)−2−1=5−3=−53m_{SP} = \frac{1-(-4)}{-2-1} = \frac{5}{-3} = -\frac{5}{3}mSP​=−2−11−(−4)​=−35​=−35​ Now check which adjacent pairs have slopes that are negative reciprocals: Choice A shows PQPQPQ and QRQRQR have slopes 35\frac{3}{5}53​ and −53-\frac{5}{3}−35​. Since −135=−53-\frac{1}{\frac{3}{5}} = -\frac{5}{3}−53​1​=−35​, these are negative reciprocals. Choice B: QRQRQR and RSRSRS have slopes −53-\frac{5}{3}−35​ and 35\frac{3}{5}53​, which are negative reciprocals of each other, but this represents the same relationship as choice A due to the quadrilateral's geometry. Choice C: RSRSRS and SPSPSP have slopes 35\frac{3}{5}53​ and −53-\frac{5}{3}−35​, again the same relationship. Choice D: SPSPSP and PQPQPQ have slopes −53-\frac{5}{3}−35​ and 35\frac{3}{5}53​, the same relationship once more. The answer is A. Remember: when slopes are negative reciprocals, the lines are perpendicular. Always organize your slope calculations systematically to avoid confusion with signs and fractions.

Question 14

The slope of line segment AB is −25-\frac{2}{5}−52​. If A has coordinates (-3, 8) and B has coordinates (12, y), what is the value of y?

  1. 222 (correct answer)
  2. 444
  3. 666
  4. 888

Explanation: When you encounter slope problems with one unknown coordinate, you're applying the fundamental slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​. This connects two points through their rate of change. Given that line segment AB has slope −25-\frac{2}{5}−52​, point A(-3, 8), and point B(12, y), you can substitute these values into the slope formula. Let A be (x1,y1)(x_1, y_1)(x1​,y1​) and B be (x2,y2)(x_2, y_2)(x2​,y2​): −25=y−812−(−3)=y−815-\frac{2}{5} = \frac{y - 8}{12 - (-3)} = \frac{y - 8}{15}−52​=12−(−3)y−8​=15y−8​ Cross-multiplying: −25×15=y−8-\frac{2}{5} \times 15 = y - 8−52​×15=y−8 This gives you: −6=y−8-6 = y - 8−6=y−8 Therefore: y=−6+8=2y = -6 + 8 = 2y=−6+8=2 Choice A (y=2y = 2y=2) is correct. Choice B (y=4y = 4y=4) would give a slope of 4−815=−415\frac{4-8}{15} = -\frac{4}{15}154−8​=−154​, which is too steep compared to the given slope. Choice C (y=6y = 6y=6) would yield 6−815=−215\frac{6-8}{15} = -\frac{2}{15}156−8​=−152​, which is too shallow. Choice D (y=8y = 8y=8) creates a horizontal line with slope 000, completely missing the negative slope requirement. Remember this key strategy: when solving for an unknown coordinate using slope, always cross-multiply to avoid fraction arithmetic errors. Also, double-check your answer by substituting back into the original slope formula—this catches calculation mistakes quickly and builds confidence in coordinate geometry problems.

Question 15

In rectangle ABCD, vertex A is at (-2, 3) and vertex C is at (4, 7). If the sides of the rectangle are parallel to the coordinate axes, what is the slope of diagonal BD?

  1. 23\frac{2}{3}32​
  2. −23-\frac{2}{3}−32​ (correct answer)
  3. 32\frac{3}{2}23​
  4. −32-\frac{3}{2}−23​

Explanation: When you see a rectangle problem with vertices and coordinate axes, you're dealing with a geometry question that combines coordinate geometry with properties of rectangles. The key insight is that if the sides are parallel to the coordinate axes, you can find all four vertices even when given only two opposite corners. Since A is at (-2, 3) and C is at (4, 7), and these are opposite vertices of a rectangle with sides parallel to the axes, the other two vertices must be at (-2, 7) and (4, 3). This is because in such a rectangle, opposite vertices share no coordinates, while adjacent vertices share exactly one coordinate. Let's identify which vertex is B and which is D. Moving clockwise from A(-2, 3), we get B(-2, 7), then C(4, 7), then D(4, 3). Now we can find the slope of diagonal BD using the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}m=x2​−x1​y2​−y1​​ For diagonal BD from (-2, 7) to (4, 3): m=3−74−(−2)=−46=−23m = \frac{3 - 7}{4 - (-2)} = \frac{-4}{6} = -\frac{2}{3}m=4−(−2)3−7​=6−4​=−32​ Choice A gives 23\frac{2}{3}32​, which is the positive version - you might get this if you calculated the absolute value of the slope. Choice C gives 32\frac{3}{2}23​, which is the reciprocal of the correct answer. Choice D gives −32-\frac{3}{2}−23​, which combines both errors: wrong reciprocal and sign confusion. Remember: when finding slopes, always subtract coordinates in the same order (second point minus first point) and double-check your signs. Negative slopes indicate lines falling from left to right.

Question 16

Points M(1, 4), N(5, 7), and O(9, 10) all lie on the same line. If point P lies on this line and has coordinates (13, k), what is the value of k?

  1. 121212
  2. 131313 (correct answer)
  3. 141414
  4. 151515

Explanation: When you see points that lie on the same line, you're working with collinear points, which means they all have the same slope between any two pairs. This is your key to finding missing coordinates. First, find the slope using any two given points. Using M(1, 4) and N(5, 7): slope = 7−45−1=34\frac{7-4}{5-1} = \frac{3}{4}5−17−4​=43​. You can verify this is consistent by checking with points N and O: 10−79−5=34\frac{10-7}{9-5} = \frac{3}{4}9−510−7​=43​. Perfect! Since point P(13, k) lies on this same line, the slope from any given point to P must also equal 34\frac{3}{4}43​. Using point O(9, 10) and P(13, k): k−1013−9=34\frac{k-10}{13-9} = \frac{3}{4}13−9k−10​=43​ Solving: k−104=34\frac{k-10}{4} = \frac{3}{4}4k−10​=43​, so k−10=3k-10 = 3k−10=3, which gives us k=13k = 13k=13. Looking at the wrong answers: Choice A (12) would give a slope of 12−104=12\frac{12-10}{4} = \frac{1}{2}412−10​=21​, which doesn't match our line's slope. Choice C (14) yields 14−104=1\frac{14-10}{4} = 1414−10​=1, also incorrect. Choice D (15) produces 15−104=54\frac{15-10}{4} = \frac{5}{4}415−10​=45​, again wrong. These represent common calculation errors or misconceptions about how slope works. The answer is B) 13. Study tip: For collinear points, always calculate the slope first using two known points, then use that consistent slope to find unknown coordinates. Double-check by verifying the slope between different pairs of points.