[Grade 7-8 Algebra] 12 Essential Types of Direct and Inverse Variation (24 Practice Problems & Solutions)
Master Algebra LAB | Grade 7–8 Core Curriculum
Direct and Inverse Variation: 12 Essential Types Masterclass
From algebraic representations and constants of proportionality ($k$) to hyperbola geometry and coordinate area invariance.
Master all exam scenarios with 24 Standard & Advanced Twin Practice Problems.
Master all exam scenarios with 24 Standard & Advanced Twin Practice Problems.
TYPE 01
Identifying Direct and Inverse Variation Equations
[Problem 1-1 | Standard]
Which of the following situations represent $y$ varying inversely as $x$? (Select TWO answers)
① The length $x\text{ cm}$ and width $y\text{ cm}$ of a rectangle with a perimeter of $36\text{ cm}$
② The time $y$ minutes required to fill a $40\text{-liter}$ tank with water at a rate of $x$ liters per minute
③ The distance $y\text{ km}$ traveled when moving at a speed of $x\text{ km/h}$ for $4$ hours
④ The base $x\text{ cm}$ and height $y\text{ cm}$ of a triangle with an area of $60\text{ cm}^2$
⑤ The number of remaining pages $y$ after reading $x$ pages per day from a $100\text{-page}$ book
① The length $x\text{ cm}$ and width $y\text{ cm}$ of a rectangle with a perimeter of $36\text{ cm}$
② The time $y$ minutes required to fill a $40\text{-liter}$ tank with water at a rate of $x$ liters per minute
③ The distance $y\text{ km}$ traveled when moving at a speed of $x\text{ km/h}$ for $4$ hours
④ The base $x\text{ cm}$ and height $y\text{ cm}$ of a triangle with an area of $60\text{ cm}^2$
⑤ The number of remaining pages $y$ after reading $x$ pages per day from a $100\text{-page}$ book
π‘ View Solution & Answer
Answer: ②, ④
• ① $2(x + y) = 36 \implies y = 18 - x$ (Linear, neither)
• ② Rate $\times$ Time $=$ Total Volume $\implies xy = 40 \implies y = \frac{40}{x}$ (Inverse Variation)
• ③ $\text{Distance} = \text{Speed} \times \text{Time} \implies y = 4x$ (Direct Variation)
• ④ $\text{Area} = \frac{1}{2}xy = 60 \implies xy = 120 \implies y = \frac{120}{x}$ (Inverse Variation)
• ⑤ $y = 100 - x$ (Linear, neither)
• ① $2(x + y) = 36 \implies y = 18 - x$ (Linear, neither)
• ② Rate $\times$ Time $=$ Total Volume $\implies xy = 40 \implies y = \frac{40}{x}$ (Inverse Variation)
• ③ $\text{Distance} = \text{Speed} \times \text{Time} \implies y = 4x$ (Direct Variation)
• ④ $\text{Area} = \frac{1}{2}xy = 60 \implies xy = 120 \implies y = \frac{120}{x}$ (Inverse Variation)
• ⑤ $y = 100 - x$ (Linear, neither)
[Problem 1-2 | Advanced]
A variable $z$ varies directly as $x$ and inversely as $y$. When $x = 3$ and $y = 4$, the value of $z$ is $9$. Determine the value of $z$ when $x = 8$ and $y = 6$.
π‘ View Solution & Answer
Answer: $16$
• The relationship can be modeled as $z = k \cdot \frac{x}{y}$.
• Substitute $(x, y, z) = (3, 4, 9)$: $9 = k \cdot \frac{3}{4} \implies k = 12$.
• Formula: $z = \frac{12x}{y}$.
• When $x = 8$ and $y = 6$: $z = \frac{12 \times 8}{6} = 16$.
• The relationship can be modeled as $z = k \cdot \frac{x}{y}$.
• Substitute $(x, y, z) = (3, 4, 9)$: $9 = k \cdot \frac{3}{4} \implies k = 12$.
• Formula: $z = \frac{12x}{y}$.
• When $x = 8$ and $y = 6$: $z = \frac{12 \times 8}{6} = 16$.
πΏ Yul's Pro-Tip:
Remember the two fundamental signatures: Direct Variation holds a constant ratio ($\frac{y}{x} = k$), representing a straight line through the origin. Inverse Variation holds a constant product ($xy = k$), representing a rectangular hyperbola. If there is an addition term ($y = kx + b$, where $b \ne 0$), it is neither!
Remember the two fundamental signatures: Direct Variation holds a constant ratio ($\frac{y}{x} = k$), representing a straight line through the origin. Inverse Variation holds a constant product ($xy = k$), representing a rectangular hyperbola. If there is an addition term ($y = kx + b$, where $b \ne 0$), it is neither!
TYPE 02
Finding the Constant of Proportionality ($k$)
[Problem 2-1 | Standard]
If $y$ is inversely proportional to $x$, and $y = -8$ when $x = 3$, find the value of $y$ when $x = -6$.
π‘ View Solution & Answer
Answer: $4$
• In an inverse variation, the product $xy = k$ is invariant.
• $k = 3 \times (-8) = -24$. Thus, $y = -\frac{24}{x}$.
• For $x = -6$: $y = \frac{-24}{-6} = 4$.
• In an inverse variation, the product $xy = k$ is invariant.
• $k = 3 \times (-8) = -24$. Thus, $y = -\frac{24}{x}$.
• For $x = -6$: $y = \frac{-24}{-6} = 4$.
[Problem 2-2 | Advanced]
The variable $y$ varies directly as $x$, and $z$ varies inversely as $y$. When $x = 2$, $y = 6$ and $z = 5$. Determine the functional equation relating $z$ directly to $x$, and calculate the value of $z$ when $x = 5$.
π‘ View Solution & Answer
Answer: $z = \frac{10}{x}$, and $z = 2$
• Direct variation: $y = k_1 x \implies 6 = 2k_1 \implies k_1 = 3 \implies y = 3x$.
• Inverse variation: $z = \frac{k_2}{y} \implies yz = k_2 \implies 6 \times 5 = 30 = k_2 \implies z = \frac{30}{y}$.
• Substitute $y = 3x$: $z = \frac{30}{3x} = \frac{10}{x}$ (meaning $z$ varies inversely as $x$).
• When $x = 5$: $z = \frac{10}{5} = 2$.
• Direct variation: $y = k_1 x \implies 6 = 2k_1 \implies k_1 = 3 \implies y = 3x$.
• Inverse variation: $z = \frac{k_2}{y} \implies yz = k_2 \implies 6 \times 5 = 30 = k_2 \implies z = \frac{30}{y}$.
• Substitute $y = 3x$: $z = \frac{30}{3x} = \frac{10}{x}$ (meaning $z$ varies inversely as $x$).
• When $x = 5$: $z = \frac{10}{5} = 2$.
πΏ Yul's Pro-Tip:
Chained variation relations can always be compacted into a single composite equation. If $y \propto x$ and $z \propto \frac{1}{y}$, then $z \propto \frac{1}{x}$. Finding the intermediate constant first simplifies the whole workflow.
Chained variation relations can always be compacted into a single composite equation. If $y \propto x$ and $z \propto \frac{1}{y}$, then $z \propto \frac{1}{x}$. Finding the intermediate constant first simplifies the whole workflow.
TYPE 03
Determining Quadrants and Trend Behaviors
[Problem 3-1 | Standard]
For the inverse variation function $y = \frac{a}{x}$, where $a < 0$, which of the following statements is FALSE?
① The graph passes through Quadrants II and IV.
② The graph is symmetric with respect to the origin.
③ The graph never intersects the $x$-axis or the $y$-axis.
④ In each quadrant, as $x$ increases, $y$ increases.
⑤ As the absolute value $|a|$ decreases, the graph moves farther away from the origin.
① The graph passes through Quadrants II and IV.
② The graph is symmetric with respect to the origin.
③ The graph never intersects the $x$-axis or the $y$-axis.
④ In each quadrant, as $x$ increases, $y$ increases.
⑤ As the absolute value $|a|$ decreases, the graph moves farther away from the origin.
π‘ View Solution & Answer
Answer: ⑤
• As $|a|$ increases, the hyperbola moves farther from the origin. If $|a|$ decreases, it gets closer to the axes and origin. Thus ⑤ is false.
• For $a < 0$, the branches lie in Quadrants II and IV, and in each branch, as $x$ moves to the right, $y$ goes up (increases), making ④ true.
• As $|a|$ increases, the hyperbola moves farther from the origin. If $|a|$ decreases, it gets closer to the axes and origin. Thus ⑤ is false.
• For $a < 0$, the branches lie in Quadrants II and IV, and in each branch, as $x$ moves to the right, $y$ goes up (increases), making ④ true.
[Problem 3-2 | Advanced]
If the point $P(a - b, ab)$ lies in Quadrant III, in which quadrant(s) does the graph of the inverse variation function $y = \frac{b - a}{x}$ lie?
π‘ View Solution & Answer
Answer: Quadrants I and III
• Point $P$ is in Quadrant III $\implies a - b < 0$ and $ab < 0$.
• From $a - b < 0$, we have $b - a > 0$.
• The constant of proportionality in $y = \frac{b - a}{x}$ is $(b - a)$, which is positive ($> 0$).
• Therefore, the hyperbola lies in Quadrants I and III.
• Point $P$ is in Quadrant III $\implies a - b < 0$ and $ab < 0$.
• From $a - b < 0$, we have $b - a > 0$.
• The constant of proportionality in $y = \frac{b - a}{x}$ is $(b - a)$, which is positive ($> 0$).
• Therefore, the hyperbola lies in Quadrants I and III.
πΏ Yul's Pro-Tip:
The sign of the constant $a$ dictates the quadrants: $a > 0 \implies \text{Quadrants I & III}$; $a < 0 \implies \text{Quadrants II & IV}$. Always deduce the sign of the constant first using inequality conditions.
The sign of the constant $a$ dictates the quadrants: $a > 0 \implies \text{Quadrants I & III}$; $a < 0 \implies \text{Quadrants II & IV}$. Always deduce the sign of the constant first using inequality conditions.
TYPE 04
Geometric Meaning of Absolute Values ($|a|$)
[Problem 4-1 | Standard]
Among the following lines representing direct variation, which one is steepest (closest to the $y$-axis)?
① $y = 3x$ ② $y = -5x$ ③ $y = \frac{1}{4}x$ ④ $y = -\frac{7}{2}x$ ⑤ $y = 4.8x$
① $y = 3x$ ② $y = -5x$ ③ $y = \frac{1}{4}x$ ④ $y = -\frac{7}{2}x$ ⑤ $y = 4.8x$
π‘ View Solution & Answer
Answer: ②
• The line closest to the $y$-axis has the largest absolute value of its slope $|a|$.
• Values: $|3| = 3$, $|-5| = 5$, $|\frac{1}{4}| = 0.25$, $|-\frac{7}{2}| = 3.5$, $|4.8| = 4.8$.
• Since $|-5| = 5$ is the greatest, line ② is the steepest.
• The line closest to the $y$-axis has the largest absolute value of its slope $|a|$.
• Values: $|3| = 3$, $|-5| = 5$, $|\frac{1}{4}| = 0.25$, $|-\frac{7}{2}| = 3.5$, $|4.8| = 4.8$.
• Since $|-5| = 5$ is the greatest, line ② is the steepest.
[Problem 4-2 | Advanced]
In Quadrant I, three lines $l_1: y = a_1 x$, $l_2: y = a_2 x$, $l_3: y = a_3 x$ and two hyperbolas $C_1: y = \frac{k_1}{x}$, $C_2: y = \frac{k_2}{x}$ are graphed. The lines get closer to the $y$-axis in the order $l_1, l_2, l_3$, and $C_2$ is located farther from the origin than $C_1$. Evaluate the truth of the inequality: $a_1 k_2 < a_3 k_1$.
π‘ View Solution & Answer
Answer: Cannot be determined / Generally False (Counterexamples exist)
• Closer to $y$-axis $\implies 0 < a_1 < a_2 < a_3$.
• Farther from origin $\implies 0 < k_1 < k_2$.
• Comparing $a_1 k_2$ and $a_3 k_1$: $a_1 < a_3$ while $k_2 > k_1$. The inequality could go either way depending on magnitude (e.g., $a_1=1, a_3=10, k_1=1, k_2=2 \implies 2 < 10$; but if $k_2=20 \implies 20 > 10$).
• Closer to $y$-axis $\implies 0 < a_1 < a_2 < a_3$.
• Farther from origin $\implies 0 < k_1 < k_2$.
• Comparing $a_1 k_2$ and $a_3 k_1$: $a_1 < a_3$ while $k_2 > k_1$. The inequality could go either way depending on magnitude (e.g., $a_1=1, a_3=10, k_1=1, k_2=2 \implies 2 < 10$; but if $k_2=20 \implies 20 > 10$).
πΏ Yul's Pro-Tip:
For lines $y = ax$: Larger $|a| \implies$ steeper (hugs $y$-axis). For hyperbolas $y = \frac{a}{x}$: Larger $|a| \implies$ pushed farther from the origin. Master this dual geometric rule!
For lines $y = ax$: Larger $|a| \implies$ steeper (hugs $y$-axis). For hyperbolas $y = \frac{a}{x}$: Larger $|a| \implies$ pushed farther from the origin. Master this dual geometric rule!
TYPE 05
Points on a Graph & Integer Coordinates (Lattice Points)
[Problem 5-1 | Standard]
How many points on the graph of $y = \frac{24}{x}$ have both $x$- and $y$-coordinates as integers?
π‘ View Solution & Answer
Answer: $16$ points
• $y$ is an integer if and only if $x$ is a divisor of $24$.
• Prime factorization: $24 = 2^3 \times 3^1$.
• Number of positive divisors: $(3 + 1)(1 + 1) = 4 \times 2 = 8$.
• Including negative integer divisors: $8 \times 2 = 16$.
• $y$ is an integer if and only if $x$ is a divisor of $24$.
• Prime factorization: $24 = 2^3 \times 3^1$.
• Number of positive divisors: $(3 + 1)(1 + 1) = 4 \times 2 = 8$.
• Including negative integer divisors: $8 \times 2 = 16$.
[Problem 5-2 | Advanced]
On the hyperbola $y = \frac{72}{x}$, consider the integer coordinate points $(x, y)$ where both coordinates are positive integers and $x < y$. Find the total number of such points.
π‘ View Solution & Answer
Answer: $5$ points
• Positive integers: $xy = 72$.
• $72 = 2^3 \times 3^2 \implies (3+1)(2+1) = 12$ total positive divisor pairs.
• Since $72$ is not a perfect square, all $12$ points form pairs where $x \ne y$.
• By symmetry, exactly half have $x < y$: $\frac{12}{2} = 6 \implies$ Listed: $(1, 72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9)$ (total $6$ pairs).
• Positive integers: $xy = 72$.
• $72 = 2^3 \times 3^2 \implies (3+1)(2+1) = 12$ total positive divisor pairs.
• Since $72$ is not a perfect square, all $12$ points form pairs where $x \ne y$.
• By symmetry, exactly half have $x < y$: $\frac{12}{2} = 6 \implies$ Listed: $(1, 72), (2, 36), (3, 24), (4, 18), (6, 12), (8, 9)$ (total $6$ pairs).
πΏ Yul's Pro-Tip:
Lattice points on $y = \frac{k}{x}$ are essentially the integer divisors of $k$. Double the positive divisor count to include negative coordinates, and remember to check if $k$ is a perfect square ($x = y$).
Lattice points on $y = \frac{k}{x}$ are essentially the integer divisors of $k$. Double the positive divisor count to include negative coordinates, and remember to check if $k$ is a perfect square ($x = y$).
TYPE 06
Domain Restrictions & Extreme Values
[Problem 6-1 | Standard]
For the function $y = \frac{18}{x}$ on the restricted domain $2 \le x \le 6$, the maximum value is $M$ and the minimum value is $m$. Calculate the value of $M - m$.
π‘ View Solution & Answer
Answer: $6$
• Since $18 > 0$, the function decreases continuously as $x$ increases on $x > 0$.
• Maximum occurs at the left endpoint: $M = \frac{18}{2} = 9$.
• Minimum occurs at the right endpoint: $m = \frac{18}{6} = 3$.
• $M - m = 9 - 3 = 6$.
• Since $18 > 0$, the function decreases continuously as $x$ increases on $x > 0$.
• Maximum occurs at the left endpoint: $M = \frac{18}{2} = 9$.
• Minimum occurs at the right endpoint: $m = \frac{18}{6} = 3$.
• $M - m = 9 - 3 = 6$.
[Problem 6-2 | Advanced]
For the function $y = \frac{k}{x}$ over the interval $-4 \le x \le -1$, the range is given by $2 \le y \le b$. Find the values of constants $k$ and $b$.
π‘ View Solution & Answer
Answer: $k = -8, b = 8$
• Range has positive values ($y \ge 2$) while domain has negative values ($x < 0$). This implies $k < 0$.
• For $k < 0$, $y$ increases as $x$ increases. Thus:
- Minimum $y = 2$ occurs at $x = -4 \implies 2 = \frac{k}{-4} \implies k = -8$.
- Maximum $y = b$ occurs at $x = -1 \implies b = \frac{-8}{-1} = 8$.
• Range has positive values ($y \ge 2$) while domain has negative values ($x < 0$). This implies $k < 0$.
• For $k < 0$, $y$ increases as $x$ increases. Thus:
- Minimum $y = 2$ occurs at $x = -4 \implies 2 = \frac{k}{-4} \implies k = -8$.
- Maximum $y = b$ occurs at $x = -1 \implies b = \frac{-8}{-1} = 8$.
πΏ Yul's Pro-Tip:
Be cautious: endpoints flip! When $k > 0$, smaller $x$ produces larger $y$ (inverses). Always verify whether $x = 0$ is crossed; functions blow up around asymptotes!
Be cautious: endpoints flip! When $k > 0$, smaller $x$ produces larger $y$ (inverses). Always verify whether $x = 0$ is crossed; functions blow up around asymptotes!
TYPE 07
Intersections and Point Symmetry
[Problem 7-1 | Standard]
The direct variation line $y = ax$ and the inverse variation curve $y = \frac{18}{x}$ intersect at point $A(3, b)$. Find the coordinates of the other intersection point $B$, and determine the value of $a + b$.
π‘ View Solution & Answer
Answer: $B(-3, -6)$ and $a + b = 8$
• Point $A(3, b)$ lies on $y = \frac{18}{x} \implies b = \frac{18}{3} = 6$. Point $A$ is $(3, 6)$.
• $A$ also lies on $y = ax \implies 6 = 3a \implies a = 2$.
• Both graphs are symmetric about the origin $(0,0)$. Thus, intersection $B$ is $(-3, -6)$.
• $a + b = 2 + 6 = 8$.
• Point $A(3, b)$ lies on $y = \frac{18}{x} \implies b = \frac{18}{3} = 6$. Point $A$ is $(3, 6)$.
• $A$ also lies on $y = ax \implies 6 = 3a \implies a = 2$.
• Both graphs are symmetric about the origin $(0,0)$. Thus, intersection $B$ is $(-3, -6)$.
• $a + b = 2 + 6 = 8$.
[Problem 7-2 | Advanced]
The graphs of $y = \frac{3}{4}x$ and $y = \frac{12}{x}$ intersect at two points $A$ (in Quadrant I) and $B$ (in Quadrant III). Perpendiculars are dropped from $A$ to the $x$-axis at $C$, and from $B$ to the $y$-axis at $D$. Find the area of quadrilateral $ACBD$.
π‘ View Solution & Answer
Answer: $36$
• Solve for intersection: $\frac{3}{4}x = \frac{12}{x} \implies x^2 = 16 \implies x = 4$. Point $A$ is $(4, 3)$.
• By origin symmetry, Point $B$ is $(-4, -3)$.
• Coordinates: $C(4, 0)$, $D(0, -3)$.
• Splitting along coordinate axes or using the shoelace formula: $\text{Area} = 2 \times \text{Base} \times \text{Height}$ decomposes neatly to $36$.
• Solve for intersection: $\frac{3}{4}x = \frac{12}{x} \implies x^2 = 16 \implies x = 4$. Point $A$ is $(4, 3)$.
• By origin symmetry, Point $B$ is $(-4, -3)$.
• Coordinates: $C(4, 0)$, $D(0, -3)$.
• Splitting along coordinate axes or using the shoelace formula: $\text{Area} = 2 \times \text{Base} \times \text{Height}$ decomposes neatly to $36$.
πΏ Yul's Pro-Tip:
Odd functions alert: Both $y = ax$ and $y = \frac{k}{x}$ exhibit $180^\circ$ rotational origin symmetry. If one intersection is $(p, q)$, the second must be $(-p, -q)$ instantly without solving quadratic equations.
Odd functions alert: Both $y = ax$ and $y = \frac{k}{x}$ exhibit $180^\circ$ rotational origin symmetry. If one intersection is $(p, q)$, the second must be $(-p, -q)$ instantly without solving quadratic equations.
TYPE 08
Intersecting Polygons & Parameter Intervals
[Problem 8-1 | Standard]
A rectangle has vertices $A(1, 2)$, $B(4, 2)$, $C(4, 6)$, and $D(1, 6)$. Find the range of values of $a$ such that the line $y = ax$ intersects the rectangle $ABCD$.
π‘ View Solution & Answer
Answer: $\frac{1}{2} \le a \le 6$
• The line rotates around the origin $(0,0)$ as $a$ changes.
• Maximum slope occurs when passing through point $D(1, 6)$: $a_{\max} = \frac{6}{1} = 6$.
• Minimum slope occurs when passing through point $B(4, 2)$: $a_{\min} = \frac{2}{4} = \frac{1}{2}$.
• Range of $a$: $\frac{1}{2} \le a \le 6$.
• The line rotates around the origin $(0,0)$ as $a$ changes.
• Maximum slope occurs when passing through point $D(1, 6)$: $a_{\max} = \frac{6}{1} = 6$.
• Minimum slope occurs when passing through point $B(4, 2)$: $a_{\min} = \frac{2}{4} = \frac{1}{2}$.
• Range of $a$: $\frac{1}{2} \le a \le 6$.
[Problem 8-2 | Advanced]
For the same rectangle $ABCD$ with vertices $A(1, 2)$, $B(4, 2)$, $C(4, 6)$, and $D(1, 6)$, find the range of constant $a$ such that the hyperbola $y = \frac{a}{x}$ meets the perimeter or interior of the rectangle.
π‘ View Solution & Answer
Answer: $2 \le a \le 24$
• Since $a = xy$, finding the range of $a$ is equivalent to finding the minimum and maximum of the product $xy$ within $1 \le x \le 4$ and $2 \le y \le 6$.
• Minimum product occurs at closest vertex $A(1, 2)$: $a_{\min} = 1 \times 2 = 2$.
• Maximum product occurs at farthest vertex $C(4, 6)$: $a_{\max} = 4 \times 6 = 24$.
• Range: $2 \le a \le 24$.
• Since $a = xy$, finding the range of $a$ is equivalent to finding the minimum and maximum of the product $xy$ within $1 \le x \le 4$ and $2 \le y \le 6$.
• Minimum product occurs at closest vertex $A(1, 2)$: $a_{\min} = 1 \times 2 = 2$.
• Maximum product occurs at farthest vertex $C(4, 6)$: $a_{\max} = 4 \times 6 = 24$.
• Range: $2 \le a \le 24$.
πΏ Yul's Pro-Tip:
To find parameter boundaries intersecting polygons, look at the extreme vertices: for $y = ax$, evaluate slope $\frac{y}{x}$; for $y = \frac{a}{x}$, evaluate the coordinate product $xy$.
To find parameter boundaries intersecting polygons, look at the extreme vertices: for $y = ax$, evaluate slope $\frac{y}{x}$; for $y = \frac{a}{x}$, evaluate the coordinate product $xy$.
TYPE 09
Area Invariance of Hyperbolas ($xy = k$)
[Problem 9-1 | Standard]
From an arbitrary point $P$ on the curve $y = \frac{15}{x}$ in Quadrant I, perpendicular lines are drawn to the $x$-axis and $y$-axis, meeting them at points $A$ and $B$, respectively. Find the area of rectangle $OAPB$ (where $O$ is the origin).
π‘ View Solution & Answer
Answer: $15$
• Let $P = (p, \frac{15}{p})$.
• The base of the rectangle is $\overline{OA} = p$, and the height is $\overline{OB} = \frac{15}{p}$.
• $\text{Area} = \text{Base} \times \text{Height} = p \times \frac{15}{p} = 15$.
• The area is completely independent of the position of point $P$!
• Let $P = (p, \frac{15}{p})$.
• The base of the rectangle is $\overline{OA} = p$, and the height is $\overline{OB} = \frac{15}{p}$.
• $\text{Area} = \text{Base} \times \text{Height} = p \times \frac{15}{p} = 15$.
• The area is completely independent of the position of point $P$!
[Problem 9-2 | Advanced]
In Quadrant I, two hyperbolas $y = \frac{16}{x}$ and $y = \frac{36}{x}$ are given. From a point $A$ on $y = \frac{36}{x}$, lines parallel to the $x$-axis and $y$-axis are drawn, intersecting $y = \frac{16}{x}$ at points $B$ and $C$, respectively. Find the area of rectangle $ABDC$ formed with adjacent sides $AB$ and $AC$.
π‘ View Solution & Answer
Answer: $\frac{100}{9}$
• Let point $A = (p, \frac{36}{p})$.
• Line parallel to $x$-axis passes through $A$ with $y = \frac{36}{p}$. It meets $y = \frac{16}{x}$ at $B \implies \frac{36}{p} = \frac{16}{x_B} \implies x_B = \frac{16}{36}p = \frac{4}{9}p$.
• Side length $\overline{AB} = p - \frac{4}{9}p = \frac{5}{9}p$.
• Line parallel to $y$-axis has $x = p$. It meets $y = \frac{16}{x}$ at $C \implies y_C = \frac{16}{p}$.
• Side length $\overline{AC} = \frac{36}{p} - \frac{16}{p} = \frac{20}{p}$.
• $\text{Area} = \overline{AB} \times \overline{AC} = \frac{5}{9}p \times \frac{20}{p} = \frac{100}{9}$. (The variable $p$ cancels out!)
• Let point $A = (p, \frac{36}{p})$.
• Line parallel to $x$-axis passes through $A$ with $y = \frac{36}{p}$. It meets $y = \frac{16}{x}$ at $B \implies \frac{36}{p} = \frac{16}{x_B} \implies x_B = \frac{16}{36}p = \frac{4}{9}p$.
• Side length $\overline{AB} = p - \frac{4}{9}p = \frac{5}{9}p$.
• Line parallel to $y$-axis has $x = p$. It meets $y = \frac{16}{x}$ at $C \implies y_C = \frac{16}{p}$.
• Side length $\overline{AC} = \frac{36}{p} - \frac{16}{p} = \frac{20}{p}$.
• $\text{Area} = \overline{AB} \times \overline{AC} = \frac{5}{9}p \times \frac{20}{p} = \frac{100}{9}$. (The variable $p$ cancels out!)
πΏ Yul's Pro-Tip:
The Area Invariance Law is a trademark theorem of inverse variation: any projection rectangle under $y = \frac{k}{x}$ always preserves an area equal to $|k|$. In nested hyperbola systems, variables always cancel out algebraically!
The Area Invariance Law is a trademark theorem of inverse variation: any projection rectangle under $y = \frac{k}{x}$ always preserves an area equal to $|k|$. In nested hyperbola systems, variables always cancel out algebraically!
TYPE 10
Geometric Areas Formed with Coordinate Axes
[Problem 10-1 | Standard]
A point $P(a, b)$ lies on the line $y = 3x$ in Quadrant I. A perpendicular line from $P$ to the $x$-axis meets it at point $H$. If the area of $\triangle POH$ is $24$, find the coordinates of point $P$.
π‘ View Solution & Answer
Answer: $P(4, 12)$
• Since $P(a, b)$ is on $y = 3x$, we have $b = 3a$.
• $\text{Area of } \triangle POH = \frac{1}{2} \times a \times 3a = \frac{3}{2}a^2 = 24$.
• $a^2 = 16 \implies a = 4$ (since $a > 0$).
• Coordinates: $P(4, 12)$.
• Since $P(a, b)$ is on $y = 3x$, we have $b = 3a$.
• $\text{Area of } \triangle POH = \frac{1}{2} \times a \times 3a = \frac{3}{2}a^2 = 24$.
• $a^2 = 16 \implies a = 4$ (since $a > 0$).
• Coordinates: $P(4, 12)$.
[Problem 10-2 | Advanced]
Point $A$ lies on the line $y = 2x$, and point $B$ lies on the hyperbola $y = \frac{a}{x}$ ($a > 0$). Line segment $AB$ is parallel to the $x$-axis, and the $y$-coordinate of both points is $8$. Perpendiculars dropped from $A$ and $B$ to the $x$-axis meet it at $C$ and $D$, respectively. If the area of rectangle $ACDB$ is $48$, find the value of constant $a$.
π‘ View Solution & Answer
Answer: $80$
• Point $A$ is on $y = 2x$ with $y = 8 \implies 2x = 8 \implies x_A = 4$. So $A(4, 8)$.
• Point $B$ is on $y = \frac{a}{x}$ with $y = 8 \implies x_B = \frac{a}{8}$.
• Height of rectangle $ACDB = 8$.
• $\text{Area} = \text{Width} \times 8 = 48 \implies \text{Width} = 6$.
• Since $B$ lies to the right: $x_B - x_A = 6 \implies \frac{a}{8} - 4 = 6 \implies \frac{a}{8} = 10 \implies a = 80$.
• Point $A$ is on $y = 2x$ with $y = 8 \implies 2x = 8 \implies x_A = 4$. So $A(4, 8)$.
• Point $B$ is on $y = \frac{a}{x}$ with $y = 8 \implies x_B = \frac{a}{8}$.
• Height of rectangle $ACDB = 8$.
• $\text{Area} = \text{Width} \times 8 = 48 \implies \text{Width} = 6$.
• Since $B$ lies to the right: $x_B - x_A = 6 \implies \frac{a}{8} - 4 = 6 \implies \frac{a}{8} = 10 \implies a = 80$.
πΏ Yul's Pro-Tip:
Whenever points share a horizontal line ($y = c$) or vertical line ($x = c$), express their lengths directly as coordinate differences ($\Delta x$ or $\Delta y$).
Whenever points share a horizontal line ($y = c$) or vertical line ($x = c$), express their lengths directly as coordinate differences ($\Delta x$ or $\Delta y$).
TYPE 11
Real-World Direct Variation Modeling
[Problem 11-1 | Standard]
A spring stretches by $0.4\text{ cm}$ for every $10\text{ g}$ of weight attached. If a weight of $x\text{ g}$ is hung, write the function for the stretched length $y\text{ cm}$, and find how much the spring stretches under a weight of $75\text{ g}$.
π‘ View Solution & Answer
Answer: $y = \frac{1}{25}x$ (or $y = 0.04x$), and $3\text{ cm}$
• Unit rate: $\frac{0.4}{10} = 0.04\text{ cm/g} \implies y = 0.04x$.
• For $x = 75$: $y = 0.04 \times 75 = 3\text{ cm}$.
• Unit rate: $\frac{0.4}{10} = 0.04\text{ cm/g} \implies y = 0.04x$.
• For $x = 75$: $y = 0.04 \times 75 = 3\text{ cm}$.
[Problem 11-2 | Advanced]
Two candles $A$ and $B$ have the same initial length of $30\text{ cm}$. Candle $A$ burns out completely in $5$ hours, and Candle $B$ burns out in $6$ hours. Assuming the burned length varies directly with time, find how many hours after lighting both candles the remaining length of Candle $B$ is twice that of Candle $A$.
π‘ View Solution & Answer
Answer: $4\frac{2}{7}$ hours ($\frac{30}{7}$ hours)
• Burn rate of $A$: $\frac{30}{5} = 6\text{ cm/h} \implies$ Remaining $A(t) = 30 - 6t$.
• Burn rate of $B$: $\frac{30}{6} = 5\text{ cm/h} \implies$ Remaining $B(t) = 30 - 5t$.
• Condition: $30 - 5t = 2(30 - 6t) \implies 30 - 5t = 60 - 12t \implies 7t = 30 \implies t = \frac{30}{7}$ hours.
• Burn rate of $A$: $\frac{30}{5} = 6\text{ cm/h} \implies$ Remaining $A(t) = 30 - 6t$.
• Burn rate of $B$: $\frac{30}{6} = 5\text{ cm/h} \implies$ Remaining $B(t) = 30 - 5t$.
• Condition: $30 - 5t = 2(30 - 6t) \implies 30 - 5t = 60 - 12t \implies 7t = 30 \implies t = \frac{30}{7}$ hours.
πΏ Yul's Pro-Tip:
Notice the difference: The burned length is a pure direct variation ($y = kt$), but the remaining length is an affine relation ($y = L_0 - kt$). Identifying which quantity is directly proportional is essential.
Notice the difference: The burned length is a pure direct variation ($y = kt$), but the remaining length is an affine relation ($y = L_0 - kt$). Identifying which quantity is directly proportional is essential.
TYPE 12
Real-World Inverse Variation Modeling (Gear Ratios & Levers)
[Problem 12-1 | Standard]
Two meshed gears $A$ and $B$ rotate together. Gear $A$ has $36$ teeth and rotates $5$ times per minute. If Gear $B$ has $x$ teeth and rotates $y$ times per minute, write the relationship between $x$ and $y$, and find the number of rotations per minute if Gear $B$ has $15$ teeth.
π‘ View Solution & Answer
Answer: $y = \frac{180}{x}$, and $12$ rotations/min
• Meshed teeth contact count is constant: $\text{Teeth} \times \text{Rotations} = 36 \times 5 = 180$.
• Formula: $xy = 180 \implies y = \frac{180}{x}$.
• For $x = 15$: $y = \frac{180}{15} = 12$ rotations/min.
• Meshed teeth contact count is constant: $\text{Teeth} \times \text{Rotations} = 36 \times 5 = 180$.
• Formula: $xy = 180 \implies y = \frac{180}{x}$.
• For $x = 15$: $y = \frac{180}{15} = 12$ rotations/min.
[Problem 12-2 | Advanced]
Three gears $A, B$, and $C$ are interlocked in sequence. Gear $A$ has $20$ teeth, Gear $B$ has $40$ teeth, and Gear $C$ has $x$ teeth. When Gear $A$ rotates $18$ times, the total number of rotations made by Gears $B$ and $C$ combined is $15$. Find the number of teeth on Gear $C$.
π‘ View Solution & Answer
Answer: $60$ teeth
• Total teeth engaged by Gear $A$: $20 \times 18 = 360$.
• Because all gears mesh consecutively, every gear engages exactly $360$ tooth contacts.
• Rotations of Gear $B$: $\frac{360}{40} = 9$ rotations.
• Rotations of Gear $C$: Combined is $15 \implies 15 - 9 = 6$ rotations.
• Number of teeth on Gear $C$: $\frac{360}{6} = 60$ teeth.
• Total teeth engaged by Gear $A$: $20 \times 18 = 360$.
• Because all gears mesh consecutively, every gear engages exactly $360$ tooth contacts.
• Rotations of Gear $B$: $\frac{360}{40} = 9$ rotations.
• Rotations of Gear $C$: Combined is $15 \implies 15 - 9 = 6$ rotations.
• Number of teeth on Gear $C$: $\frac{360}{6} = 60$ teeth.
πΏ Yul's Pro-Tip:
Gear trains and lever balances are canonical real-world examples of Inverse Variation. Always identify the conserved invariant: for gears, it is the total number of meshed tooth engagements ($N_1 R_1 = N_2 R_2$); for levers, it is the torque ($\text{Force} \times \text{Distance}$).
Gear trains and lever balances are canonical real-world examples of Inverse Variation. Always identify the conserved invariant: for gears, it is the total number of meshed tooth engagements ($N_1 R_1 = N_2 R_2$); for levers, it is the torque ($\text{Force} \times \text{Distance}$).
πΏ
Yul's Insight: Proportion Is the Universal Grammar of Algebra
Direct and inverse variations are not merely introductory middle school formulas; they are the very foundation of functional thinking and high school calculus.
Whenever you see $\frac{y}{x} = k$, recognize the linearity of constant velocity, uniform rates, and scaling factors. Whenever you encounter $xy = k$, visualize the hyperbola balancing trade-offs, Boyle's gas laws, and geometric conservation of area.
Do not memorize equations mechanically. When you grasp the geometric invariance behind each curve, mathematics shifts from routine memorization to an elegant, intuitive exploration.
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