[High School Math] Coordinate Geometry: 15 Essential Practice Problems on Section Formulas (Internal & External Division)


 Mechanical formula memorization alone is never enough to solve advanced geometric problems. This practice set is structured across 4 progressive difficulty tiers—from fundamental coordinate calculations to advanced reasoning and optimization.

πŸ“Š Tier Breakdown:

  • STEP 1. Core Foundations (Q1~Q3): Definitions on coordinate axes and quadrant restrictions
  • STEP 2. Magnitude Comparison & Intuition (Q4~Q5): Weighted averages and inequality deductions
  • STEP 3. Geometric Applications (Q6~Q10): Angle bisector theorem, parallelograms, centroids, and collinear extensions
  • STEP 4. Advanced Reasoning & Real-World Models (Q11~Q15): Locus equations, Circle of Apollonius, and area optimization

※ Solve each problem independently before clicking the answer button to review the step-by-step solution!

STEP 1. Core Foundations (Q1~Q3)

[Problem 01] Given two points $A(-3)$ and $B(7)$ on a number line, let $P$ be the point that internally divides segment $AB$ in the ratio $3:2$, and $Q$ be the point that externally divides segment $AB$ in the ratio $3:2$. Find the coordinates of the midpoint $M$ of segment $PQ$.

πŸ‘‰ View Answer & Solution

Answer: $15$

Solution:

  • Internal division point $P$: $\frac{3(7) + 2(-3)}{3 + 2} = \frac{21 - 6}{5} = 3$
  • External division point $Q$: $\frac{3(7) - 2(-3)}{3 - 2} = \frac{21 + 6}{1} = 27$
  • Midpoint $M$ of $PQ$: $\frac{3 + 27}{2} = 15$

[Problem 02] For two points $A(2, -1)$ and $B(-3, 4)$ on a Cartesian plane, the point dividing segment $AB$ externally in the ratio $1:k$ ($k > 0, k \neq 1$) lies on the $y$-axis. Find the value of constant $k$ and the $y$-coordinate of this external division point.

πŸ‘‰ View Answer & Solution

Answer: $k = \frac{3}{2}$, $y\text{-coordinate} = -11$

Solution:

  • Since the point lies on the $y$-axis, its $x$-coordinate is $0$:
    $x = \frac{1(-3) - k(2)}{1 - k} = 0 \implies -3 - 2k = 0 \implies k = \frac{3}{2}$
  • $y$-coordinate of the external point:
    $\frac{1(4) - \frac{3}{2}(-1)}{1 - \frac{3}{2}} = \frac{4 + \frac{3}{2}}{-\frac{1}{2}} = -11$

[Problem 03] Given two points $A(-2, 5)$ and $B(4, 1)$, find the range of real number $t$ such that point $P$, which internally divides segment $AB$ in the ratio $t : (1-t)$ ($0 < t < 1$), lies in the First Quadrant.

πŸ‘‰ View Answer & Solution

Answer: $\frac{1}{3} < t < 1$

Solution:

  • $x$-coordinate of $P$: $4t - 2(1-t) = 6t - 2 > 0 \implies t > \frac{1}{3}$
  • $y$-coordinate of $P$: $1(t) + 5(1-t) = 5 - 4t > 0 \implies t < \frac{5}{4}$
  • Intersection with $0 < t < 1$: $\frac{1}{3} < t < 1$

STEP 2. Magnitude Comparison & Intuition (Q4~Q5)

[Problem 04] For two distinct points $A(a)$ and $B(b)$ on a number line where $a < b$, the coordinates of three points $P, Q, R$ are given as follows. Which of the following correctly describes their relative order?

$$P = \frac{3a + 2b}{5}, \quad Q = \frac{2a + 3b}{5}, \quad R = \frac{a + 4b}{5}$$

① $P < Q < R$    ② $R < Q < P$    ③ $Q < P < R$    ④ $P < R < Q$    ⑤ $R < P < Q$

πŸ‘‰ View Answer & Solution

Answer: ①

Solution:

  • $P$ divides segment $AB$ internally in the ratio $2:3$ (closer to $a$).
  • $Q$ divides segment $AB$ internally in the ratio $3:2$ (closer to $b$).
  • $R$ divides segment $AB$ internally in the ratio $4:1$ (closest to $b$).
  • Since $a < b$, we have $a < P < Q < R < b$. Therefore, ① is correct.

[Problem 05] For two positive numbers $a, b$ ($0 < a < b$), three numbers $A, B, C$ are given below. Compare the magnitudes of $A, B, C$.

$$A = \frac{a+b}{2}, \quad B = \frac{a^2+b^2}{a+b}, \quad C = \frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}$$

πŸ‘‰ View Answer & Solution

Answer: $C < A < B$

Solution:

  • $A$: $1:1$ internal division point of $a$ and $b$ (Arithmetic Mean).
  • $B = \frac{a \cdot a + b \cdot b}{a+b}$: Internal division point of $a, b$ in the ratio $b:a$. Because $b > a$, the weight on $b$ is greater, making $B > A$.
  • $C = \sqrt{ab}$: Geometric Mean. Since $a \neq b$, by AM-GM inequality, $C < A$.
  • Therefore, $C < A < B$.

STEP 3. Geometric Applications (Q6~Q10)

[Problem 06] In triangle $ABC$ with vertices $A(1, 5)$, $B(-2, 1)$, and $C(4, 1)$, the bisector of $\angle A$ intersects side $BC$ at point $D(p, q)$. Find the value of $p + q$.

πŸ‘‰ View Answer & Solution

Answer: $2$

Solution:

  • $\overline{AB} = \sqrt{(-3)^2 + (-4)^2} = 5$, $\overline{AC} = \sqrt{3^2 + (-4)^2} = 5$
  • By the Angle Bisector Theorem, point $D$ internally divides segment $BC$ in the ratio $\overline{AB} : \overline{AC} = 5:5 = 1:1$ (Midpoint).
  • $D = \left(\frac{-2+4}{2}, \frac{1+1}{2}\right) = (1, 1) \implies p + q = 1 + 1 = 2$

[Problem 07] Three vertices of parallelogram $ABCD$ are $A(1, 3)$, $B(-2, -1)$, and $C(4, 1)$. Find the coordinates of the fourth vertex $D$.

πŸ‘‰ View Answer & Solution

Answer: $(7, 5)$

Solution:

  • In a parallelogram, diagonals $AC$ and $BD$ share the same midpoint.
  • Midpoint of $AC$: $\left(\frac{1+4}{2}, \frac{3+1}{2}\right) = \left(\frac{5}{2}, 2\right)$
  • Letting $D(x, y)$: $\frac{-2+x}{2} = \frac{5}{2} \implies x = 7$ and $\frac{-1+y}{2} = 2 \implies y = 5$. Thus, $D(7, 5)$.

[Problem 08] Let $D, E, F$ be points internally dividing sides $AB, BC, CA$ of triangle $ABC$ in the ratio $2:1$, respectively. If the centroid of triangle $DEF$ is $(3, -2)$, find the sum of coordinates of triangle $ABC$, $(x_1+x_2+x_3, y_1+y_2+y_3)$.

πŸ‘‰ View Answer & Solution

Answer: $(9, -6)$

Solution:

  • A triangle formed by dividing each side in the same ratio shares its centroid with the original triangle $ABC$.
  • $G(3, -2) = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
  • Therefore, $(\sum x, \sum y) = 3 \times (3, -2) = (9, -6)$.

[Problem 09] Point $C$ lies on the extended line segment $AB$ connecting $A(1, 2)$ and $B(5, 8)$. If $2\overline{AB} = 3\overline{BC}$ and the $x$-coordinate of $C$ is greater than that of $B$, find the coordinates of point $C$.

πŸ‘‰ View Answer & Solution

Answer: $\left(\frac{23}{3}, 12\right)$

Solution:

  • $\overline{AB} : \overline{BC} = 3 : 2$. Since $C$ lies beyond point $B$, point $C$ externally divides segment $AB$ in the ratio $(3+2) : 2 = 5 : 2$.
  • $C = \left(\frac{5(5)-2(1)}{5-2}, \frac{5(8)-2(2)}{5-2}\right) = \left(\frac{23}{3}, \frac{36}{3}\right) = \left(\frac{23}{3}, 12\right)$

[Problem 10] Given an equilateral triangle $OAB$ with vertices $O(0, 0)$, $A(6, 0)$, and $B(3, 3\sqrt{3})$. Let $P$ be the point dividing segment $OA$ internally in the ratio $2:1$, and $Q$ be the point dividing segment $AB$ internally in the ratio $1:2$. Find the length of segment $PQ$.

πŸ‘‰ View Answer & Solution

Answer: $2$

Solution:

  • Point $P$: Internally divides $OA$ in ratio $2:1 \implies P(4, 0)$
  • Point $Q$: Internally divides $AB$ in ratio $1:2 \implies Q\left(\frac{1(3)+2(6)}{3}, \frac{1(3\sqrt{3})+2(0)}{3}\right) = (5, \sqrt{3})$
  • $\overline{PQ} = \sqrt{(5-4)^2 + (\sqrt{3}-0)^2} = \sqrt{1+3} = 2$

STEP 4. Advanced Reasoning & Real-World Models (Q11~Q15)

[Problem 11] For two points $A(0, 4)$ and $B(6, -2)$, point $P$ divides segment $AB$ internally in the ratio $t:(1-t)$ ($0 \le t \le 1$). A line perpendicular to segment $AB$ passing through $P$ intersects the $y$-axis at $(0, k)$. Find the sum of the maximum and minimum values of $k$.

πŸ‘‰ View Answer & Solution

Answer: $-4$

Solution:

  • Slope of $AB$: $\frac{-2-4}{6-0} = -1 \implies$ Perpendicular line slope: $1$
  • Coordinates of $P$: $(6t, 4-6t)$
  • Equation of line: $y - (4-6t) = 1(x - 6t) \implies y = x + 4 - 12t$
  • $y$-intercept $k = 4 - 12t$ ($0 \le t \le 1$): Maximum is $4$ ($t=0$), Minimum is $-8$ ($t=1$)
  • Sum: $4 + (-8) = -4$

[Problem 12] In triangle $ABC$, point $D$ on side $BC$ divides segment $BC$ in the ratio $4:3$, and point $P$ on segment $AD$ satisfies $\overline{AP} : \overline{PD} = 7:2$. Find the ratio of the area of $\triangle PBC$ to that of $\triangle ABC$ in simplest integer ratio.

πŸ‘‰ View Answer & Solution

Answer: $2 : 9$

Solution:

  • Since both triangles share the base $BC$, the ratio of their areas equals the ratio of their heights.
  • Along segment $AD$, the height of $\triangle PBC$ is proportional to $\frac{\overline{PD}}{\overline{AD}}$ relative to $\triangle ABC$.
  • $\triangle PBC : \triangle ABC = \overline{PD} : \overline{AD} = 2 : (7+2) = 2 : 9$

[Problem 13] Two warehouses of a logistics company are located at $A(20, 10)$ and $B(80, 70)$. The daily cargo volumes are $300$ tons for $A$ and $600$ tons for $B$. If a transfer hub $H$ is to be established along the line connecting the warehouses (segment $AB$) to minimize the weighted squared distance sum $f(H) = 300\overline{AH}^2 + 600\overline{BH}^2$, find the coordinates of hub $H$.

πŸ‘‰ View Answer & Solution

Answer: $(60, 50)$

Solution:

  • The point minimizing the weighted sum $m\overline{AH}^2 + n\overline{BH}^2$ internally divides segment $AB$ in the ratio $n:m = 600:300 = 2:1$ (Weighted Average).
  • $H = \left(\frac{2(80)+1(20)}{3}, \frac{2(70)+1(10)}{3}\right) = (60, 50)$

[Problem 14] For two points $A(-2, 0)$ and $B(4, 0)$, point $P$ satisfies $\overline{PA} : \overline{PB} = 1:2$. Find the maximum distance between any point on the locus of $P$ and the point $C(0, 6)$.

πŸ‘‰ View Answer & Solution

Answer: $2\sqrt{13} + 4$

Solution:

  • Circle of Apollonius: Has diameter endpoints at the $1:2$ internal point $(0, 0)$ and $1:2$ external point $(-8, 0)$.
  • Center: $(-4, 0)$, Radius: $R = 4$
  • Distance from $C(0, 6)$ to Center $(-4, 0)$: $d = \sqrt{(-4)^2 + 6^2} = \sqrt{52} = 2\sqrt{13}$
  • Maximum distance $= d + R = 2\sqrt{13} + 4$

[Problem 15] Given points $A(1, 2)$ and $B(5, 5)$, point $P$ internally divides segment $AB$ in the ratio $t:(1-t)$ ($0 < t < 1$). For point $Q(0, k)$ on the $y$-axis, the product of the two possible values of $k$ that make the area of triangle $OPQ$ equal to $3$ is $-4$. Find the value of real number $t$. (Where $O$ is the origin.)

πŸ‘‰ View Answer & Solution

Answer: $\frac{1}{2}$

Solution:

  • $P(4t+1, 3t+2)$. With base $|k|$ and height $4t+1 > 0$:
    $\text{Area} = \frac{1}{2}|k|(4t+1) = 3 \implies |k| = \frac{6}{4t+1} \implies k = \pm \frac{6}{4t+1}$
  • Product of the two $k$ values: $-\left(\frac{6}{4t+1}\right)^2 = -4 \implies \frac{6}{4t+1} = 2$
  • $4t + 1 = 3 \implies 4t = 2 \implies t = \frac{1}{2}$

πŸ“ [Self-Test] Twin Practice Worksheet (15 Variant Problems)

Now that you've reviewed the core solutions, test your mastery with these 15 twin variant problems featuring the exact same structures and difficulty levels with modified numerical values!

[Variant] STEP 1. Core Foundations (Q1~Q3)

[Variant 01] Given two points $A(-5)$ and $B(5)$ on a number line, let $P$ be the point that internally divides segment $AB$ in the ratio $3:1$, and $Q$ be the point that externally divides segment $AB$ in the ratio $3:1$. Find the coordinates of the midpoint $M$ of segment $PQ$.

πŸ‘‰ View Answer & Solution

Answer: $\frac{25}{4} \ (6.25)$

Solution:

  • Internal division point $P$: $\frac{3(5) + 1(-5)}{3 + 1} = \frac{10}{4} = \frac{5}{2}$
  • External division point $Q$: $\frac{3(5) - 1(-5)}{3 - 1} = \frac{20}{2} = 10$
  • Midpoint $M$: $\frac{\frac{5}{2} + 10}{2} = \frac{25}{4}$

[Variant 02] For two points $A(3, -2)$ and $B(-2, 5)$ on a Cartesian plane, the point dividing segment $AB$ externally in the ratio $2:k$ ($k > 0, k \neq 2$) lies on the $y$-axis. Find the value of constant $k$ and the $y$-coordinate of this external division point.

πŸ‘‰ View Answer & Solution

Answer: $k = 3$, $y\text{-coordinate} = -9$

Solution:

  • Since it lies on the $y$-axis, the $x$-coordinate is $0$:
    $\frac{2(-2) - k(3)}{2 - k} = 0 \implies -4 - 3k = 0 \implies k = 3$
  • $y$-coordinate of external point:
    $\frac{2(5) - 3(-2)}{2 - 3} = \frac{10 + 6}{-1} = -9$

[Variant 03] Given two points $A(-3, 7)$ and $B(5, 2)$, find the range of real number $t$ such that point $P$, which internally divides segment $AB$ in the ratio $t : (1-t)$ ($0 < t < 1$), lies in the First Quadrant.

πŸ‘‰ View Answer & Solution

Answer: $\frac{3}{8} < t < 1$

Solution:

  • $x$-coordinate of $P$: $5t - 3(1-t) = 8t - 3 > 0 \implies t > \frac{3}{8}$
  • $y$-coordinate of $P$: $2t + 7(1-t) = 7 - 5t > 0 \implies t < \frac{7}{5}$
  • Intersection with $0 < t < 1$: $\frac{3}{8} < t < 1$

[Variant] STEP 2. Magnitude Comparison & Intuition (Q4~Q5)

[Variant 04] For two distinct points $A(a)$ and $B(b)$ on a number line where $a < b$, the coordinates of three points $P, Q, R$ are given as follows. Which of the following correctly describes their relative order?

$$P = \frac{5a + 2b}{7}, \quad Q = \frac{3a + 4b}{7}, \quad R = \frac{a + 6b}{7}$$

① $P < Q < R$    ② $R < Q < P$    ③ $Q < P < R$    ④ $P < R < Q$    ⑤ $R < P < Q$

πŸ‘‰ View Answer & Solution

Answer: ①

Solution:

  • $P, Q, R$ divide segment $AB$ internally in the ratios $2:5$, $4:3$, and $6:1$ respectively.
  • Since $a < b$, greater weight on $b$ yields a larger value. Thus, $a < P < Q < R < b$.

[Variant 05] For two positive numbers $a, b$ ($0 < a < b$), three numbers $X, Y, Z$ are given below. Compare the magnitudes of $X, Y, Z$.

$$X = \frac{a+2b}{3}, \quad Y = \frac{a^2+2b^2}{a+2b}, \quad Z = \sqrt{ab}$$

πŸ‘‰ View Answer & Solution

Answer: $Z < X < Y$

Solution:

  • $X$: $2:1$ weighted average of $a$ and $b$.
  • $Y$: Weighted heavier toward $b$, making $Y > X$.
  • $Z$: Geometric Mean, which is smaller than the arithmetic-weighted averages. Thus, $Z < X < Y$.

[Variant] STEP 3. Geometric Applications (Q6~Q10)

[Variant 06] In triangle $ABC$ with vertices $A(2, 6)$, $B(-4, -2)$, and $C(5, 2)$, the bisector of $\angle A$ intersects side $BC$ at point $D(p, q)$. Find the value of $p + q$.

πŸ‘‰ View Answer & Solution

Answer: $\frac{8}{3}$

Solution:

  • $\overline{AB} = \sqrt{(-6)^2 + (-8)^2} = 10$, $\overline{AC} = \sqrt{3^2 + (-4)^2} = 5$
  • Ratio is $10:5 = 2:1$. Point $D$ internally divides $BC$ in ratio $2:1$.
  • $D = \left(\frac{2(5) + 1(-4)}{3}, \frac{2(2) + 1(-2)}{3}\right) = \left(2, \frac{2}{3}\right) \implies p + q = 2 + \frac{2}{3} = \frac{8}{3}$

[Variant 07] Three vertices of parallelogram $ABCD$ are $A(2, 4)$, $B(-3, 0)$, and $C(5, 2)$. Find the coordinates of the fourth vertex $D$.

πŸ‘‰ View Answer & Solution

Answer: $(10, 6)$

Solution:

  • Midpoint of diagonal $AC$: $\left(\frac{2+5}{2}, \frac{4+2}{2}\right) = \left(\frac{7}{2}, 3\right)$
  • Equating with midpoint of $BD$: $\frac{-3+x}{2} = \frac{7}{2} \implies x = 10$, and $\frac{0+y}{2} = 3 \implies y = 6$. Thus, $D(10, 6)$.

[Variant 08] Let $D, E, F$ be points internally dividing sides $AB, BC, CA$ of triangle $ABC$ in the ratio $3:1$, respectively. If the centroid of triangle $DEF$ is $(4, -1)$, find the sum of coordinates of triangle $ABC$, $(x_1+x_2+x_3, y_1+y_2+y_3)$.

πŸ‘‰ View Answer & Solution

Answer: $(12, -3)$

Solution:

  • Triangles formed by dividing each side in the same ratio share their centroid with the original triangle.
  • $(\sum x, \sum y) = 3 \times (4, -1) = (12, -3)$.

[Variant 09] Point $C$ lies on the extended line segment $AB$ connecting $A(2, 1)$ and $B(6, 9)$. If $3\overline{AB} = 2\overline{BC}$ and the $x$-coordinate of $C$ is greater than that of $B$, find the coordinates of point $C$.

πŸ‘‰ View Answer & Solution

Answer: $(12, 21)$

Solution:

  • $\overline{AB} : \overline{BC} = 2 : 3$. Point $C$ externally divides segment $AB$ in the ratio $(2+3) : 3 = 5 : 3$.
  • $C = \left(\frac{5(6)-3(2)}{5-3}, \frac{5(9)-3(1)}{5-3}\right) = \left(\frac{24}{2}, \frac{42}{2}\right) = (12, 21)$

[Variant 10] Given an equilateral triangle $OAB$ with vertices $O(0, 0)$, $A(8, 0)$, and $B(4, 4\sqrt{3})$. Let $P$ be the point dividing segment $OA$ internally in the ratio $3:1$, and $Q$ be the point dividing segment $AB$ internally in the ratio $1:3$. Find the length of segment $PQ$.

πŸ‘‰ View Answer & Solution

Answer: $2$

Solution:

  • $P$: Internally divides $OA$ in $3:1 \implies P(6, 0)$
  • $Q$: Internally divides $AB$ in $1:3 \implies Q\left(\frac{1(4)+3(8)}{4}, \frac{1(4\sqrt{3})+3(0)}{4}\right) = (7, \sqrt{3})$
  • $\overline{PQ} = \sqrt{(7-6)^2 + (\sqrt{3}-0)^2} = \sqrt{1+3} = 2$

[Variant] STEP 4. Advanced Reasoning & Real-World Models (Q11~Q15)

[Variant 11] For two points $A(0, 6)$ and $B(8, -2)$, point $P$ divides segment $AB$ internally in the ratio $t:(1-t)$ ($0 \le t \le 1$). A line perpendicular to segment $AB$ passing through $P$ intersects the $y$-axis at $(0, k)$. Find the sum of the maximum and minimum values of $k$.

πŸ‘‰ View Answer & Solution

Answer: $-4$

Solution:

  • Slope of $AB$: $\frac{-2-6}{8-0} = -1 \implies$ Perpendicular slope: $1$
  • Point $P$: $(8t, 6-8t)$. Line equation: $y = x + 6 - 16t$
  • $y$-intercept $k = 6 - 16t$ ($0 \le t \le 1$): Max is $6$ ($t=0$), Min is $-10$ ($t=1$).
  • Sum: $6 + (-10) = -4$

[Variant 12] In triangle $ABC$, point $D$ on side $BC$ divides segment $BC$ in the ratio $3:2$, and point $P$ on segment $AD$ satisfies $\overline{AP} : \overline{PD} = 5:3$. Find the ratio of the area of $\triangle PBC$ to that of $\triangle ABC$ in simplest integer ratio.

πŸ‘‰ View Answer & Solution

Answer: $3 : 8$

Solution:

  • Area ratio equals height ratio along segment $AD$: $\overline{PD} : \overline{AD} = 3 : (5+3) = 3 : 8$.

[Variant 13] Two warehouses are located at $A(10, 20)$ and $B(70, 80)$ with daily cargo volumes of $200$ tons for $A$ and $400$ tons for $B$. If a transfer hub $H$ is to be established along segment $AB$ to minimize the weighted squared distance sum $f(H) = 200\overline{AH}^2 + 400\overline{BH}^2$, find the coordinates of hub $H$.

πŸ‘‰ View Answer & Solution

Answer: $(50, 60)$

Solution:

  • Minimizing point internally divides $AB$ in the ratio $400:200 = 2:1$.
  • $H = \left(\frac{2(70)+1(10)}{3}, \frac{2(80)+1(20)}{3}\right) = (50, 60)$

[Variant 14] For two points $A(-3, 0)$ and $B(3, 0)$, point $P$ satisfies $\overline{PA} : \overline{PB} = 2:1$. Find the maximum distance between any point on the locus of $P$ and the point $C(0, 8)$.

πŸ‘‰ View Answer & Solution

Answer: $\sqrt{89} + 4$

Solution:

  • Circle of Apollonius: Diameter endpoints at $2:1$ internal point $(1, 0)$ and $2:1$ external point $(9, 0)$.
  • Center: $(5, 0)$, Radius: $R = 4$
  • Distance from $C(0, 8)$ to Center $(5, 0)$: $d = \sqrt{5^2 + 8^2} = \sqrt{89}$
  • Maximum distance $= d + R = \sqrt{89} + 4$

[Variant 15] Given points $A(2, 1)$ and $B(8, 4)$, point $P$ internally divides segment $AB$ in the ratio $t:(1-t)$ ($0 < t < 1$). For point $Q(0, k)$ on the $y$-axis, the product of the two possible values of $k$ that make the area of triangle $OPQ$ equal to $6$ is $-9$. Find the value of real number $t$. (Where $O$ is the origin.)

πŸ‘‰ View Answer & Solution

Answer: $\frac{1}{3}$

Solution:

  • $P(6t+2, 3t+1)$. Area equation yields $k = \pm \frac{6}{3t+1}$
  • Product of $k$ values: $-\left(\frac{6}{3t+1}\right)^2 = -9 \implies \frac{6}{3t+1} = 3$
  • $3t + 1 = 2 \implies 3t = 1 \implies t = \frac{1}{3}$

✍️ Final Insight from Yul

Mathematics is not about mechanically memorizing formulas to plug in numbers. True mastery begins with visualizing the geometric proportional relationships behind internal and external division points. Once the core principle is understood, computational methods follow naturally.

The concepts of segment ratios and division points do not end on the coordinate plane—they serve as powerful foundations extending directly into advanced vector locus problems and 3D spatial geometry. Build from the fundamentals, and develop mathematical intuition that lasts.

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