[Coordinate Geometry: Practice] Distance from Point to Line: 15 Master Problems & Diagnostic Variations
When encountering challenging coordinate geometry problems, the common pitfall is to rush into brute-force algebra: parameterizing points on lines, expanding cumbersome distance radicals, or attempting calculus derivatives where pure geometry suffices. In the process, minutes evaporate, sign errors multiply, and straightforward geometric relations turn into algebraic quagmires. Others hurriedly apply formulas without structural understanding, missing a single sign and compromising an entire multi-step solution.
The distance formula from a point to a line is far more than an introductory algebra exercise. It serves as the universal orthogonal ruler of advanced mathematics—the foundational bridge connecting circle tangency ($d = r$), curve optimization in AP Calculus, and vector projections in linear algebra.
In this master practice guide, we move past brute-force arithmetic. Utilizing 3-second parallel strip evaluations, acute/obtuse angle bisector discrimination, the origin-shifted Shoelace shortcut, and Cauchy-Schwarz geometric bounds, we equip you with the intuition needed to solve 15 high-level challenge problems and diagnostic variations with speed, precision, and geometric elegance. Attempt each problem independently first, then expand the toggle to audit your step-by-step reasoning.
1. Polygons Between Parallel Lines: Never pick arbitrary test points. Extract the orthogonal height immediately via $d = \dfrac{|c - c'|}{\sqrt{a^2 + b^2}}$ to compute edge lengths in 3 seconds.
2. Angle Bisectors: Formulate the equidistant locus $\dfrac{|a_1x+b_1y+c_1|}{\sqrt{a_1^2+b_1^2}} = \dfrac{|a_2x+b_2y+c_2|}{\sqrt{a_2^2+b_2^2}}$, then discriminate between acute and obtuse bisectors using the normal dot product sign ($a_1 a_2 + b_1 b_2$).
3. Circle-Line Extremal Distances: Strictly avoid quadratic discriminants ($D=0$). Use the geometric shortcut "Center distance $\pm$ radius ($d \pm r$)" to solve tangency and range bounds mentally.
Theme 1. Parallel Line Separations & Regular Polygon Geometry (Q01 ~ Q04)
[Problem 01] Equilateral Triangle Inscribed Between Parallel Lines
Two parallel lines are given by $l_1 : 3x - 4y + 7 = 0$ and $l_2 : 3x - 4y - 13 = 0$. An equilateral triangle has one side lying entirely along $l_1$, while its opposite vertex lies on $l_2$. Calculate the area of this equilateral triangle.
π View Solution & Key Steps
Answer: $\dfrac{16\sqrt{3}}{3}$
[Step-by-Step Solution]
1) The orthogonal separation $h$ between $l_1$ and $l_2$ is the height of the triangle:
$h = \dfrac{|7 - (-13)|}{\sqrt{3^2 + (-4)^2}} = \dfrac{20}{\sqrt{25}} = \dfrac{20}{5} = 4$
2) Let $a$ be the side length. By the altitude formula $h = \dfrac{\sqrt{3}}{2}a$:
$4 = \dfrac{\sqrt{3}}{2}a \implies a = \dfrac{8}{\sqrt{3}}$
3) Equilateral triangle area $S$:
$S = \dfrac{\sqrt{3}}{4}a^2 = \dfrac{\sqrt{3}}{4} \times \dfrac{64}{3} = \mathbf{\dfrac{16\sqrt{3}}{3}}$.
Inscribed polygon problems bounded by parallel tracks collapse once you realize that the line separation $d$ represents the geometric height ($h$) of the figure. Bypass arbitrary test points and use $\dfrac{|c - c'|}{\sqrt{a^2 + b^2}}$ directly.
A regular hexagon is inscribed between parallel lines $l_1 : 2x + y - 4 = 0$ and $l_2 : 2x + y + 6 = 0$ such that two opposite edges lie along $l_1$ and $l_2$. Calculate the exact area of this regular hexagon.
Check Variation Solution
Explanation: Distance $H = \dfrac{|-4 - 6|}{\sqrt{2^2 + 1^2}} = \dfrac{10}{\sqrt{5}} = 2\sqrt{5}$. For a regular hexagon, the distance between opposite parallel edges is $H = \sqrt{3}a \implies a = \dfrac{2\sqrt{5}}{\sqrt{3}}$. Area $S = 6 \times \left(\dfrac{\sqrt{3}}{4}a^2\right) = \dfrac{3\sqrt{3}}{2} \times \dfrac{20}{3} = 10\sqrt{3}$.
[Problem 02] Parallel Line Separation with Scaled Coefficients & Square Construction
The perpendicular distance between lines $l_1 : 2x - y + 3 = 0$ and $l_2 : 4x - 2y + k = 0$ is $2\sqrt{5}$. Find the product of all possible real values of constant $k$.
π View Solution & Key Steps
Answer: $-364$
[Step-by-Step Solution]
1) Normalize linear coefficients: Multiply $l_1$ by $2$ to obtain $4x - 2y + 6 = 0$.
2) Apply parallel distance formula between $4x - 2y + 6 = 0$ and $4x - 2y + k = 0$:
$d = \dfrac{|k - 6|}{\sqrt{4^2 + (-2)^2}} = \dfrac{|k - 6|}{\sqrt{20}} = \dfrac{|k - 6|}{2\sqrt{5}}$
3) Equate to given distance $2\sqrt{5}$:
$\dfrac{|k - 6|}{2\sqrt{5}} = 2\sqrt{5} \implies |k - 6| = 20$
$k - 6 = 20 \implies k = 26$
$k - 6 = -20 \implies k = -14$
The product of all values is $26 \times (-14) = \mathbf{-364}$.
Avoid the coefficient disparity trap! Never write $|k - 3|$. You must scale the linear coefficients $(a, b)$ to match perfectly before subtracting constant terms.
A square $ABCD$ has two parallel edges lying on lines $x - 3y + 2 = 0$ and $x - 3y + k = 0$ ($k > 2$). If the area of the square is $10$, determine the value of constant $k$.
Check Variation Solution
Explanation: Edge length $d = \sqrt{10}$. Distance formula: $d = \dfrac{|k - 2|}{\sqrt{1^2 + (-3)^2}} = \dfrac{|k - 2|}{\sqrt{10}} = \sqrt{10} \implies |k - 2| = 10$. Since $k > 2$, $k - 2 = 10 \implies k = 12$.
[Problem 03] Locus of Points Equidistant from Two Parallel Lines
Find the equation of the locus of all points $P(x, y)$ that are equidistant from the parallel lines $l_1 : 2x + 3y - 5 = 0$ and $l_2 : 2x + 3y + 9 = 0$.
π View Solution & Key Steps
Answer: $2x + 3y + 2 = 0$
[Step-by-Step Solution]
[Geometric 1-Second Rule]
The line equidistant from two parallel lines is parallel to both and lies precisely halfway between them, taking the arithmetic mean of their constant terms.
Mean of constants: $\dfrac{-5 + 9}{2} = \dfrac{4}{2} = 2$.
Thus, the equation is $\mathbf{2x + 3y + 2 = 0}$.
(Algebraic verification: $\dfrac{|2x+3y-5|}{\sqrt{13}} = \dfrac{|2x+3y+9|}{\sqrt{13}} \implies 2x+3y-5 = -(2x+3y+9) \implies 4x+6y+4=0 \implies 2x+3y+2=0$.)
The midline between parallel lines is simply $\dfrac{c_1 + c_2}{2}$. For an internal distance ratio of $m : n$, use the section formula $\dfrac{n c_1 + m c_2}{m + n}$ to write the answer mentally.
Let $L$ be the line lying between $3x - 4y + 1 = 0$ and $3x - 4y - 14 = 0$ whose distance ratio to them is $1 : 2$. Find the intersection of $L$ with the line passing through $(1, 2)$ perpendicular to $L$.
Check Variation Solution
Explanation: Weighted constant: $c = \dfrac{2(1) + 1(-14)}{3} = -4 \implies L: 3x - 4y - 4 = 0$. Perpendicular line through $(1, 2)$ is $4x + 3y - 10 = 0$. Solving the $2\times 2$ system yields $(x, y) = \left(\dfrac{52}{25}, \dfrac{14}{25}\right)$.
[Problem 04] Maximum Parallel Strip Width Traversing a Rectangle
A rectangle $OABC$ has vertices $O(0, 0)$, $A(4, 0)$, $B(4, 3)$, and $C(0, 3)$. Two parallel lines with slope $m$ ($m > 0$) pass through vertices $O$ and $B$ respectively. Determine the maximum possible distance between these two lines.
π View Solution & Key Steps
Answer: $5$
[Step-by-Step Solution]
1) The lines are parameterized as $l_1 : mx - y = 0$ and $l_2 : mx - y - (4m - 3) = 0$.
2) The orthogonal distance $d$ between lines passing through fixed points $O$ and $B$ is bounded by the hypotenuse $\overline{OB}$ in the projection right triangle:
$d \le \overline{OB} = \sqrt{4^2 + 3^2} = 5$.
Equality holds when segment $OB$ is perpendicular to the parallel lines.
Thus, the maximum possible distance is $\mathbf{5}$.
The distance between two parallel lines pivoting through fixed points achieves its maximum when the segment joining the pivots becomes the normal vector. The Euclidean length of the segment is the upper bound.
Find the minimum width (perpendicular separation) of a parallel strip that completely covers $\triangle ABC$ with vertices $A(1, 4)$, $B(5, 2)$, and $C(3, 7)$.
Check Variation Solution
Explanation: The minimum strip width corresponds to the minimum altitude of $\triangle ABC$, which occurs across its longest edge $\overline{AB} = \sqrt{4^2 + (-2)^2} = 2\sqrt{5}$. Triangle area $S = 8$. Minimum altitude $h_{\text{min}} = \dfrac{2S}{\overline{AB}} = \dfrac{16}{2\sqrt{5}} = \dfrac{8\sqrt{5}}{5}$.
Theme 2. Triangle Area & The Shoelace Formula (Q05 ~ Q08)
Finding the area of a coordinate triangle by calculating side lengths and line altitudes is algebraically tedious. The fastest analytical tool is the Shoelace Formula (Gauss's Area Algorithm).
List coordinates $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ vertically, repeating the first vertex at the bottom:
↘ ↗
x₂ y₂
↘ ↗
x₃ y₃
↘ ↗
x₁ y₁
π Subtract the sum of upward products ($\nearrow$) from downward products ($\searrow$), take the absolute value, and divide by $2$:
$S = \dfrac{1}{2} \Big| (x_1 y_2 + x_2 y_3 + x_3 y_1) - (y_1 x_2 + y_2 x_3 + y_3 x_1) \Big|$
Computing 6 separate products under timed conditions invites sign errors. Translate the most complex vertex to $(0, 0)$, and the formula simplifies to a single cross-multiplication:
1. Shift $B(-1, 1)$ to $(0, 0)$ ($\to$ add $+1$ to $x$, subtract $-1$ from $y$ across all points)
2. New coordinates: $A'(3, 4)$, $B'(0, 0)$, $C'(5, -4)$
3. Outer product $-$ inner product:
$S = \dfrac{1}{2}|(3)(-4) - (4)(5)| = \dfrac{1}{2}|-12 - 20| = \dfrac{1}{2}|-32| = \mathbf{16}$ (Computed in 5 seconds!)
[Problem 05] Formal Derivation of Triangle Area via Altitude Distance
Find the area of $\triangle ABC$ with vertices $A(2, 5)$, $B(-1, 1)$, and $C(4, -3)$ using the standard curricular protocol: compute the length of base $\overline{BC}$, find the equation of line $BC$, and determine the altitude from vertex $A$.
π View Solution & Key Steps
Answer: $16$
[Step-by-Step Solution]
1) Base length $\overline{BC}$:
$\overline{BC} = \sqrt{(4 - (-1))^2 + (-3 - 1)^2} = \sqrt{25 + 16} = \sqrt{41}$
2) Line $BC$ equation:
Slope $m = \dfrac{-3 - 1}{4 - (-1)} = -\dfrac{4}{5} \implies y - 1 = -\dfrac{4}{5}(x + 1) \implies 4x + 5y - 1 = 0$
3) Altitude $h$ from $A(2, 5)$ to $4x + 5y - 1 = 0$:
$h = \dfrac{|4(2) + 5(5) - 1|}{\sqrt{4^2 + 5^2}} = \dfrac{32}{\sqrt{41}}$
4) Area $S$:
$S = \dfrac{1}{2} \times \overline{BC} \times h = \dfrac{1}{2} \times \sqrt{41} \times \dfrac{32}{\sqrt{41}} = \mathbf{16}$.
On free-response exams, writing the Shoelace formula without derivation may forfeit partial credit. Follow the 4-step rubric: Base length $\to$ Line in general form $\to$ Point-to-line altitude $h$ $\to$ $\dfrac{1}{2} \times \text{base} \times h$.
Quadrilateral $ABCD$ has vertices $A(0, 2)$, $B(4, 4)$, $C(5, 1)$, and $D(1, -1)$ with $AB \parallel CD$. Compute the orthogonal altitude between the parallel bases and determine the exact area of trapezoid $ABCD$.
Check Variation Solution
Explanation: Line $AB: x - 2y + 4 = 0$, Line $CD: x - 2y - 3 = 0$. Height $h = \dfrac{|4 - (-3)|}{\sqrt{5}} = \dfrac{7}{\sqrt{5}}$. Base lengths are $AB = 2\sqrt{5}$ and $CD = \sqrt{20} = 2\sqrt{5}$ (parallelogram case: Area $= 15$).
[Problem 06] Triangle Area Minimization with a Moving Vertex Along a Line
For fixed points $A(1, 2)$ and $B(5, 4)$, and point $P$ constrained to line $l : x - 2y + 8 = 0$, determine the minimum possible area of $\triangle PAB$.
π View Solution & Key Steps
Answer: $5$
[Step-by-Step Solution]
1) Slope of line $AB$: $m = \dfrac{4 - 2}{5 - 1} = \dfrac{1}{2}$.
Line $l : x - 2y + 8 = 0$ also has slope $\dfrac{1}{2}$. Hence, line $AB$ and line $l$ are parallel.
2) Base length: $\overline{AB} = \sqrt{(5 - 1)^2 + (4 - 2)^2} = \sqrt{20} = 2\sqrt{5}$.
3) Line $AB$ equation: $x - 2y + 3 = 0$.
4) Constant orthogonal altitude $h$ between $x - 2y + 3 = 0$ and $x - 2y + 8 = 0$:
$h = \dfrac{|8 - 3|}{\sqrt{1^2 + (-2)^2}} = \dfrac{5}{\sqrt{5}} = \sqrt{5}$.
By Cavalieri's principle, the area is invariant for any point $P$ on $l$:
$S = \dfrac{1}{2} \times 2\sqrt{5} \times \sqrt{5} = \mathbf{5}$.
Whenever a vertex moves along a line, first compare the slope of the line with the fixed base. If slopes match, the area is constant across the entire line; if slopes differ, the area reaches $0$ at the intersection.
Given $A(2, 1)$ and $B(6, 3)$, find the sum of the $x$-coordinates of all points $P$ on line $l : 2x - y + 1 = 0$ such that the area of $\triangle PAB$ equals $10$.
Check Variation Solution
Explanation: Base $\overline{AB} = 2\sqrt{5}$, required altitude $h = 2\sqrt{5}$. Line $AB: x - 2y = 0$. For $P(t, 2t+1)$, distance to $x - 2y = 0$ is $\dfrac{|t - 2(2t+1)|}{\sqrt{5}} = 2\sqrt{5} \implies |3t + 2| = 10$. Solutions are $t = \dfrac{8}{3}$ and $t = -4$. Sum $= \dfrac{8}{3} - 4 = -\dfrac{4}{3}$.
[Problem 07] Derivation of the Determinant Area Formula via Origin Translation
Prove that the area of $\triangle ABC$ with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ equals $S = \dfrac{1}{2}|(x_1 - x_3)(y_2 - y_3) - (x_2 - x_3)(y_1 - y_3)|$ by translating vertex $C$ to the origin and applying the point-to-line distance formula.
π View Solution & Key Steps
[Analytical Proof]
1) Translate $C$ to origin $(0, 0)$: $A'(X_1, Y_1) = (x_1 - x_3, y_1 - y_3)$, $B'(X_2, Y_2) = (x_2 - x_3, y_2 - y_3)$, $C'(0, 0)$.
2) Line $C'B'$ passing through origin: $Y_2 X - X_2 Y = 0$.
3) Base length: $\overline{C'B'} = \sqrt{X_2^2 + Y_2^2}$.
4) Altitude $h$ from $A'(X_1, Y_1)$ to $Y_2 X - X_2 Y = 0$:
$h = \dfrac{|Y_2 X_1 - X_2 Y_1|}{\sqrt{Y_2^2 + (-X_2)^2}} = \dfrac{|X_1 Y_2 - X_2 Y_1|}{\sqrt{X_2^2 + Y_2^2}}$.
5) Area $S = \dfrac{1}{2} \times \overline{C'B'} \times h = \mathbf{\dfrac{1}{2}|X_1 Y_2 - X_2 Y_1|}$.
Substituting original coordinates completes the proof.
The Shoelace formula is algebraically equivalent to the 2D cross product of displacement vectors $\dfrac{1}{2}|\vec{u} \times \vec{v}|$. Origin translation eliminates extraneous arithmetic terms immediately.
Calculate the area of convex quadrilateral $ABCD$ with vertices $A(1, 3)$, $B(5, 1)$, $C(6, 6)$, and $D(2, 7)$ by translating vertex $A$ to the origin and partitioning into two triangles.
Check Variation Solution
Explanation: Shift $A(1, 3)$ to $(0, 0) \implies B'(4, -2), C'(5, 3), D'(1, 4)$. Split along diagonal $A'C'$:
$\text{Area}(\triangle A'B'C') = \dfrac{1}{2}|4(3) - (-2)(5)| = 11$.
$\text{Area}(\triangle A'C'D') = \dfrac{1}{2}|5(4) - 3(1)| = \dfrac{17}{2}$. Total Area $= 11 + 8.5 \approx 20$ (via polygon matrix).
[Problem 08] Minimum Sum of Distances to Three Enclosing Boundary Lines
Point $P$ lies strictly inside the triangle bounded by lines $l_1 : y = 0$, $l_2 : 3x - 4y = 0$, and $l_3 : 3x + 4y - 12 = 0$. Characterize the conditions under which the sum of the distances from $P$ to all three boundary lines is minimized.
π View Solution & Key Steps
[Key Analytical Finding]
Inside the triangle, evaluate the signed distances:
$d_1 = y$, $d_2 = \dfrac{3x - 4y}{5}$, $d_3 = \dfrac{12 - (3x + 4y)}{5}$.
Summing these distances:
$S = d_1 + d_2 + d_3 = y + \dfrac{3x - 4y + 12 - 3x - 4y}{5} = y + \dfrac{12 - 8y}{5} = \mathbf{\dfrac{12 - 3y}{5}}$.
The horizontal coordinate $x$ cancels out completely. Thus, the sum depends solely on $y$, decreasing monotonically as $y$ increases toward the upper vertex.
Do not be deterred by three absolute value expressions. Due to geometric symmetry in boundary equations, one of the variables frequently cancels out, transforming the problem into a 1D linear programming optimization.
A point $P(x, y)$ in the first quadrant moves such that its distance to the $x$-axis and its distance to line $4x + 3y - 12 = 0$ maintain a ratio of $1 : 2$. Find the equations of the locus of point $P$.
Check Variation Solution
Explanation: Distances satisfy $2 d_1 = d_2 \implies 2y = \dfrac{|4x + 3y - 12|}{5} \implies |4x + 3y - 12| = 10y$. Resolving the $\pm$ signs produces the two ray trajectories.
Theme 3. Angle Bisectors & Acute vs. Obtuse Discrimination (Q09 ~ Q12)
[Problem 09] Identifying the Obtuse Angle Bisector Between Intersecting Lines
Two lines are given by $l_1 : 2x - y + 4 = 0$ and $l_2 : x - 2y - 1 = 0$. Determine the equation of the bisector of the obtuse angle formed by these two lines.
π View Solution & Key Steps
Answer: $x + y + 5 = 0$
[Step-by-Step Solution]
1) Equidistant locus identity ($d_1 = d_2$):
$\dfrac{|2x - y + 4|}{\sqrt{5}} = \dfrac{|x - 2y - 1|}{\sqrt{5}} \implies |2x - y + 4| = |x - 2y - 1|$
- Positive sign: $2x - y + 4 = x - 2y - 1 \implies x + y + 5 = 0$ (slope $-1$)
- Negative sign: $2x - y + 4 = -(x - 2y - 1) \implies x - y + 1 = 0$ (slope $1$)
2) Discriminate acute vs. obtuse angle:
Normal vectors are $\vec{n}_1 = (2, -1)$ and $\vec{n}_2 = (1, -2)$.
Dot product: $\vec{n}_1 \cdot \vec{n}_2 = 2(1) + (-1)(-2) = 4 > 0$.
Because the dot product is positive, the normal vectors enclose an acute angle, which means the obtuse angle bisector corresponds to the opposite-sign combination: $\mathbf{x + y + 5 = 0}$.
To isolate the obtuse bisector without sketching, check the normal vector dot product $a_1 a_2 + b_1 b_2$. If positive, the sum formulation ($+$) gives the acute bisector and the difference formulation ($-$) gives the obtuse bisector (after aligning constants).
Find the equation of the angle bisector containing the origin $(0, 0)$ for lines $3x + 4y - 5 = 0$ and $5x - 12y + 13 = 0$.
Check Variation Solution
Explanation: Normalize constant signs to positive: $-3x - 4y + 5 = 0$ and $5x - 12y + 13 = 0$. The region containing $(0, 0)$ is selected by matching the positive signs: $\dfrac{-3x - 4y + 5}{5} = \dfrac{5x - 12y + 13}{13} \implies 64x - 8y = 0 \implies 8x - y = 0$.
[Problem 10] Incenter Coordinates via Angle Bisector Equidistance
Find the coordinates of the incenter $I$ of the triangle enclosed by lines $l_1 : 4x - 3y = 0$, $l_2 : y = 0$, and $l_3 : 3x + 4y - 15 = 0$ using distance equality.
π View Solution & Key Steps
Answer: $(2, 1)$
[Step-by-Step Solution]
Let $I(a, b)$ with $b > 0$. Distance to $l_2 (y = 0)$ is $b$.
1) Distance to $l_1 (4x - 3y = 0)$:
$\dfrac{4a - 3b}{5} = b \implies 4a - 3b = 5b \implies 4a = 8b \implies a = 2b \quad \cdots ①$
2) Distance to $l_3 (3x + 4y - 15 = 0)$:
$\dfrac{15 - (3a + 4b)}{5} = b \implies 15 - 3a - 4b = 5b \implies 3a + 9b = 15 \implies a + 3b = 5 \quad \cdots ②$
Substitute $①$ into $②$: $2b + 3b = 5 \implies b = 1 \implies a = 2$.
Hence, the incenter is $\mathbf{(2, 1)}$.
The incenter $I$ is equidistant from all three lines by inradius $r$. When one boundary line aligns with an axis ($y=0$), the coordinate directly equates to $r$, collapsing the system into a 20-second linear substitution.
Find the coordinates of the excenter in the first quadrant for the triangle bounded by $x = 0$, $y = 0$, and $3x + 4y - 12 = 0$.
Check Variation Solution
Explanation: Center is $(R, R)$. Distance to $3x + 4y - 12 = 0$ is $\dfrac{|7R - 12|}{5} = R \implies 7R - 12 = 5R \implies 2R = 12 \implies R = 6$. Thus, the excenter is $(6, 6)$.
[Problem 11] Interior Angle Bisector Theorem & Section Formula
In $\triangle ABC$ with vertices $A(1, 5)$, $B(-2, 1)$, and $C(4, 1)$, the interior bisector of $\angle A$ intersects side $BC$ at point $D$. Find the coordinates of point $D$.
π View Solution & Key Steps
Answer: $(1, 1)$
[Step-by-Step Solution]
Apply the Angle Bisector Theorem ($\overline{AB} : \overline{AC} = \overline{BD} : \overline{DC}$).
1) Side lengths:
$\overline{AB} = \sqrt{(-2 - 1)^2 + (1 - 5)^2} = \sqrt{9 + 16} = 5$
$\overline{AC} = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{9 + 16} = 5$
2) Since $\overline{AB} = \overline{AC} = 5$, $\triangle ABC$ is isosceles.
Hence, $D$ is the midpoint of $BC$ ($1 : 1$ internal division):
$D = \left(\dfrac{-2 + 4}{2}, \dfrac{1 + 1}{2}\right) = \mathbf{(1, 1)}$.
When triangle vertices are known, never compute angle bisector line equations. Convert the problem into synthetic side-ratio partitioning ($\overline{AB} : \overline{AC}$) and use the midpoint or section formula.
For vertices $A(0, 3)$, $B(-3, -1)$, and $C(5, -1)$, the exterior angle bisector of $\angle A$ intersects line $BC$ at point $E$. Determine the coordinates of $E$.
Check Variation Solution
Explanation: Apply the Exterior Angle Bisector Theorem: $\overline{AB} : \overline{AC} = \overline{BE} : \overline{CE}$. $\overline{AB} = 5$, $\overline{AC} = \sqrt{25 + 16} = \sqrt{41}$ (adjusted to integer ratios such as $5:3$ for external section division).
[Problem 12] Orthogonality of Complementary Angle Bisectors
Find the product of the slopes of the two lines that bisect the angles formed at the intersection of $x + 2y - 4 = 0$ and $2x + y + 1 = 0$.
π View Solution & Key Steps
Answer: $-1$
[Step-by-Step Solution]
[1-Second Theoretical Property]
The angle bisectors of two intersecting straight lines are always mutually perpendicular ($90^\circ$).
Because they are orthogonal, the product of their slopes must be $\mathbf{-1}$.
(Verification: Solving $|x+2y-4| = |2x+y+1|$ gives $x - y + 5 = 0$ with slope $1$, and $x + y - 1 = 0$ with slope $-1$. Product: $1 \times (-1) = -1$.)
Any pair of intersecting straight lines generates two supplementary angles ($2\alpha + 2\beta = 180^\circ \implies \alpha + \beta = 90^\circ$). Consequently, their bisectors are unconditionally perpendicular ($m_1 m_2 = -1$).
Let $L_1$ and $L_2$ be the two angle bisectors between $3x - 4y + 1 = 0$ and $4x + 3y - 7 = 0$. Find the distance between the two points where $L_1$ and $L_2$ intersect the horizontal line $y = 2$.
Check Variation Solution
Explanation: Both denominators equal $5$. Line $L_1: x + 7y - 8 = 0$; setting $y=2$ yields $x = -6$. Line $L_2: 7x - y - 6 = 0$; setting $y=2$ yields $x = \dfrac{8}{7}$. Distance between intercepts $= \dfrac{8}{7} - (-6) = \dfrac{50}{7}$.
Theme 4. Competitive Exam 4-Point Killer Integrations (Q13 ~ Q15)
[Problem 13] Extremal Distances from Points on a Circle to a Line
Let $P$ be an arbitrary point lying on the circle $(x - 3)^2 + (y + 1)^2 = 4$, and consider the line $l : 3x - 4y + 7 = 0$. Let $M$ and $m$ denote the maximum and minimum distances from $P$ to line $l$, respectively. Calculate $M \times m$.
π View Solution & Key Steps
Answer: $12$
[Step-by-Step Solution]
1) Center $C(3, -1)$, radius $r = 2$.
2) Distance $d$ from center $C(3, -1)$ to line $3x - 4y + 7 = 0$:
$d = \dfrac{|3(3) - 4(-1) + 7|}{\sqrt{3^2 + (-4)^2}} = \dfrac{20}{5} = 4$.
3) Bounds: $M = d + r = 4 + 2 = 6$, $m = d - r = 4 - 2 = 2$.
4) Product: $M \times m = (d + r)(d - r) = d^2 - r^2 = 4^2 - 2^2 = 16 - 4 = \mathbf{12}$.
Do not parameterize using trigonometric identities. The product of the maximum and minimum distances from a circle to an external line is identically the power-like difference of squares $d^2 - r^2$.
Point $P$ moves along circle $x^2 + y^2 = 9$. Find the maximum area of $\triangle PAB$ where $A(-4, 0)$ and $B(0, -4)$ are fixed vertices.
Check Variation Solution
Explanation: Line $AB: x + y + 4 = 0$, base $\overline{AB} = 4\sqrt{2}$. Distance from center $(0, 0)$ to line $AB$ is $d = 2\sqrt{2}$. Radius $r = 3 \implies h_{\text{max}} = 2\sqrt{2} + 3$. Maximum area $= \dfrac{1}{2} \times 4\sqrt{2} \times (2\sqrt{2} + 3) = 8 + 6\sqrt{2}$.
[Problem 14] Maximum Distance from the Origin to a Parameterized Line Family
For parameter $k \in \mathbb{R}$, consider the line family $(2k + 1)x + (k - 1)y - 4k - 5 = 0$. Determine the maximum distance $d$ from the origin $O(0, 0)$ to this line family.
π View Solution & Key Steps
Answer: $\sqrt{13}$
[Step-by-Step Solution]
1) Factor by parameter $k$ to locate the fixed pivot point $A$:
$(x - y - 5) + k(2x + y - 4) = 0 \implies \begin{cases} x - y = 5 \\ 2x + y = 4 \end{cases} \implies A(3, -2)$.
2) Geometric projection maximization:
In the right triangle formed by origin $O$, pivot $A(3, -2)$, and the altitude foot $H$, altitude $d = \overline{OH}$ is bounded by hypotenuse $\overline{OA}$:
$d \le \overline{OA} = \sqrt{3^2 + (-2)^2} = \sqrt{13}$.
The maximum is attained when line $OA$ serves as the perpendicular normal vector to the line.
Do not differentiate the distance formula! The maximum distance from an external point to a pivoting line family is always the Euclidean distance to its fixed pivot.
Find the value of $m$ that maximizes the distance between line $y - 3 = m(x - 1)$ and point $P(4, 7)$, and state the maximum distance.
Check Variation Solution
Explanation: Pivot is $A(1, 3)$. Maximum distance is $\overline{AP} = \sqrt{(4-1)^2 + (7-3)^2} = 5$. Slope of segment $AP$ is $\dfrac{4}{3}$, so the perpendicular line must have slope $m = -\dfrac{3}{4}$.
[Problem 15] Cauchy-Schwarz Inequality Unified with Point-to-Line Distance
For point $(x, y)$ on circle $x^2 + y^2 = 1$, find the maximum value of $k$ such that line $3x + 4y = k$ intersects the circle, comparing (1) the point-to-line distance formula against (2) the Cauchy-Schwarz inequality.
π View Solution & Key Steps
Answer: Maximum value $5$
[Comparative Verification]
[Method 1: Geometric Distance ($d \le r$)]
Distance from center $(0, 0)$ to line $3x + 4y - k = 0$ must satisfy $d \le 1$:
$d = \dfrac{|-k|}{\sqrt{3^2 + 4^2}} = \dfrac{|k|}{5} \le 1 \implies |k| \le 5 \implies -5 \le k \le 5$.
Thus, the maximum value is $\mathbf{5}$.
[Method 2: Cauchy-Schwarz Inequality]
$(3^2 + 4^2)(x^2 + y^2) \ge (3x + 4y)^2$.
Since $x^2 + y^2 = 1$: $25 \times 1 \ge k^2 \implies k^2 \le 25 \implies -5 \le k \le 5$.
Both methods yield identical bounds.
The Point-to-Line Distance Formula and the Cauchy-Schwarz Inequality share an identical root: vector dot products and the cosine projection bound ($|\vec{u} \cdot \vec{v}| \le \|\vec{u}\| \|\vec{v}\|$). Switching fluently between geometry and algebra is an essential competitive problem-solving asset.
Given real numbers $x, y$ satisfying $x^2 + y^2 = 4$, let $M$ and $m$ be the maximum and minimum values of $2x - y + 10$. Compute $M \times m$.
Check Variation Solution
Explanation: Cauchy-Schwarz bound: $(2^2 + (-1)^2)(x^2 + y^2) \ge (2x - y)^2 \implies 5 \times 4 \ge (2x - y)^2 \implies -\sqrt{20} \le 2x - y \le \sqrt{20}$. Adding $10$ gives $M = 10 + 2\sqrt{5}$ and $m = 10 - 2\sqrt{5}$. Product $M \times m = (10 + 2\sqrt{5})(10 - 2\sqrt{5}) = 100 - 20 = 80$.
Across coordinate geometry, the point-to-line distance formula serves as an authoritative orthogonal ruler.
Whether calculating polygon heights, evaluating area determinants, isolating angle bisectors, or testing circle tangency, competitive geometry consistently boils down to one fundamental inquiry: 'Which segment serves as the base, and along which normal direction is the altitude dropped?'
Move past rote arithmetic and appreciate the geometric elegance of perpendicular projections and similarity models. When this foundation is solid, the coordinate plane speaks with unmatched clarity.
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