[Coordinate Geometry: Practice] Lines with Parameters: 15 Master Problems & Diagnostic Variations
In our previous conceptual guide, [Linear Equations: Theory], we explored how grouping by an unknown parameter ($k$ or $m$) immediately unveils the stationary fixed pivot point and the dynamic 360-degree rotating line model.
However, real exam problems—whether on SAT Math, AP Precalculus, IB Math AA, or competitive tests—disguise this concept behind intricate geometric scenarios: triangle non-formation conditions, quadrant exclusions, segment intersections, perpendicularity, and area optimization. Here are 15 essential advanced practice problems along with Yul's Pro-Tips & 1:1 Self-Check Variation Problems. Work through each question first, then toggle open the step-by-step breakdown.
1. Contains an arbitrary parameter ($k, m$)? $\to$ Do not expand into standard form; group by the parameter to fix the stationary pivot point.
2. Passes through the intersection of two lines? $\to$ Avoid solving the system directly; express it as $l_1 + k l_2 = 0$ in a single line.
3. Meets a segment or misses a quadrant? $\to$ Pin down the pivot point and track boundary slopes visually.
Theme 1. Identities, Pivots & Rotation Models (Q01 ~ Q04)
[Problem 01] Fixed Pivot Extraction & Quadrant Exclusion
The line $(2k+1)x + (1-k)y + 3k - 2 = 0$ always passes through a fixed point $P$ regardless of the real value of $k$. Find the range of $k$ such that this line does not pass through the fourth quadrant.
π View Answer & Detailed Solution
Answer: $k \ge 1$ or $k \le -\dfrac{1}{2}$
[Solution]
1) Group terms by $k$:
$(x + y - 2) + k(2x - y + 3) = 0$
Solving $\begin{cases} x + y - 2 = 0 \\ 2x - y + 3 = 0 \end{cases}$ yields $x = -\dfrac{1}{3}, y = \dfrac{7}{3}$. Thus, pivot point $P\left(-\dfrac{1}{3}, \dfrac{7}{3}\right)$ lies in Quadrant II.
2) For a line pivoting in Quadrant II to miss Quadrant IV completely, its slope must be nonnegative ($\text{slope} \ge 0$), its $y$-intercept must be nonnegative ($\ge 0$), and its $x$-intercept must be nonpositive ($\le 0$).
Slope: $\dfrac{2k+1}{k-1} \ge 0 \implies k > 1$ or $k \le -\dfrac{1}{2}$.
When $k=1$, the equation becomes $3x + 1 = 0 \implies x = -\dfrac{1}{3}$ (a vertical line in Quadrants II and III), which avoids Quadrant IV. Combining these gives $k \ge 1$ or $k \le -\dfrac{1}{2}$.
Do not treat quadrant-exclusion problems purely algebraically with multiple inequalities. Plot the pivot point first, pin down the pencil of lines, and visually inspect which boundary slopes prevent the rotating ray from sweeping across the target quadrant.
π― [Self-Check Variation 01]
The line $(k+2)x + (1-k)y - 4k - 5 = 0$ passes through a fixed point $Q$ for all real $k$. Find the range of $k$ such that the line does not pass through Quadrant III.
Check Variation Solution
Explanation: Grouping gives $(2x + y - 5) + k(x - y - 4) = 0$, yielding pivot $Q(3, -1)$ in Quadrant IV. Rotating through positive intercepts and checking nonpositive slopes gives $k \le -2$ or $k > 1$.
[Problem 02] Pivot Inside a Polygon Boundary Analysis
Let rectangle $OABC$ have vertices $O(0, 0)$, $A(4, 0)$, $B(4, 3)$, and $C(0, 3)$. Find the range of real values of $m$ such that the line $mx - y - 2m + 1 = 0$ intersects the perimeter of rectangle $OABC$ at two distinct points.
π View Answer & Detailed Solution
Answer: All real numbers $m$
[Solution]
Rewriting in point-slope form: $y - 1 = m(x - 2)$. The line pivots at $P(2, 1)$.
Since $0 < 2 < 4$ and $0 < 1 < 3$, the point $P(2, 1)$ lies strictly inside the interior of rectangle $OABC$.
Any straight line passing through an interior point of a convex polygon (like a rectangle) must pierce its boundary at exactly two distinct points, regardless of orientation or slope.
Thus, the condition holds for all real numbers $m$.
Whenever a problem asks about intersections between a parameterized line and a polygon, check first whether the pivot point is internal or external. If internal, every line intersects the boundary twice. If external, boundary slopes connecting the vertices determine the range.
π― [Self-Check Variation 02]
Given rectangle $OABC$ with $O(0, 0), A(3, 0), B(3, 2), C(0, 2)$, find the range of $m$ such that the line $mx - y - 5m + 1 = 0$ intersects the perimeter of $OABC$.
Check Variation Solution
Explanation: Pivot point is $P(5, 1)$ outside the rectangle. Connecting $P$ to vertices: to $B(3, 2) \implies m = -\dfrac{1}{2}$; to $O(0, 0) \implies m = \dfrac{1}{5}$. The envelope of slopes sweeping across the rectangle is $-\dfrac{1}{2} \le m \le \dfrac{1}{5}$.
[Problem 03] Simultaneous Intersection with Two Segments
Given four points $A(1, 3)$, $B(2, 5)$, $C(4, 2)$, and $D(5, -1)$, find the range of real values of $k$ such that the line $kx - y - 2k + 1 = 0$ intersects both segment $AB$ and segment $CD$ simultaneously.
π View Answer & Detailed Solution
Answer: $-\dfrac{2}{3} \le k \le \dfrac{1}{2}$
[Solution]
The line is $y - 1 = k(x - 2)$, pivoting at $P(2, 1)$.
1) Intersecting segment $CD$: The slope from $P(2, 1)$ to $C(4, 2)$ is $\dfrac{2-1}{4-2} = \dfrac{1}{2}$, and to $D(5, -1)$ is $\dfrac{-1-1}{5-2} = -\dfrac{2}{3}$. Hence $-\dfrac{2}{3} \le k \le \dfrac{1}{2}$.
2) Segment $AB$ lies in the opposite direction ($x \le 2, y \ge 1$). Since a full geometric line extends in both opposite directions from pivot $P$, extending the line backwards across pivot $P$ intersects segment $AB$ throughout this entire angular sweep.
Therefore, the simultaneous intersection interval is $-\dfrac{2}{3} \le k \le \dfrac{1}{2}$.
Remember that a line is not a one-directional ray. It extends infinitely in both opposite directions through the pivot. When one side passes through a target segment, the opposite wing sweeps through vertical opposite angles across the pivot.
π― [Self-Check Variation 03]
Given segment $AB$ connecting $A(-2, 3), B(-1, 5)$ and segment $CD$ connecting $C(3, 1), D(4, -2)$, find the range of $m$ for which $y - 2 = m(x - 1)$ intersects both segments.
Check Variation Solution
Explanation: Slope from pivot $P(1, 2)$ to $A(-2, 3)$ is $-\dfrac{1}{3}$, and to $D(4, -2)$ is $-\dfrac{4}{3}$. Both wings intersect the opposite segments in the range $-\dfrac{4}{3} \le m \le -\dfrac{1}{3}$.
[Problem 04] Maximum Distance from a Fixed Point to a Rotating Line
The line $(k+2)x + (2k-1)y - 5k = 0$ always passes through point $P$ for any real $k$. Let $d$ be the distance from the origin $O(0, 0)$ to this line. Find the maximum value of $d$ and the corresponding value of $k$.
π View Answer & Detailed Solution
Answer: Maximum distance $\sqrt{5}$, when $k = \dfrac{1}{2}$
[Solution]
Grouping by $k$: $(2x - y) + k(x + 2y - 5) = 0 \implies P(1, 2)$ is the fixed pivot point.
Geometric Theorem: For any line rotating around pivot $P$, the perpendicular distance $d$ to any external point $O$ forms a right triangle where segment $OP$ is the hypotenuse. Hence, $d \le OP$.
The maximum distance occurs when the line is perpendicular to segment $OP$.
$d_{\max} = OP = \sqrt{1^2 + 2^2} = \sqrt{5}$.
The slope of $OP$ is $2$, so the perpendicular line has slope $-\dfrac{1}{2}$. Solving $-\dfrac{k+2}{2k-1} = -\dfrac{1}{2}$ yields $k = \dfrac{1}{2}$.
Never differentiate cumbersome algebraic distance formulas with square roots. The maximum distance from a point $A$ to any line through pivot $P$ is always the geometric length $AP$ itself, achieved precisely when the line is perpendicular to $AP$.
π― [Self-Check Variation 04]
The line $(2k+1)x + (k-1)y - 7k - 2 = 0$ passes through fixed point $P$ for all $k$. Find the maximum distance from $A(1, -1)$ to this line.
Check Variation Solution
Explanation: Grouping gives pivot $P(3, 1)$. The maximum distance is simply the length of segment $AP = \sqrt{(3-1)^2 + (1 - (-1))^2} = \sqrt{4 + 4} = 2\sqrt{2}$.
Theme 2. Triangle Non-Formation & Intersection Positions (Q05 ~ Q08)
[Problem 05] Conditions for Three Lines Not to Form a Triangle
Find the sum of all constants $a$ such that the three lines $l_1 : x - y + 1 = 0$, $l_2 : x + y - 5 = 0$, and $l_3 : ax - y + 2 = 0$ do not form a triangle in the plane.
π View Answer & Detailed Solution
Answer: $\dfrac{1}{2}$
[Solution] Three lines fail to form a triangle under exactly two geometric cases:
Case 1: At least two lines are parallel
Slopes are $m_1 = 1$, $m_2 = -1$, and $m_3 = a$.
- $l_3 \parallel l_1 \implies a = 1$
- $l_3 \parallel l_2 \implies a = -1$
Case 2: All three lines are concurrent (meet at a single point)
Intersecting $l_1$ and $l_2$: $\begin{cases} x - y + 1 = 0 \\ x + y - 5 = 0 \end{cases} \implies (2, 3)$.
Substituting $(2, 3)$ into $l_3$: $2a - 3 + 2 = 0 \implies a = \dfrac{1}{2}$.
Sum of all possible values: $1 + (-1) + \dfrac{1}{2} = \mathbf{\dfrac{1}{2}}$.
The classic triangle non-formation toolkit is: [2 parallel lines + 1 concurrence = 3 candidate values]. Always double-check whether two lines are coincident (identical) by comparing intercepts.
π― [Self-Check Variation 05]
Find the product of all constants $m$ such that $2x - y - 1 = 0$, $x + y - 5 = 0$, and $mx - y + 3 = 0$ do not form a triangle.
Check Variation Solution
Explanation: Parallel slopes: $m = 2, m = -1$. Concurrency at $(2, 3)$ gives $2m - 3 + 3 = 0 \implies m = 0$. The product of values is $2 \times (-1) \times 0 = 0$.
[Problem 06] Plane Partition into 6 Regions
Find the number of constants $a$ and their values such that the three distinct lines $x + 2y - 4 = 0$, $2x - y + 3 = 0$, and $ax + y - 1 = 0$ partition the plane into exactly 6 regions.
π View Answer & Detailed Solution
Answer: 3 values ($a = -2, \dfrac{1}{2}, 3$)
[Solution]
Plane partition numbers by 3 lines:
- 4 regions: All 3 lines are parallel.
- 6 regions: Exactly 2 lines are parallel, or all 3 lines are concurrent at one point.
- 7 regions: Standard triangle (general position).
1) Parallel condition: Slopes are $-\dfrac{1}{2}, 2, -a$.
$-a = -\dfrac{1}{2} \implies a = \dfrac{1}{2}$; $-a = 2 \implies a = -2$.
2) Concurrence condition: Intersecting first two lines gives $x = -\dfrac{2}{5}, y = \dfrac{11}{5}$. Substituting into the third line yields $a = 3$.
Thus, there are 3 values: $a = -2, \dfrac{1}{2}, 3$.
"Partitions the plane into 6 regions" is 100% mathematically equivalent to "fails to form a triangle (excluding the all-parallel 4-region case)". Do not get confused by vocabulary shifts.
π― [Self-Check Variation 06]
Find the sum of all values of $k$ for which $x - y = 0$, $x + y - 2 = 0$, and $kx - y + k + 1 = 0$ partition the plane into fewer than 7 regions.
Check Variation Solution
Explanation: Fewer than 7 regions means the lines fail to form a general triangle. Parallel conditions yield $k = 1, k = -1$. Concurrency at $(1, 1)$ yields $k = 0$. Sum $= 1 + (-1) + 0 = 0$.
[Problem 07] The Inherent Right-Triangle Trap
Find all real values of $k$ such that the three lines $l_1 : 3x - 4y + 5 = 0$, $l_2 : 4x + 3y - 12 = 0$, and $l_3 : kx + y - 2 = 0$ form a right triangle. (Assume the three lines are not concurrent.)
π View Answer & Detailed Solution
Answer: All real numbers $k$ except $k = -\dfrac{3}{4}$ and $k = \dfrac{4}{3}$
[Solution]
Check the dot product of the normal vectors of $l_1$ and $l_2$:
$3(4) + (-4)(3) = 12 - 12 = 0 \implies l_1 \perp l_2$!
The first two lines are already mutually perpendicular. Thus, as long as $l_3$ is not parallel to either line and does not pass through their intersection, any general third line completes a right triangle.
Parallel exclusions: $-k \ne \dfrac{3}{4} \implies k \ne -\dfrac{3}{4}$; $-k \ne -\dfrac{4}{3} \implies k \ne \dfrac{4}{3}$.
Whenever a question asks when three lines form a right triangle, always spend 1 second testing whether the two given static lines already satisfy $aa' + bb' = 0$ before forcing the parameterized line to be perpendicular to them.
π― [Self-Check Variation 07]
Find the positive constant $a$ such that $x - 2y = 0$, $2x + y - 4 = 0$, and $ax - y + 1 = 0$ fail to form a right triangle.
Check Variation Solution
Explanation: The first two lines are already perpendicular ($1(2) + (-2)(1) = 0$). To fail to form a right triangle, the lines must fail to form a triangle altogether. For parallel slopes, the positive value is $a = \dfrac{1}{2}$.
[Problem 08] Intersection Point Confined to Quadrant I
Find the range of real values of $m$ such that the intersection of the two lines $x + 2y - 2 = 0$ and $mx - y + 2m + 1 = 0$ lies strictly in the first quadrant.
π View Answer & Detailed Solution
Answer: $-\dfrac{1}{4} < m < 0$
[Solution]
$l_1 : x + 2y - 2 = 0$ has $x$-intercept $A(2, 0)$ and $y$-intercept $B(0, 1)$.
$l_2 : m(x + 2) - (y - 1) = 0$ rotates around pivot $P(-2, 1)$.
For the intersection to lie in Quadrant I, $l_2$ must pierce the open segment $AB$ connecting the intercepts.
- Passing through $B(0, 1)$: $m = 0$.
- Passing through $A(2, 0)$: $m(2 + 2) - (0 - 1) = 0 \implies m = -\dfrac{1}{4}$.
Inspecting the sweep from pivot $P(-2, 1)$, the slope range is $-\dfrac{1}{4} < m < 0$.
Do not solve for $(x, y)$ in terms of $m$ to solve fractional inequalities $x > 0, y > 0$. Think geometrically: the parameterized line sweeps across the intercept segment of the fixed line.
π― [Self-Check Variation 08]
Find the range of $k$ such that the intersection of $2x + y - 4 = 0$ and $kx - y - k + 3 = 0$ lies in Quadrant I.
Check Variation Solution
Explanation: Intercepts are $(2, 0)$ and $(0, 4)$. Pivot point is $P(1, 3)$. Evaluating boundaries at both intercepts gives the range $-3 < k < 1$.
Theme 3. Pencil of Lines & Geometric Loci (Q09 ~ Q12)
[Problem 09] Line Through Intersection Perpendicular to Another Line
Find the equation of the line passing through the intersection of $2x - 3y + 4 = 0$ and $3x + y - 5 = 0$, perpendicular to the line $4x - 2y + 1 = 0$.
π View Answer & Detailed Solution
Answer: $x + 2y - 5 = 0$
[Solution]
Set up the pencil of lines: $(2x - 3y + 4) + k(3x + y - 5) = 0$
Rearranging: $(2 + 3k)x + (-3 + k)y + (4 - 5k) = 0$.
Using the perpendicular condition $aa' + bb' = 0$ with $4x - 2y + 1 = 0$:
$4(2 + 3k) + (-2)(-3 + k) = 0 \implies 10k + 14 = 0 \implies k = -\dfrac{7}{5}$.
Multiplying by $5$ and substituting back gives $x + 2y - 5 = 0$.
Whenever finding the intersection point directly results in tedious fractions, setting up $l_1 + kl_2 = 0$ saves you from unnecessary computation errors.
π― [Self-Check Variation 09]
Find the equation of the line passing through the intersection of $x - 2y + 3 = 0$ and $2x + y - 4 = 0$, parallel to $3x + y - 2 = 0$.
Check Variation Solution
Explanation: The intersection is $(1, 2)$. With slope $m = -3$, the line is $y - 2 = -3(x - 1) \implies 3x + y - 5 = 0$.
[Problem 10] Line Through Intersection at a Given Distance
Find the equations of all lines passing through the intersection of $x - y + 1 = 0$ and $x + y - 3 = 0$ that are at distance $\sqrt{2}$ from the origin $O(0, 0)$.
π View Answer & Detailed Solution
Answer: $y = 2$ or $(-2+\sqrt{6})x - y + 4-\sqrt{6} = 0$ (equations with slopes $m = -2 \pm \sqrt{6}$)
[Solution]
Intersection is $P(1, 2)$. Line through $P$: $mx - y - m + 2 = 0$.
Distance to origin is $\sqrt{2}$:
$\dfrac{|-m + 2|}{\sqrt{m^2 + 1}} = \sqrt{2} \implies (2 - m)^2 = 2(m^2 + 1) \implies m^2 + 4m - 2 = 0$.
Solving yields two real slopes $m = -2 \pm \sqrt{6}$.
If your quadratic equation for slope $m$ yields only 1 solution when geometric intuition indicates there must be 2, the missing second line is the vertical line ($x = x_1$) whose slope is undefined.
π― [Self-Check Variation 10]
Find the product of the slopes of the two lines passing through $(2, 0)$ whose distance from the origin is $1$.
Check Variation Solution
Explanation: $\dfrac{|-2m|}{\sqrt{m^2+1}} = 1 \implies 4m^2 = m^2 + 1 \implies m^2 = \dfrac{1}{3}$. The product of the two slopes is $-\dfrac{1}{3}$.
[Problem 11] The "Excluded Line" Multiple-Choice Trap
Consider two lines $l_1 : 2x + y - 4 = 0$ and $l_2 : x - 2y + 3 = 0$. Which of the following statements regarding $(2x + y - 4) + k(x - 2y + 3) = 0$ are true?
A. It passes through the intersection of $l_1$ and $l_2$ for any real $k$.
B. The lines $l_1$ and $l_2$ are mutually perpendicular.
C. There exists a real value of $k$ that exactly represents the line $x - 2y + 3 = 0$.
π View Answer & Detailed Solution
Answer: A and B only
[Solution]
- A (True): Linear identity guarantees concurrence at the intersection.
- B (True): $2(1) + 1(-2) = 0 \implies l_1 \perp l_2$.
- C (False): No finite real $k$ can eliminate the leading term $(2x + y - 4)$ to isolate $l_2 = 0$.
Statements claiming that $l_1 + kl_2 = 0$ represents "all lines passing through the intersection" are technically false. The single line multiplied by $k$ is permanently excluded.
π― [Self-Check Variation 11]
Name the single line that the family $k(3x - y + 1) + (x + 2y - 4) = 0$ cannot represent for any real $k$.
Check Variation Solution
Explanation: The expression multiplied directly by parameter $k$ cannot be isolated on its own.
[Problem 12] Geometric Locus of Orthogonal Lines
As real parameter $m$ varies, find the equation of the locus traced by the intersection point $P(x, y)$ of the two lines $mx - y + 1 = 0$ and $x + my - 3 = 0$. (State any excluded point.)
π View Answer & Detailed Solution
Answer: $\left(x - \dfrac{3}{2}\right)^2 + \left(y - \dfrac{1}{2}\right)^2 = \dfrac{5}{2}$ (excluding $(0, 1)$)
[Solution]
- First line: $y - 1 = mx \implies$ passes through fixed point $A(0, 1)$.
- Second line: $x - 3 = -my \implies$ passes through fixed point $B(3, 0)$.
Product of slopes: $m \times \left(-\dfrac{1}{m}\right) = -1 \implies$ they are always perpendicular ($\angle APB = 90^\circ$).
By Thales's Theorem, the locus of point $P$ is a circle with diameter $AB$.
Center is the midpoint $\left(\dfrac{3}{2}, \dfrac{1}{2}\right)$ and radius squared is $\dfrac{5}{2}$. The vertical line through $A(0, 1)$ cannot be represented, so $(0, 1)$ is excluded.
Whenever you notice two perpendicular lines rotating around two fixed points, recognize the inscribed right angle ($90^\circ$) immediately: the locus is always a circle with the segment between the pivots as its diameter.
π― [Self-Check Variation 12]
Find the circumference of the locus of intersection $P$ of $kx - y = 0$ and $x + ky - 4 = 0$ as $k$ varies.
Check Variation Solution
Explanation: Pivots are $(0, 0)$ and $(4, 0)$, forming a circle of diameter $4$. Circumference is $\pi d = 4\pi$.
Theme 4. High-Scoring Killer Applications (Q13 ~ Q15)
[Problem 13] Area Bisector Line Passing Through a Pivot
Consider $\triangle ABC$ with vertices $A(2, 6)$, $B(-2, 2)$, and $C(4, -2)$. A line $l$ passing through pivot $P(0, 1)$ bisects the area of $\triangle ABC$. Find the slope of line $l$.
π View Answer & Detailed Solution
Answer: $\dfrac{1}{3}$
[Solution]
Total area of $\triangle ABC$ is $24$, so each partitioned region must have area $12$.
Line $AB$ has equation $y = x + 4$. Point $P(0, 1)$ does not lie on vertex $A$.
Setting line $l: y - 1 = mx$, we locate intersection points with sides $AC$ and $BC$. Setting the partitioned area ratio equal to $\dfrac{1}{2}$ yields the slope $m = \mathbf{\dfrac{1}{3}}$.
When a line bisects a triangle's area from a vertex, it simply connects to the opposite midpoint. But when it passes through an arbitrary pivot point, formulate the area using side-length ratios or coordinate determinants.
π― [Self-Check Variation 13]
Given $\triangle OAB$ with $O(0, 0), A(6, 0), B(0, 4)$, find the slope of the line through $(0, 2)$ that bisects the area of $\triangle OAB$.
Check Variation Solution
Explanation: Point $(0, 2)$ is already the midpoint of side $OB$. Therefore, the area bisector must pass directly through opposite vertex $A(6, 0)$. Slope is $\dfrac{0 - 2}{6 - 0} = -\dfrac{1}{3}$.
[Problem 14] Slopes of Tangents from an External Point
Two tangent lines are drawn from $P(3, 1)$ to the circle $x^2 + y^2 = 2$. If their slopes are $m_1$ and $m_2$, find the sum $m_1 + m_2$ and product $m_1 m_2$.
π View Answer & Detailed Solution
Answer: Sum: $\dfrac{6}{7}$, Product: $-\dfrac{1}{7}$
[Solution]
Equation of line through $P(3, 1)$: $mx - y - 3m + 1 = 0$.
Setting the distance from origin equal to radius $r = \sqrt{2}$:
$\dfrac{|-3m + 1|}{\sqrt{m^2 + 1}} = \sqrt{2} \implies (3m - 1)^2 = 2(m^2 + 1) \implies 7m^2 - 6m - 1 = 0$.
By Vieta's formulas: $m_1 + m_2 = \dfrac{6}{7}$ and $m_1 m_2 = -\dfrac{1}{7}$.
Do not waste time solving the quadratic equation to find each irrational slope individually. Apply Vieta's formulas directly to the quadratic relation in $m$.
π― [Self-Check Variation 14]
Find the product of the slopes of two tangent lines drawn from $(2, 3)$ to the circle $x^2 + y^2 = 1$.
Check Variation Solution
Explanation: $\dfrac{|-2m + 3|}{\sqrt{m^2 + 1}} = 1 \implies (2m - 3)^2 = m^2 + 1 \implies 3m^2 - 12m + 8 = 0$. Product $= \dfrac{8}{3}$.
[Problem 15] Rotating Line & Maximum Distance Difference
Consider the parameterized line $l : (k+1)x + (2k-1)y - 4k - 1 = 0$ and two points $A(2, 5)$ and $B(6, 1)$. For a moving point $Q$ on line $l$, find the value of $k$ that maximizes the difference $|AQ - BQ|$.
▲ By the Triangle Inequality $|AQ - BQ| \le AB$, equality occurs when $A, B, Q$ are collinear
π View Answer & Detailed Solution
Answer: $k = 0$
[Solution]
1) Grouping gives $(x - y - 1) + k(x + 2y - 4) = 0 \implies$ stationary pivot $P(2, 1)$.
2) In any triangle $ABQ$, the triangle inequality states $|AQ - BQ| \le AB$.
Equality occurs precisely when points $A, B$, and $Q$ are collinear.
The line through $A(2, 5)$ and $B(6, 1)$ is $x + y - 7 = 0$. For line $l$ to contain the collinear point $Q$, evaluating the system yields $k = \mathbf{0}$.
The minimum sum of distances ($AQ + BQ$) requires reflecting one point across the line. Conversely, the maximum difference of distances ($|AQ - BQ|$) requires NO reflection; it is achieved on the straight line connecting $A$ and $B$.
π― [Self-Check Variation 15]
Given $A(1, 4), B(5, 2)$ and moving point $P$ on the $x$-axis, find the maximum value of $|AP - BP|$.
Check Variation Solution
Explanation: Maximum difference is simply the direct length $AB = \sqrt{(5-1)^2 + (2-4)^2} = \sqrt{16 + 4} = 2\sqrt{5}$.
When confronted with complicated linear equations tangled with unknown parameters, resist the impulse to expand blindly into standard form.
The instant you factor out the parameter and anchor a single stationary pivot point on the Cartesian grid, the problem transforms from abstract algebra into an intuitive rotating stick model.
Top performance in advanced mathematics does not come from memorizing dozens of disconnected formulas; it comes from identifying the one unshakeable invariant point within a sea of moving geometry. Make sure to sketch each pivot point with your own hand, observe the rotational sweep, and make this geometric intuition your own.
- 12+ Correct: Top tier mastery. Your rotational intuition and parameter-factoring habits are solid.
- 8 to 11 Correct: Solid algebraic foundation. Watch out for classic pitfalls like excluded lines and triangle non-formation conditions.
- Under 7: Revisit our theory post
[Linear Equations: Theory] to master the 3-step pivot anchoring routine before retrying.
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