[Coordinate Geometry: Practice] Perpendicular Lines: 15 Master Problems & Diagnostic Variations
With ten minutes left on the clock in a timed exam, a student encounters a high-stakes coordinate geometry problem. Faced with finding the vertices of an inclined rhombus or determining the intersection of orthogonal lines, the instinctive, panicked impulse is to assign four separate unknowns ($a, b, c, d$), set up slopes in slope-intercept forms, and dive into a 4-variable linear system. Minutes evaporate, negative signs flip during fraction arithmetic, and an entire 4-point response is lost to computational fatigue.
The perpendicularity condition between two straight lines ($m_1 m_2 = -1$ or $aa' + bb' = 0$) is far more than an isolated identity in introductory geometry. It is the master geometric blueprint that instantly constructs circles' normal lines and tangents, establishes orthogonal trajectory and distance optimization in AP Calculus, and anchors the vector dot product ($\vec{u} \cdot \vec{v} = 0$) in linear algebra—serving as the universal orthogonal benchmark across advanced mathematics.
In this master practice guide, we dismantle brute-force algebraic labor. Equipped with "1-line perpendicular bisectors via midpoints and negative reciprocals, rhombus orthogonal diagonal decompositions, and Thales's circle locus for $90^\circ$ inscribed angles," you will develop the insight needed to solve 15 competitive challenge problems and their 1:1 diagnostic variations with unmatched speed and clarity. Attempt each problem independently first, then expand the solution toggle to audit your step-by-step reasoning.
1. Perpendicular Bisectors: Never solve systems of distances. Formulate the line in 1 line using "midpoint coordinates + negative reciprocal slope ($-\dfrac{1}{m}$)".
2. Rhombus & Orthogonal Diagonals: Because diagonals bisect each other orthogonally, eliminate variables simultaneously via "common midpoint + perpendicular slope product ($-1$)".
3. Orthogonal Intersection Locus: Never eliminate parameters by brute force. Recognize the geometric signature of "Thales's theorem: $90^\circ$ inscribed angles form a circle over the diameter".
Theme 1. Perpendicular Bisectors & Advanced Line Reflections (Q01 ~ Q04)
[Problem 01] Perpendicular Bisector & Minimum Sum of Squared Distances
For two fixed points $A(-1, 5)$ and $B(3, 1)$, let $P$ be an arbitrary moving point on the perpendicular bisector of segment $AB$. For fixed points $C(2, 6)$ and $D(6, 2)$, determine the minimum possible value of $\overline{PC}^2 + \overline{PD}^2$.
π View Solution & Key Steps
Answer: $20$
[Step-by-Step Solution]
1) Perpendicular bisector of $AB$:
- Midpoint $M$: $\left(\dfrac{-1+3}{2}, \dfrac{5+1}{2}\right) = (1, 3)$
- Slope of $AB$: $\dfrac{1-5}{3-(-1)} = -1 \implies$ Perpendicular slope is $1$.
- Bisector line $l$: $y - 3 = 1(x - 1) \implies x - y + 2 = 0$.
2) Apollonius's (Median) Theorem transformation:
Let $N$ be the midpoint of segment $CD$: $N\left(\dfrac{2+6}{2}, \dfrac{6+2}{2}\right) = (4, 4)$.
By the median theorem: $\overline{PC}^2 + \overline{PD}^2 = 2(\overline{PN}^2 + \overline{CN}^2)$.
$\overline{CN}^2 = (4-2)^2 + (4-6)^2 = 4 + 4 = 8$ (constant).
The expression is minimized when distance $\overline{PN}$ is minimized, which occurs when $P$ is the orthogonal projection of $N(4, 4)$ onto line $l$:
$\overline{PN}_{\text{min}} = \dfrac{|4 - 4 + 2|}{\sqrt{1^2 + (-1)^2}} = \dfrac{2}{\sqrt{2}} = \sqrt{2} \implies \overline{PN}^2_{\text{min}} = 2$.
Therefore, the minimum value is $2(2 + 8) = \mathbf{20}$.
While parameterizing $P(t, t+2)$ into a quadratic function works, converting the sum of squared distances into Apollonius's median theorem $2(\overline{PN}^2 + \overline{CN}^2)$ and applying the point-to-line distance formula eliminates algebra and prevents computational errors.
Let $P$ be any point on the perpendicular bisector of $A(1, 7)$ and $B(5, -1)$. Find the minimum value of $\overline{PC}^2 + \overline{PD}^2$ for $C(-2, 1)$ and $D(2, 5)$.
Check Variation Solution
Explanation: Midpoint $M(3, 3)$, slope $AB = -2 \implies$ Bisector $l: x - 2y + 3 = 0$. Midpoint of $CD$ is $N(0, 3)$, $\overline{CN}^2 = 8$. Distance from $N$ to $l$ is $d = \dfrac{|0 - 6 + 3|}{\sqrt{5}} = \dfrac{3}{\sqrt{5}} \implies d^2 = \dfrac{9}{5}$. Minimum value $= 2\left(\dfrac{9}{5} + 8\right) = \dfrac{98}{5}$.
[Problem 02] Reflection Across an Oblique Line & Slope Inversion
The line $l_1 : 2x - y + 1 = 0$ is reflected across the line $m : x + y - 2 = 0$. If the equation of the reflected line is expressed in the form $ax + by + 1 = 0$, compute the product of constants $ab$.
π View Solution & Key Steps
Answer: $-\dfrac{2}{9}$
[Step-by-Step Solution]
The line of reflection is $m : x + y = 2 \implies x = 2 - y, y = 2 - x$.
Because the slope of $m$ is $-1$, any point $(x, y)$ reflected across $x + y = 2$ transforms via direct coordinate exchange: $x' = 2 - y, y' = 2 - x \implies x = 2 - y', y = 2 - x'$.
Substitute into $l_1 : 2x - y + 1 = 0$:
$2(2 - y') - (2 - x') + 1 = 0 \implies x' - 2y' + 3 = 0$.
Normalize the constant term to $1$ by dividing across by $3$:
$\dfrac{1}{3}x - \dfrac{2}{3}y + 1 = 0 \implies a = \dfrac{1}{3}, b = -\dfrac{2}{3}$.
Therefore, $ab = \left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right) = \mathbf{-\dfrac{2}{9}}$.
When reflecting across lines with slope $\pm 1$ ($y = x + k$ or $y = -x + k$), do not solve systems of midpoints and slopes. Directly solve for $x$ and $y$ from the reflection axis equation and substitute them into the target line.
Find the equation of the line resulting from reflecting $l : 3x - 4y + 5 = 0$ across the oblique axis $m : x - 2y + 1 = 0$.
Check Variation Solution
Explanation: Intersection of $l$ and $m$ is fixed pivot $(3, 2)$. Take another point on $l$, such as $(-3, -1)$, reflect it across $m$ using standard midpoint and perpendicular slope conditions to get $(x_1, y_1)$, then construct the line connecting it with $(3, 2)$.
[Problem 03] Optical Reflection & Angle of Incidence-Reflection Symmetry
A light ray emitted from point $A(2, 5)$ reflects off a mirror lying along line $l : x - 2y + 3 = 0$ at point $P$ and arrives at point $B(7, 0)$. Determine the exact coordinates of the incidence point $P$ that minimizes the total path length $AP + PB$.
π View Solution & Key Steps
Answer: $(1, 2)$
[Step-by-Step Solution]
Reflect point $A(2, 5)$ across line $l : x - 2y + 3 = 0$ to find virtual image $A'(a, b)$:
1) Perpendicular condition: Line $l$ has slope $\dfrac{1}{2} \implies$ slope of $AA'$ is $-2$.
$\dfrac{b - 5}{a - 2} = -2 \implies 2a + b = 9 \quad \cdots ①$
2) Midpoint condition: Midpoint $\left(\dfrac{a+2}{2}, \dfrac{b+5}{2}\right)$ lies on $l$.
$\dfrac{a+2}{2} - 2\left(\dfrac{b+5}{2}\right) + 3 = 0 \implies a - 2b - 2 = 0 \quad \cdots ②$
Solving $①$ and $②$ gives $a = 4, b = 1 \implies A'(4, 1)$.
The minimal path is the straight line connecting $A'(4, 1)$ and $B(7, 0)$:
Slope: $m = \dfrac{0 - 1}{7 - 4} = -\dfrac{1}{3} \implies x + 3y - 7 = 0$.
Intersecting with mirror line $l : x - 2y + 3 = 0$ yields $y = 2, x = 1$.
Hence, the reflection point is $\mathbf{(1, 2)}$.
In geometric reflection problems, select the point that yields integer arithmetic when setting up midpoint equations. The optimal contact point $P$ is always the intersection between the mirror line and the segment connecting the reflected image to the destination.
A ray from $A(3, 4)$ in the first quadrant reflects sequentially off the $x$-axis at point $P$ and off line $y = x$ at point $Q$, returning to $A$. Determine the minimum perimeter of $\triangle APQ$.
Check Variation Solution
Explanation: Reflect $A(3, 4)$ across the $x$-axis to get $A_1(3, -4)$, and across line $y = x$ to get $A_2(4, 3)$. The minimum perimeter equals the distance $\overline{A_1 A_2} = \sqrt{(4-3)^2 + (3 - (-4))^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2}$.
[Problem 04] Circumcenter of an Obtuse Triangle via Perpendicular Bisectors
Find the coordinates of the circumcenter $O'$ of $\triangle ABC$ with vertices $A(1, 4)$, $B(-2, 1)$, and $C(4, -1)$ by setting up linear equations of perpendicular bisectors.
π View Solution & Key Steps
Answer: $\left(\dfrac{5}{4}, \dfrac{3}{4}\right)$
[Step-by-Step Solution]
The circumcenter is the concurrent intersection of the perpendicular bisectors of the sides.
1) Perpendicular bisector of $AB$ ($l_1$):
- Midpoint: $\left(-\dfrac{1}{2}, \dfrac{5}{2}\right)$, slope $m_{AB} = \dfrac{1-4}{-2-1} = 1 \implies$ perpendicular slope is $-1$.
- Line $l_1 : y - \dfrac{5}{2} = -1\left(x + \dfrac{1}{2}\right) \implies x + y - 2 = 0 \quad \cdots ①$
2) Perpendicular bisector of $BC$ ($l_2$):
- Midpoint: $(1, 0)$, slope $m_{BC} = \dfrac{-1-1}{4-(-2)} = -\dfrac{1}{3} \implies$ perpendicular slope is $3$.
- Line $l_2 : y - 0 = 3(x - 1) \implies 3x - y - 3 = 0 \quad \cdots ②$
Adding $①$ and $②$: $4x - 5 = 0 \implies x = \dfrac{5}{4}, y = \dfrac{3}{4}$.
Thus, the circumcenter is $\mathbf{\left(\dfrac{5}{4}, \dfrac{3}{4}\right)}$.
Setting up quadratic distance equalities ($PA^2 = PB^2 = PC^2$) introduces algebraic clutter. Identify two sides with convenient integer slopes, construct their perpendicular bisectors in 1 line, and solve the resulting $2\times 2$ system.
For vertices $A(0, 5)$, $B(-3, -4)$, and $C(5, 0)$, compute the circumcenter coordinates $(a, b)$ and circumradius $R$, and state the value of $a + b + R^2$.
Check Variation Solution
Explanation: Bisector of $AB: x + 3y = 0$; bisector of $AC: x - y = 0$. Intersecting gives circumcenter $P(0, 0)$. Circumradius $R = \overline{PA} = 5 \implies R^2 = 25$. Value $= 0 + 0 + 25 = 25$.
Theme 2. Rhombus & Orthogonal Diagonal Decompositions (Q05 ~ Q08)
[Problem 05] Rhombus Diagonal Perpendicular Bisector Property
Quadrilateral $ABCD$ with vertices $A(1, 4)$, $B(a, b)$, $C(5, 6)$, and $D(c, d)$ forms a rhombus. If the line passing through opposite vertices $B$ and $D$ is given by $x + ky + m = 0$, calculate the sum of constants $k + m$.
π View Solution & Key Steps
Answer: $-5$
[Step-by-Step Solution]
The diagonals of a rhombus bisect each other at right angles ($90^\circ$). Thus, line $BD$ is the perpendicular bisector of diagonal $AC$.
1) Midpoint of diagonal $AC$: $M(3, 5)$.
2) Slope of diagonal $AC$: $m_{AC} = \dfrac{6 - 4}{5 - 1} = \dfrac{1}{2} \implies$ slope of line $BD$ is $-2$.
3) Line $BD$: $y - 5 = -2(x - 3) \implies 2x + y - 11 = 0 \implies x + \dfrac{1}{2}y - \dfrac{11}{2} = 0$.
Matching coefficients: $k = \dfrac{1}{2}, m = -\dfrac{11}{2} \implies k + m = \dfrac{1}{2} - \dfrac{11}{2} = \mathbf{-5}$.
Because rhombus diagonals are mutually perpendicular and bisect each other, diagonal $BD$ is fully determined by diagonal $AC$'s midpoint and negative reciprocal slope in 3 seconds, without ever calculating vertices $B$ or $D$.
Rhombus $ABCD$ has vertices $A(-1, 2)$ and $C(3, 4)$. If vertex $B$ lies on the $x$-axis, determine the coordinates of vertex $D$.
Check Variation Solution
Explanation: Midpoint of $AC$ is $M(1, 3)$, slope is $\dfrac{1}{2} \implies$ Diagonal $BD$ has slope $-2$: $y - 3 = -2(x - 1) \implies y = -2x + 5$. Since $B$ is on the $x$-axis, $y = 0 \implies B\left(\dfrac{5}{2}, 0\right)$. Because $M$ is also the midpoint of $BD$: $\dfrac{5/2 + x_D}{2} = 1 \implies x_D = -\dfrac{1}{2}, y_D = 6$.
[Problem 06] Square Vertex Determination via $90^\circ$ Rotational Vectors
Square $ABCD$ has two adjacent vertices $A(1, 2)$ and $B(4, 3)$ on the coordinate plane. If vertices $C$ and $D$ lie entirely in the first quadrant, determine the exact coordinates of vertex $C$.
π View Solution & Key Steps
Answer: $(3, 6)$
[Step-by-Step Solution]
The displacement vector from $A(1, 2)$ to $B(4, 3)$ is $\vec{v} = \langle 3, 1 \rangle$.
In a square, adjacent edge vector $\vec{BC}$ has identical length and is oriented at $90^\circ$ counterclockwise relative to $\vec{v}$.
Rotating $\langle 3, 1 \rangle$ by $90^\circ$ counterclockwise yields $\langle -1, 3 \rangle$.
Adding this displacement to vertex $B(4, 3)$ gives:
$C = (4 - 1, 3 + 3) = \mathbf{(3, 6)}$.
Use the 2D orthogonal rotation vector rule: a vector $\langle a, b \rangle$ rotated by $90^\circ$ becomes $\langle -b, a \rangle$. This avoids slope systems and solves polygon constructions mentally.
Opposite vertices of square $ABCD$ are $A(1, 1)$ and $C(5, 5)$. Determine the coordinates of vertices $B$ and $D$, and find the square's area.
Check Variation Solution
Explanation: Center is $M(3, 3)$. Half-diagonal vector is $\langle 2, 2 \rangle$. Perpendicular half-diagonal vectors are $\langle -2, 2 \rangle$ and $\langle 2, -2 \rangle$. Adding to center yields $(1, 5)$ and $(5, 1)$. Diagonal length is $4\sqrt{2} \implies$ side length is $4$, area is $16$.
[Problem 07] Comprehensive Classification of Right Triangle Vertex Orientations
Find the product of all possible real values of $k$ such that $\triangle ABC$ with vertices $A(-1, 2)$, $B(3, 4)$, and $C(k, 0)$ is a right triangle.
π View Solution & Key Steps
Answer: $0$
[Step-by-Step Solution]
Check three possible right-angle vertices:
1) $\angle A = 90^\circ$: $m_{AB} \cdot m_{AC} = -1 \implies \left(\dfrac{1}{2}\right)\left(\dfrac{-2}{k+1}\right) = -1 \implies k = 0$.
2) $\angle B = 90^\circ$: $m_{AB} \cdot m_{BC} = -1 \implies \left(\dfrac{1}{2}\right)\left(\dfrac{-4}{k-3}\right) = -1 \implies k = 5$.
3) $\angle C = 90^\circ$: $m_{AC} \cdot m_{BC} = -1 \implies k^2 - 2k + 5 = 0$ ($D < 0$, no real solutions).
Possible real values: $k = 0, 5$. Product: $0 \times 5 = \mathbf{0}$.
Unless the right-angled vertex is explicitly stated, always analyze all three candidate vertices ($\angle A=90^\circ, \angle B=90^\circ, \angle C=90^\circ$) to guard against missing roots.
Find the sum of all real values of $k$ such that vertices $A(1, 3)$, $B(5, 5)$, and $C(3, k)$ form a right triangle.
Check Variation Solution
Explanation: Slope $m_{AB} = \dfrac{1}{2}$.
1. $\angle A = 90^\circ \implies m_{AC} = -2 \implies k = -1$.
2. $\angle B = 90^\circ \implies m_{BC} = -2 \implies k = 9$.
3. $\angle C = 90^\circ \implies m_{AC} \cdot m_{BC} = -1 \implies k^2 - 8k + 11 = 0$ (discriminant $>0$, sum of roots is $8$).
Sum of all roots: $-1 + 9 + 8 = 16$.
[Problem 08] Intercept-Form Orthogonality & Enclosed Triangle Area
The lines $l_1 : \dfrac{x}{a} + \dfrac{y}{3} = 1$ and $l_2 : \dfrac{x}{4} - \dfrac{y}{b} = 1$ are mutually perpendicular. If the triangle enclosed by $l_1$ and the coordinate axes has an area of $6$, find the value of $a + b$ for positive constants $a$ and $b$.
π View Solution & Key Steps
Answer: $\dfrac{28}{3}$
[Step-by-Step Solution]
1) Area of triangle under $l_1$: $S = \dfrac{1}{2} \cdot a \cdot 3 = 6 \implies a = 4$.
2) Intercept form perpendicular identity $pp' + qq' = 0$:
Here $p = a = 4, q = 3$ and $p' = 4, q' = -b$.
$4(4) + 3(-b) = 0 \implies 16 - 3b = 0 \implies b = \dfrac{16}{3}$.
Therefore, $a + b = 4 + \dfrac{16}{3} = \mathbf{\dfrac{28}{3}}$.
For intercept forms $\dfrac{x}{p} + \dfrac{y}{q} = 1$ and $\dfrac{x}{p'} + \dfrac{y}{q'} = 1$, remember the orthogonal intercept identity: $pp' + qq' = 0$. It bypasses slope conversion completely.
Lines $\dfrac{x}{p} + \dfrac{y}{q} = 1$ and $\dfrac{x}{q} - \dfrac{y}{p} = 1$ are perpendicular. If the locus of their intersection point encloses a circle of area $25\pi$, determine the value of $p^2 + q^2$.
Check Variation Solution
Explanation: Both lines pass through pivots on the axes and are orthogonal, tracing a Thales circle centered at the origin with radius $R = 5$. By geometric symmetry, $p^2 + q^2 = 2R^2 = 50$.
Theme 3. Triangle Orthocenters & Thales's Locus (Q09 ~ Q12)
[Problem 09] High-Precision Orthocenter Coordinates via Axis Alignment
Determine the exact coordinates of the orthocenter $H$ of $\triangle OAB$ with vertices $O(0, 0)$, $A(6, 0)$, and $B(2, 4)$.
π View Solution & Key Steps
Answer: $(2, 2)$
[Step-by-Step Solution]
1) Altitude from $B(2, 4)$ to side $OA$ (the $x$-axis):
Since $OA$ lies on $y = 0$, the perpendicular line is vertical: $x = 2 \quad \cdots ①$
2) Altitude from $A(6, 0)$ to side $OB$:
Slope of $OB$ is $m_{OB} = \dfrac{4 - 0}{2 - 0} = 2 \implies$ altitude has slope $-\dfrac{1}{2}$.
Equation: $y - 0 = -\dfrac{1}{2}(x - 6) \implies y = -\dfrac{1}{2}x + 3 \quad \cdots ②$
Substitute $x = 2$ into $②$: $y = -\dfrac{1}{2}(2) + 3 = 2$.
Thus, the orthocenter is $\mathbf{(2, 2)}$.
When computing the orthocenter, always align one side along an axis ($y=0$ or $x=0$). This yields one altitude instantly as a vertical line ($x = x_B$), reducing the calculation to a simple 1-variable evaluation.
In $\triangle OAB$ with vertices $O(0, 0)$, $A(8, 0)$, and $B(2, 6)$, let $H$ be the orthocenter and $G$ be the centroid. Calculate the length of segment $HG$.
Check Variation Solution
Explanation: Altitude from $B$ is $x = 2$. Altitude from $A$ to $OB$ (slope 3) is $y = -\dfrac{1}{3}(x - 8) \implies H(2, 2)$. Centroid $G$ is $\left(\dfrac{10}{3}, 2\right)$. Because $y$-coordinates match, $\overline{HG} = \dfrac{10}{3} - 2 = \dfrac{4}{3}$.
[Problem 10] Orthogonal Intersection Locus & Perimeter Evaluation
For parameter $m \in \mathbb{R}$, consider lines $l_1 : mx - y + 2m = 0$ and $l_2 : x + my - 4 = 0$. If $P$ is their point of intersection, find the total length of the curve traced by $P$.
π View Solution & Key Steps
Answer: $6\pi$
[Step-by-Step Solution]
1) Fixed pivots: $l_1 : m(x + 2) - y = 0 \implies A(-2, 0)$; $l_2$ passes through $B(4, 0)$.
2) Normal vector dot product: $m(1) + (-1)(m) = 0$.
Because the lines are always mutually perpendicular ($\angle APB = 90^\circ$), by Thales's theorem, point $P$ traces a circle with diameter $AB$.
Diameter: $d = 4 - (-2) = 6 \implies$ Radius $R = 3$.
Total trajectory circumference: $C = 2\pi R = 2\pi(3) = \mathbf{6\pi}$.
Whenever two parameterized lines have coefficient dot product $aa' + bb' = 0$, they are orthogonal lines pivoting through fixed points, tracing a Thales circle over the segment joining their pivots.
Lines $mx - y - 3m = 0$ and $x + my + 1 = 0$ intersect at point $P$. Determine the maximum distance between $P$ and the fixed point $A(1, 5)$.
Check Variation Solution
Explanation: Fixed pivots are $(3, 0)$ and $(-1, 0)$. Diameter is $4 \implies$ circle center is $C(1, 0)$ with radius $R = 2$. Distance from $A(1, 5)$ to center $C(1, 0)$ is $d = 5$. Maximum distance $= d + R = 5 + 2 = 7$.
[Problem 11] Orthic Projection Foot onto the Hypotenuse
In the right triangle formed by lines $x = 0$, $y = 0$, and $3x + 4y - 12 = 0$, determine the exact coordinates of the projection foot dropped from the origin onto the hypotenuse.
π View Solution & Key Steps
Answer: $\left(\dfrac{36}{25}, \dfrac{48}{25}\right)$
[Step-by-Step Solution]
Hypotenuse normal vector is $\vec{n} = (3, 4)$.
The perpendicular line passing through the origin is given by $4x - 3y = 0 \implies y = \dfrac{4}{3}x$.
Substitute into $3x + 4y = 12$:
$3x + 4\left(\dfrac{4}{3}x\right) = 12 \implies \dfrac{25}{3}x = 12 \implies x = \dfrac{36}{25}$.
Then $y = \dfrac{4}{3}\left(\dfrac{36}{25}\right) = \dfrac{48}{25}$.
Thus, the altitude foot is $\mathbf{\left(\dfrac{36}{25}, \dfrac{48}{25}\right)}$.
The line perpendicular to $ax + by + c = 0$ passing through the origin is instantly written as $bx - ay = 0$ by swapping coefficients and inverting one sign.
Find the projection foot $H$ dropped from $P(2, 4)$ onto line $2x - y + 5 = 0$, and evaluate the altitude segment length $PH$.
Check Variation Solution
Explanation: Perpendicular line through $P$ is $x + 2y - 10 = 0$. Intersecting with $2x - y + 5 = 0$ yields $H(0, 5)$. Distance $PH = \sqrt{(0-2)^2 + (5-4)^2} = \sqrt{5}$.
[Problem 12] AM-GM Minimization of Triangle Area Formed by Orthogonal Lines
Two mutually perpendicular lines $l_1$ and $l_2$ pass through fixed point $P(1, 2)$. If they intersect the positive $x$-axis and $y$-axis respectively, determine the minimum area of the triangle formed by the origin and these intercepts.
π View Solution & Key Steps
Answer: $4$
[Step-by-Step Solution]
Let $l_1$ have slope $m > 0$; then $l_2$ has slope $-\dfrac{1}{m}$.
Evaluating the $x$-intercept of $l_1$ and $y$-intercept of $l_2$ and applying the Arithmetic Mean - Geometric Mean (AM-GM) inequality yields the minimal area bound of $\mathbf{4}$.
Whenever orthogonal slopes $m$ and $-\dfrac{1}{m}$ appear in optimization questions, they form complementary reciprocal terms that collapse under the AM-GM inequality ($x + \dfrac{1}{x} \ge 2$).
Line $l_1$ passes through $(2, 3)$ with slope $m$, and $l_2$ passes through $(3, 2)$ with slope $-\dfrac{1}{m}$ ($m > 0$). Find the minimum value of the sum of their $y$-intercepts.
Check Variation Solution
Explanation: $y$-intercepts are $3 - 2m$ and $2 + \dfrac{3}{m}$. Setting up the sum and applying AM-GM bounds under appropriate positivity conditions gives $5 + 2\sqrt{6}$.
Theme 4. Competitive Exam 4-Point Killer Integrations (Q13 ~ Q15)
[Problem 13] Orthogonal Tangents to a Circle & The Director Circle
Two tangent lines drawn from external point $P(a, 3)$ to circle $(x-1)^2 + y^2 = 8$ are perpendicular to each other. Calculate the positive real value of constant $a$.
π View Solution & Key Steps
Answer: $1 + \sqrt{7}$
[Step-by-Step Solution]
When two tangents to a circle are orthogonal, the center, two tangency points, and $P$ form a square of side $r$.
The distance from center $C(1, 0)$ to point $P$ must equal the square diagonal $r\sqrt{2}$ (the Director Circle theorem):
$\overline{CP} = r\sqrt{2} = 2\sqrt{2} \times \sqrt{2} = 4$.
$\overline{CP}^2 = (a - 1)^2 + 3^2 = 16 \implies (a - 1)^2 = 7 \implies a = 1 \pm \sqrt{7}$.
Since $a > 0$, the answer is $\mathbf{1 + \sqrt{7}}$.
Do not use discriminant $D=0$ for tangent slopes! Orthogonal tangents always trace a concentric Director Circle with radius $R = r\sqrt{2}$.
The locus of point $P$ from which two perpendicular tangents can be drawn to $x^2 + y^2 = 18$ intersects line $x - y + k = 0$ at exactly one point. Find the positive value of $k$.
Check Variation Solution
Explanation: Director circle radius is $R = \sqrt{18}\cdot\sqrt{2} = 6$. The line is tangent to $x^2 + y^2 = 36 \implies \dfrac{|k|}{\sqrt{2}} = 6 \implies k = 6\sqrt{2}$.
[Problem 14] Extremal Distance Bounds Over an Orthogonal Intersection Locus
For parameter $k \in \mathbb{R}$, let $P$ be the intersection of lines $l_1 : kx - y + 2k = 0$ and $l_2 : x + ky - 4 = 0$. Let $M$ and $m$ denote the maximum and minimum distances between $P$ and $A(1, 4)$. Calculate $M^2 + m^2$.
π View Solution & Key Steps
Answer: $50$
[Step-by-Step Solution]
The locus of $P$ is a circle with diameter endpoints $(-2, 0)$ and $(4, 0)$.
Center $C(1, 0)$, radius $R = 3$.
Distance from $A(1, 4)$ to center $C(1, 0)$ is $d = 4$.
$M = d + R = 4 + 3 = 7$, $m = d - R = 4 - 3 = 1$.
Therefore, $M^2 + m^2 = 7^2 + 1^2 = 49 + 1 = \mathbf{50}$.
Once you recognize that the orthogonal locus is a circle, the extreme distances to an external point are immediately $d + R$ and $d - R$.
Find the maximum distance between line $3x - 4y + 18 = 0$ and the locus of the intersection of $kx - y = 0$ and $x + ky - 8 = 0$.
Check Variation Solution
Explanation: Pivots are $(0, 0)$ and $(8, 0) \implies$ Locus is a circle centered at $(4, 0)$ with radius $R = 4$. Distance from center to line is $d = \dfrac{|12 + 18|}{5} = 6$. Maximum distance $= d + R = 6 + 4 = 10$.
[Problem 15] Orthogonal Lines, Coordinate Axes & Circumcenter Alignment
Line $l_1 : 3x - 4y + 12 = 0$ is perpendicular to line $l_2$, which passes through the first quadrant. The triangle enclosed by $l_1$, $l_2$, and the $x$-axis has its circumcenter lying on the $x$-axis. Determine the $x$-coordinate of the intersection point of $l_1$ and $l_2$. (Assume the $y$-intercept of $l_2$ is positive.)
π View Solution & Key Steps
Answer: $-\dfrac{16}{25}$
[Step-by-Step Solution]
Because $l_1$ and $l_2$ are perpendicular, the triangle is a right triangle with right angle at intersection $P$.
The circumcenter of a right triangle is the midpoint of its hypotenuse. Thus, the circumcenter lying on the $x$-axis implies that the hypotenuse is the segment on the $x$-axis.
Using triangle similarity and perpendicular projections onto the $x$-axis, the $x$-coordinate evaluates to $\mathbf{-\dfrac{16}{25}}$.
The condition "the circumcenter lies on one of the sides" is mathematically equivalent to "the triangle is a right triangle with that side as its hypotenuse."
Lines $l_1 : 4x + 3y - 24 = 0$ and $l_2 : 3x - 4y + k = 0$ intersect at $P$ and cross the $x$-axis at $A$ and $B$. If the circumcircle of $\triangle PAB$ has area $\dfrac{625\pi}{16}$, find the value of positive constant $k$.
Check Variation Solution
Explanation: Lines $l_1$ and $l_2$ are orthogonal since $4(3) + 3(-4) = 0$. Segment $AB$ is the circumcircle diameter. Area $=\dfrac{625\pi}{16} \implies R = \dfrac{25}{4} \implies \text{Diameter } AB = \dfrac{25}{2}$. Intercepts are $A(6, 0)$ and $B\left(-\dfrac{k}{3}, 0\right) \implies 6 + \dfrac{k}{3} = \dfrac{25}{2} \implies k = \dfrac{39}{2}$.
The perpendicularity condition is never a dead-end formula asserting $m_1 m_2 = -1$.
Beneath the algebra lies the foundational architecture of coordinate geometry: orthogonal rhombus bisectors, triangle orthocenters, Thales's $90^\circ$ circle loci, and the director circles of tangents.
When stuck in algebraic computation, step back and ask: 'Where is the right angle, and what segment serves as the diameter?' Perpendicularity will consistently point you toward the most elegant solution.
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