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Lesson 4 of 5

Straight Lines · Lesson 4 of 5

Geometric Applications of Line Equations

“Combine slopes, intersections, midpoint conditions and distances to reason about reflections, areas, loci and routes.”

Learning Objectives

• Test concurrency by finding a common intersection. • Distinguish perpendicular distance from distance along a specified direction. • Construct a reflected point and explain a reflected path. • Find areas bounded by lines and a line determined by a midpoint condition. • Form distance loci with all absolute-value branches and restrictions. • Combine earlier methods in multi-step geometric applications.

Build the Geometry Before the Algebra

A longer problem usually combines familiar ingredients rather than requiring a new formula. First identify the unknown point or line, then list the relationships it must satisfy. An intersection gives simultaneous equations; perpendicularity gives a direction; reflection adds a midpoint condition; a locus asks for every possible point satisfying a condition.

Name unknown coordinates clearly and keep different points distinct. Use a sketch to show which segment lies on which line, then translate each geometric statement. After solving, substitute into the original conditions. This final check catches roots introduced by squaring, forgotten sign choices and a line that has the correct slope but passes through the wrong point.

Intersections and Concurrent Lines

Two nonparallel lines meet at the point satisfying both equations. Three or more lines are concurrent when they share one point. To test concurrency, find the intersection of two lines first, then require the remaining line to contain it.

Definition
Concurrent lines

Three or more lines passing through a common point. Distinct parallel lines cannot be part of a concurrent set.

Example — Determine a concurrency parameter

Problem
Find k if 2x+y−3=0, 5x+ky−3=0 and 3x−y−2=0 are concurrent.

  1. 1.Solve the first and third: 2x+y=3 and 3x−y=2.
  2. 2.Adding gives 5x=5, so x=1. Then y=1.
  3. 3.The middle line must contain (1,1): 5+k−3=0, giving k=−2.
  4. 4.Substitute (1,1) into all three equations with k=−2; each is satisfied.

A general concurrency relationship follows without memorising a determinant. Suppose the lines y=mᵢx+cᵢ share (u,v). Then cᵢ=v−mᵢu for i=1,2,3. Substitute these expressions into m₁(c₂−c₃)+m₂(c₃−c₁)+m₃(c₁−c₂). All terms cancel, giving zero. This proves the necessary condition in the source exercise. If all three slopes are equal, the same identity can hold for distinct parallel lines, so the identity by itself is not sufficient in that degenerate case.

Example — An axis-parallel line through an intersection

Problem
Find the vertical line through the intersection of x−7y+5=0 and 3x+y=0.

  1. 1.From the second equation y=−3x.
  2. 2.Substitute into the first: x+21x+5=0, so x=−5/22.
  3. 3.Then y=15/22. A vertical line preserves the intersection’s x-coordinate.
  4. 4.The requested equation is x=−5/22, not y=15/22.

For equal-intercept lines, use x+y=a through the computed intersection. For a line through a fixed point making an angle θ with a known finite slope m, solve the two branches of (s−m)/(1+sm)=±tan θ. They give s=(m±tan θ)/(1∓m tan θ), provided the denominators are nonzero. A zero denominator signals a vertical direction, which must be represented by x=constant rather than discarded.

Distance Along a Chosen Direction

The shortest distance is perpendicular, but a route may be constrained to another direction. Then the endpoint is where the allowed route meets the target line. Find that intersection and use the ordinary two-point distance; the perpendicular formula would answer a different question.

Example — An oblique route to a line

Problem
Find the distance from P=(4,1) to 4x−y=0 measured along a line inclined at 135°.

  1. 1.The allowed direction has slope tan 135°=−1. Through P its line is y−1=−(x−4), or x+y=5.
  2. 2.The target is y=4x. Solve with x+y=5 to get 5x=5, hence Q=(1,4).
  3. 3.The route length is PQ=√((1−4)²+(4−1)²)=3√2 units.
  4. 4.The perpendicular distance is |16−1|/√17=15/√17, which is smaller. The two values measure different segments.
An oblique routexyOP (4,1)Q (1,4)
An oblique route— The route x+y=5 through P meets y=4x at Q=(1,4). The coloured route is not perpendicular to the target.

If a specified route is parallel to a distinct target line, it never reaches the target and no finite route length exists. If the point already lies on the target, the distance is zero. These are geometric cases to decide before blindly solving a system with no unique intersection.

Example — Choose a direction from a length constraint

Problem
Through P=(−1,2), choose a line meeting x+y=4 at a point 3 units from P.

  1. 1.Let the meeting point be Q=(t,4−t). The length condition is (t+1)²+(2−t)²=9.
  2. 2.Expanding gives 2t²−2t+5=9, hence t²−t−2=0.
  3. 3.Therefore t=2 or t=−1. The possible points are (2,2) and (−1,5).
  4. 4.The corresponding lines through P are y=2 and x=−1. Their directions are horizontal and vertical.
  5. 5.Each route has length 3; a slope-only method could accidentally lose the vertical solution.

Reflection Uses a Perpendicular Bisector

A reflected point lies on the other side of the mirror line at the same perpendicular distance. The original point and its image are joined by a segment perpendicular to the mirror. The mirror crosses that segment at its midpoint, which supplies two independent conditions.

Example — Reflect a point across a line

Problem
Reflect P=(1,2) in x−3y+4=0.

  1. 1.Let the image be Q=(h,k). The mirror has slope 1/3, so PQ must have slope −3.
  2. 2.Use the perpendicular line through P: k−2=−3(h−1), giving 3h+k=5.
  3. 3.The midpoint ((h+1)/2,(k+2)/2) lies on the mirror: (h+1)/2−3(k+2)/2+4=0.
  4. 4.Multiply by 2 and collect terms: h−3k=−3.
  5. 5.Solve with 3h+k=5: k=5−3h, so h−15+9h=−3. Thus h=6/5 and k=7/5.
  6. 6.The image is (6/5,7/5). The two mirror expressions at P and Q are −1 and 1, confirming opposite sides and equal distances.
Mirror as perpendicular bisectorxyOP (1,4)Q (4,1)H
Mirror as perpendicular bisector— Across y=x, P=(1,4) and Q=(4,1) share perpendicular midpoint H=(2.5,2.5). This simpler picture illustrates the same construction.

Another method is to locate the perpendicular foot H first, then use Q=2H−P coordinate by coordinate. If P is already on the mirror, H=P and Q=P. Horizontal and vertical mirrors are easy special cases: reflection in y=0 changes (x,y) to (x,−y); reflection in x=0 changes it to (−x,y).

Example — A reflected ray on the x-axis

Problem
A ray from (1,2) reflects at A on the x-axis and then passes through (5,3). Find A.

  1. 1.Reflect the starting point P=(1,2) across the x-axis to P′=(1,−2).
  2. 2.Unfolding the reflection turns the two ray segments into the straight segment from P′ to Q=(5,3).
  3. 3.Its slope is (3+2)/(5−1)=5/4 and its equation is y+2=(5/4)(x−1).
  4. 4.At the mirror y=0, solve 2=(5/4)(x−1), giving x=13/5. Hence A=(13/5,0).
  5. 5.The horizontal mirror and the reflection construction ensure equal angles with the perpendicular normal at A. The hit point lies between x=1 and x=5, as expected.

Areas Bounded by Lines

A region bounded by three lines is a triangle when the lines produce three distinct vertices. Find their pairwise intersections first. Then use a suitable base and perpendicular height, or the coordinate area formula. This links an algebraic system to an actual bounded region.

Example — A general triangle area

Problem
Find the area bounded by y=m₁x+c₁, y=m₂x+c₂ and x=0.

  1. 1.Assume m₁≠m₂ so the sloping lines intersect. Their y-axis crossings are P=(0,c₁), Q=(0,c₂).
  2. 2.The vertical base has length |c₁−c₂|. At the third vertex R, (m₁−m₂)x=c₂−c₁, so xR=(c₂−c₁)/(m₁−m₂).
  3. 3.The perpendicular height to the vertical base is the horizontal distance |xR|.
  4. 4.Area=½|c₁−c₂|·|(c₂−c₁)/(m₁−m₂)|=(c₁−c₂)²/(2|m₁−m₂|).
  5. 5.For a genuine triangle also require c₁≠c₂. If c₁=c₂, all three meet at one point and the area is zero.
Example — A triangle with an axis-parallel side

Problem
Find the area enclosed by y=x, y=−x and x=k.

  1. 1.The two diagonally directed lines meet at O=(0,0).
  2. 2.Their meetings with x=k are (k,k) and (k,−k). The vertical base length is 2|k|.
  3. 3.The horizontal height from O to x=k is |k|.
  4. 4.Therefore area=½(2|k|)|k|=k² square units. The result is valid for negative k too; k=0 is a degenerate triangle.

A Line Fixed by a Midpoint Condition

Sometimes the endpoints of a segment lie on two known lines, while the segment’s midpoint is given. The endpoints are not known yet, but their coordinate averages are. Express one endpoint using the midpoint and the other endpoint, then impose the line conditions.

Example — A segment bisected at a given point

Problem
Find a line whose intersections with 5x−y+4=0 and 3x+4y−4=0 have midpoint M=(1,5).

  1. 1.Let the first endpoint be A=(u,v). Since M is the midpoint, the second is B=(2−u,10−v).
  2. 2.A lies on the first line, so v=5u+4.
  3. 3.B lies on the second: 3(2−u)+4(10−v)−4=0, so 3u+4v=42.
  4. 4.Substitute v=5u+4: 3u+20u+16=42, hence u=26/23 and v=222/23.
  5. 5.The other endpoint is B=(20/23,8/23). Both have midpoint (1,5).
  6. 6.The slope from M to A is (222/23−5)/(26/23−1)=(107/23)/(3/23)=107/3.
  7. 7.Thus y−5=(107/3)(x−1), giving 107x−3y−92=0. Both endpoints satisfy this equation.

A related ratio problem begins with an intersection. For the segment from A=(−1,1) to B=(5,7), write a point as (−1+6t,1+6t). On x+y=4 this gives 12t=4, so t=1/3 and the intersection is (1,3). The internal division ratio is t:(1−t)=1:2. Here 0<t<1 verifies that the point lies on the segment, rather than on its extension.

Loci and Absolute-Value Branches

A locus is the set of all points satisfying a stated condition. When that condition uses distances from lines, absolute values usually create multiple branches. Keeping every branch is necessary to describe the complete set, and each branch may have restrictions from its assumed signs.

Definition
Locus

The complete set of positions a point can occupy while satisfying a given condition. An equation found by algebra must still describe exactly those positions, including every valid branch.

Example — Points equidistant from two intersecting lines

Problem
Find the locus equidistant from 3x−2y=5 and 3x+2y=5.

  1. 1.For a variable point (x,y), equal distances give |3x−2y−5|/√13=|3x+2y−5|/√13.
  2. 2.Cancel the positive denominator. For |U|=|V|, either U=V or U=−V.
  3. 3.The first branch gives 3x−2y−5=3x+2y−5, hence y=0.
  4. 4.The second gives 3x−2y−5=−3x−2y+5, hence x=5/3.
  5. 5.The complete locus is the union of two straight lines, y=0 and x=5/3. Each passes through the original intersection (5/3,0).
  6. 6.These are the two angle-bisector lines. Calling the full locus just one straight line would miss half of the answer.
Example — The middle of two parallel lines

Problem
Find the line equidistant from 9x+6y−7=0 and 3x+2y+6=0.

  1. 1.Match coefficients by multiplying the second equation by 3, obtaining 9x+6y+18=0.
  2. 2.Let T=9x+6y. Equal distances require |T−7|=|T+18|.
  3. 3.The same-sign branch T−7=T+18 is impossible. The opposite branch gives T−7=−T−18.
  4. 4.Thus 2T=−11, so the middle line is 9x+6y+11/2=0, or 18x+12y+11=0.
  5. 5.Its constant is the average of −7 and 18 after coefficient matching. There is one middle parallel line because the original lines are distinct and parallel.

For two intersecting lines, a constant sum of perpendicular distances does not generally give one entire straight line. For example, the source condition is |x+y−5|/√2+|3x−2y+7|/√13=10. In each region where both inner expressions have fixed signs, remove the absolute values with those signs; the result is a linear equation restricted to that region. The full locus is assembled from the resulting line pieces. The region restrictions cannot be dropped, and the source’s single-line wording needs this qualification.

Common mistake

For |U|=|V|, keep U=V and U=−V. For |U|+|V|=constant, solve separate sign regions and retain the corresponding inequalities. Squaring or removing absolute values without tracking branches can add points or discard valid ones.

Some miscellaneous questions combine these methods with algebraic or trigonometric identities. For example, a line with parameter k must still have at least one nonzero x- or y-coefficient after substitution. To be horizontal its x-coefficient is zero and its y-coefficient remains nonzero; to be vertical the reverse holds; to pass through the origin its constant is zero. These tests are more reliable than dividing by a coefficient before checking whether it vanishes.

Example — Parameter conditions with valid-line checks

Problem
Classify values of k for (k−3)x−(4−k²)y+k²−7k+6=0 to be horizontal, vertical or through the origin.

  1. 1.The coefficients are A=k−3, B=k²−4, C=k²−7k+6.
  2. 2.Horizontal requires A=0, giving k=3. Then B=5≠0, so a genuine horizontal line remains.
  3. 3.Vertical requires B=0, giving k=2 or −2. In both cases A≠0, so both are valid.
  4. 4.Passing through the origin requires C=(k−1)(k−6)=0, giving k=1 or 6. Neither makes A and B both zero.
  5. 5.The coefficient checks are part of the solution, not an optional detail.
Example — A product of perpendicular distances

Problem
For a>b>0, let F₁=(√(a²−b²),0), F₂=(−√(a²−b²),0). Show that their distances to (x/a)cos θ+(y/b)sin θ=1 have product b².

  1. 1.Let c=√(a²−b²). The line coefficients are A=cos θ/a, B=sin θ/b and C=−1.
  2. 2.The product of the two distance numerators is |(c cos θ/a−1)(−c cos θ/a−1)|=|1−c²cos²θ/a²|.
  3. 3.This expression is positive and equals (a²sin²θ+b²cos²θ)/a².
  4. 4.The common denominator squared is cos²θ/a²+sin²θ/b²=(b²cos²θ+a²sin²θ)/(a²b²).
  5. 5.Dividing the numerator product by that squared denominator gives b². This is an application of the line-distance formula; no additional curve theory is needed.
Example — The shortest route from a junction

Problem
The junction is where 2x−3y+4=0 and 3x+4y−5=0 meet. Find the line of the shortest route to 6x−7y+8=0.

  1. 1.Solve 2x−3y=−4 and 3x+4y=5. Multiply by 4 and 3 and add: 17x=−1, so x=−1/17.
  2. 2.Substitution gives y=22/17. Thus the junction is J=(−1/17,22/17).
  3. 3.The target has slope 6/7, so its perpendicular has slope −7/6.
  4. 4.Through J, y−22/17=−(7/6)(x+1/17). Multiply through to get 119x+102y−125=0.
  5. 5.This is the shortest geometric route because it is perpendicular to the target. It also minimises travel time if speed is constant and there are no route restrictions.
Example — A trigonometric two-point distance

Problem
Find the distance from the origin to the line joining (cos θ,sin θ) and (cos φ,sin φ), when these are distinct points.

  1. 1.Put u=(θ+φ)/2 and v=(θ−φ)/2. The angles are in a consistent angular unit.
  2. 2.Each endpoint satisfies x cos u+y sin u=cos v, by the cosine-difference identity. Since the points are distinct, this is their joining line.
  3. 3.The denominator for distance is √(cos²u+sin²u)=1.
  4. 4.Therefore the distance is |cos v|=|cos((θ−φ)/2)|. Distinct antipodal points give zero, as their joining line passes through the origin.
Example — Lines equally inclined to another line

Problem
Find m if y=3x+1 and 2y=x+3 are equally inclined to y=mx+4.

  1. 1.Only directions matter here. The two known slopes are 3 and 1/2.
  2. 2.For nonperpendicular cases equate the smaller-angle tangent magnitudes: |(m−3)/(1+3m)|=|(m−1/2)/(1+m/2)|.
  3. 3.The same-sign branch would equate two different line directions and produces no real m. The opposite-sign branch yields (m−3)(1+m/2)=−(m−1/2)(1+3m).
  4. 4.Expanding and collecting gives 7m²−2m−7=0.
  5. 5.Thus m=(1+5√2)/7 or (1−5√2)/7. These are the two perpendicular bisector directions; their product is −1.
  6. 6.Check separately m=−1/3 and m=−2, where a displayed denominator vanishes: neither makes both smaller angles equal. A vertical candidate also fails because the two known lines have unequal angles to the vertical.
Example — Axis-parallel legs of a right triangle

Problem
The hypotenuse has endpoints (1,3) and (−4,1). Find possible axis-parallel leg lines.

  1. 1.One leg is horizontal and one is vertical. The right-angle vertex must combine the x-coordinate of one endpoint with the y-coordinate of the other.
  2. 2.One possibility is (1,1), with legs x=1 and y=1.
  3. 3.The other is (−4,3), with legs x=−4 and y=3.
  4. 4.Both constructions connect the given endpoints as the hypotenuse and give perpendicular axis-parallel legs.

Quiz

Quick check

Three concurrent lines:

Quick check

A mirror line relative to a point and its reflection is their:

Quick check

For |U|=|V|, which branches are necessary?

Quick check

Distance along a specified direction is found by:

Quick check

The full locus equidistant from two distinct intersecting lines is:

Quick check

Reflection of (3,8) in the x-axis is:

Key Takeaways

Key Takeaways

• Use simultaneous equations for intersections and concurrency. • A constrained route length is a point-to-point distance after locating the route intersection. • Reflection combines a perpendicular direction with a midpoint on the mirror. • Areas bounded by lines come from intersection vertices and perpendicular heights. • A midpoint condition can determine unknown endpoints and then their connecting line. • Distance loci require all branches and their sign-region restrictions. • The shortest unrestricted route to a line is perpendicular to it.

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Distances from Points and Between Lines

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Chapter Summary and Practice