Straight Lines · Lesson 5 of 5
Chapter Summary and Practice
“Connect the whole chapter through a method-selection guide, mixed worked examples and reasoning checks.”
• Connect coordinates, slope, line equations, intersections and distances. • Select methods from the information supplied rather than memorised question patterns. • Recall each major formula with its conditions and special cases. • Solve mixed problems involving line constructions, areas and reflections. • Verify results using equation membership, geometry and distance checks.
One Connected Story
Coordinates describe points; slope describes direction; a line equation combines location with direction. Intersections locate shared points, and perpendicular constructions measure shortest lengths. More demanding applications use several of these ideas together, so revision should rebuild their connections rather than separate them into isolated formulas.
Start a problem by deciding whether the unknown is a point, a line, an angle, a length, an area or a set of positions. Next identify what is actually given: endpoints, a slope, intercepts, a parallel or perpendicular relationship, a midpoint or a distance condition. Translate the simplest condition first. A useful equation is one whose quantities you understand and whose assumptions are satisfied.
| Chapter idea | Core relationship | Condition or interpretation |
|---|---|---|
| Two-point distance | √((x₂−x₁)²+(y₂−y₁)²) | Length is nonnegative |
| Internal division and midpoint | Weighted coordinates; equal weights give the midpoint | Keep the ratio order consistent |
| Triangle area | ½|x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)| | Zero area indicates collinearity |
| Slope and inclination | m=(y₂−y₁)/(x₂−x₁)=tan θ | Nonvertical line; signed ratio |
| Parallel and perpendicular directions | m₁=m₂; m₁m₂=−1 | Separate coincident lines and vertical–horizontal cases |
| Smaller angle | tan θ=|(m₂−m₁)/(1+m₁m₂)| | Zero denominator means a right angle |
| Line through point and slope | y−y₀=m(x−x₀) | Use x=x₀ for a vertical line |
| Two-point line | (x₂−x₁)(y−y₁)=(y₂−y₁)(x−x₁) | The points must be distinct |
| Slope and intercepts | y=mx+c; y=m(x−d); x/a+y/b=1 | a,b nonzero for intercept form |
| General equation | Ax+By+C=0 | A,B not both zero |
| Point-to-line distance | |Ax₁+By₁+C|/√(A²+B²) | Perpendicular length |
| Parallel-line distance | |C₁−C₂|/√(A²+B²) | Match A and B first |
| Intersections and concurrency | Solve line equations simultaneously | Common point satisfies every equation |
| Reflection | Mirror is perpendicular bisector | Foot is midpoint of point and image |
| Areas and midpoint-constrained lines | Find vertices or endpoints before measuring | Check degeneracies |
| Loci and route applications | Distance branches; route intersections | Retain signs, restrictions and geometric meaning |
Direction, Position and Membership
Direction alone does not determine a line. Infinitely many parallel lines share the same slope. A point, an intercept or another location condition is needed to place the line. Conversely, one point alone allows infinitely many possible directions.
Before calculating a slope from two points, ask whether their x-coordinates agree. Before taking a negative reciprocal, ask whether the given line is horizontal or vertical. Before treating equal slopes as parallel distinct lines, check whether the equations actually describe the same line. Before calling three points collinear by slope, remember the vertical case, which is more directly recognised by equal x-coordinates.
Problem
Find the line through the midpoint of A=(−2,1), B=(4,5), perpendicular to AB, then find its intercepts.
- 1.The midpoint is M=(1,3). The slope of AB is (5−1)/(4+2)=2/3.
- 2.The perpendicular slope is −3/2. Through M, y−3=−(3/2)(x−1).
- 3.Multiply by 2: 2y−6=−3x+3, giving 3x+2y−9=0.
- 4.Set y=0 to get x-intercept 3; set x=0 to get y-intercept 9/2.
- 5.The equation is also x/3+y/(9/2)=1. Check M: 3+6−9=0.
- 6.The perpendicular direction and midpoint location show that this is the perpendicular bisector of AB.
Select an Equation Form from the Given Facts
The various forms are different views of the same line. A point–slope form makes its anchor and direction visible. Slope–intercept form makes graphing and linear modelling convenient. General form treats vertical lines and prepares the equation for distance calculations.
To move from general form to slope–intercept form, isolate y when its coefficient is nonzero. To obtain intercepts, set the other coordinate to zero and solve. To build a median, find the opposite side’s midpoint and join it to the vertex. To build a perpendicular through a dividing point, use the section formula before choosing the new direction. To test collinearity, construct a line from two distinct points and check the third.
Problem
Find the lines through (3,2) making a smaller angle of 45° with x−2y=3.
- 1.The reference line is y=x/2−3/2, so its slope is 1/2.
- 2.Solve |(s−1/2)/(1+s/2)|=1 to get s=3 or s=−1/3.
- 3.The first line is y−2=3(x−3), giving 3x−y−7=0.
- 4.The second is y−2=−(x−3)/3, giving x+3y−9=0.
- 5.Both pass through (3,2); each makes the required smaller angle. They are themselves perpendicular, since 3(−1/3)=−1.
For a physical linear model, retain the units of the slope and describe the assumption that makes the relationship linear. A rod’s length change per degree and a store’s litres per rupee are different interpretations of the same slope ratio. A model prediction outside the observed input range is an extrapolation and depends on the relationship continuing.
Distance, Height and Scale
A line equation can be multiplied by any nonzero number without changing the line. This is why a distance cannot be read from a raw substituted value alone. The coefficient denominator compensates for the equation’s scale and returns a geometric length.
Problem
For P=(4,3), find the distance to x+y−2=0, its perpendicular foot H, and its reflected image Q.
- 1.The target has slope −1, so the perpendicular through P has slope 1: y−3=x−4, or y=x−1.
- 2.Solve with x+y=2: x+x−1=2, so H=(3/2,1/2).
- 3.Distance from the formula is |4+3−2|/√2=5/√2. The two-point distance PH gives the same value.
- 4.Since H is the midpoint of P and Q, Q=2H−P=(3−4,1−3)=(−1,−2).
- 5.The line expression at P is 5 and at Q is −5, so they are on opposite sides at equal distances.
- 6.The midpoint of P and Q is indeed H, and PQ has slope 1, perpendicular to the mirror slope −1.
Problem
Find the separation and middle line of 2x−y+1=0 and 4x−2y−6=0.
- 1.Divide the second equation by 2 to obtain 2x−y−3=0. The coefficients now match.
- 2.Separation is |1−(−3)|/√5=4/√5.
- 3.For a variable point, equal distances give |2x−y+1|=|2x−y−3|.
- 4.The same-sign branch is impossible. The opposite branch gives 2(2x−y)=2, so the middle line is 2x−y−1=0.
- 5.Its separation from either original line is 2/√5, exactly half the total separation.
Mixed Geometry: Intersections First
When a problem gives several lines, identify the points they create before measuring a region or describing a path. The intersection of two equations gives a vertex, junction or mirror hit point. Its interpretation decides what you do next.
Problem
Find the area enclosed by x=0, y=0 and 3x+4y=12, then the distance from the origin to the sloping side.
- 1.The axis vertices are O=(0,0), A=(4,0), B=(0,3).
- 2.Using OA as base and OB as height gives area=½×4×3=6 square units.
- 3.The sloping side has length AB=√(16+9)=5. Using AB as base, its perpendicular height from O must satisfy 6=½×5×d.
- 4.Therefore d=12/5. The distance formula independently gives |−12|/√(3²+4²)=12/5.
- 5.The area is the same whichever base–height pair is used; the height must always be perpendicular to the chosen base.
Concurrency asks whether all relevant equations accept the same point. A midpoint condition links two unknown endpoints by their coordinate averages. A reflection adds a perpendicular bisector. A shortest path to a line uses the perpendicular direction; a prescribed-direction path instead uses the intersection along that chosen line. These are different geometric conditions, even when the algebra uses similar simultaneous equations.
Verify Branches and Special Cases
A plausible equation is not a complete answer until all its conditions have been checked. Absolute values can give two branches; a division can exclude a valid vertical direction; a construction can collapse to a single point. Verification should return to the original geometry, not just repeat the last algebraic step.
| Tempting shortcut | Why it fails | Reliable check |
|---|---|---|
| Set vertical slope to zero | Zero describes a horizontal direction | Check whether Δx=0 |
| Use one sign of an angle or distance equation | A second valid direction or location may exist | Solve both sign branches |
| Subtract unmatched parallel-line constants | The equations may use different scales | Match both variable coefficients |
| Call an equidistant intersecting-line locus one line | There are two angle-bisector lines | Use both U=V and U=−V |
| Remove absolute values in a distance sum globally | Signs change between regions | Retain each region’s sign inequalities |
| Use intercept form for a line through the origin | Nonzero intercept denominators are required | Use point–slope or general form |
| Use perpendicular distance for an oblique route | The route endpoint is constrained | Find its actual intersection |
| Find triangle area without checking vertices | Parallel or concurrent boundaries can degenerate | Check three distinct intersections |
There is no separate practice-problem block in these lessons. The mixed worked examples and the checks below consolidate the chapter; the dedicated solved-exercise collection can provide the full exercise sequence. When reviewing an answer there, explain why its starting equation matches the given facts before following the calculation.
Quiz
The line through (0,2) parallel to 3x−y+5=0 is:
The perpendicular bisector of (0,0) and (4,0) is:
Distance between 3x+4y+1=0 and 6x+8y−8=0 is:
A point reflected across x=2 changes (5,1) to:
Which claim always holds for a point on Ax+By+C=0?
The triangle enclosed by x=0, y=0 and x+y=6 has area:
The locus equidistant from x=0 and y=0 is:
A point-to-line distance formula is useful for finding:
Key Takeaways
• Coordinates, slope and line equations describe location and direction together. • Choose a form from the given information, then verify equation membership. • Perpendicularity links right angles, shortest distances, altitudes and reflection. • Parallel-line distances require matching equation scales. • Intersections and midpoint relations turn geometric conditions into simultaneous equations. • Distance and angle problems may have multiple branches and valid vertical cases. • A complete answer includes conditions, units and checks against the original geometry.
Previous · Lesson 4
Geometric Applications of Line Equations
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