Straight Lines ISC Class 11: Equations, Distances and Locus

What Straight Lines is about

Weightage: Straight Lines belongs to the unit Coordinate Geometry, which CISCE weights at 20 of the 80 marks in the ISC Class XI Mathematics Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Mathematics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Recall: section formula, area and collinearity

The point dividing the join of A(x₁, y₁) and B(x₂, y₂) internally in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)); externally it is ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)). The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is ½|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|. Three points are collinear exactly when this area is zero.

Worked example

Find the points dividing the join of A(1, 2) and B(7, 10) in the ratio 1 : 3 (i) internally (ii) externally.

Try it first, then show the solutionSolution

1. (i) ((1 × 7 + 3 × 1)/4, (1 × 10 + 3 × 2)/4) = (2.5, 4)
2. (ii) ((7 − 3)/(1 − 3), (10 − 6)/(1 − 3)) = (−2, −2)
3. Check (ii): AP = √(9 + 16) = 5 and PB = √(81 + 144) = 15, ratio 1 : 3 ✓

Answer: (2.5, 4); (−2, −2)

Incentre of a triangle

The incentre is where the internal angle bisectors meet; it is the centre of the inscribed circle. If the sides opposite A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) have lengths a = BC, b = CA, c = AB, the incentre is I = ((ax₁ + bx₂ + cx₃)/(a + b + c), (ay₁ + by₂ + cy₃)/(a + b + c)): each vertex is weighted by the length of the opposite side. The inradius is r = Area/s, where s is the semi-perimeter.

Worked example

Find the incentre and inradius of the triangle A(0, 0), B(4, 0), C(0, 3).

Try it first, then show the solutionSolution

1. a = BC = 5, b = CA = 3, c = AB = 4; a + b + c = 12
2. I = ((5 × 0 + 3 × 4 + 4 × 0)/12, (5 × 0 + 3 × 0 + 4 × 3)/12) = (1, 1)
3. Area = 6, s = 6, r = 1 (consistent: (1, 1) is 1 unit from both axes)

Answer: I(1, 1), r = 1

Shifting of origin

If the origin is moved to O′(h, k) without turning the axes, a point with old coordinates (x, y) has new coordinates (X, Y) where x = X + h, y = Y + k (equivalently X = x − h, Y = y − k). Substituting these into an equation gives the equation of the same curve in the new system. Choosing (h, k) well removes first-degree terms and reveals the shape.

Worked example

Find the new equation of x² + y² − 4x + 6y − 3 = 0 when the origin is shifted to (2, −3).

Try it first, then show the solutionSolution

1. Put x = X + 2, y = Y − 3
2. (X + 2)² + (Y − 3)² − 4(X + 2) + 6(Y − 3) − 3 = X² + Y² + (4 + 9 − 8 − 18 − 3) = X² + Y² − 16
3. New equation X² + Y² = 16

Answer: X² + Y² = 16 (a circle of radius 4 about the new origin)

Slope and the angle between two lines

The slope (gradient) of a non-vertical line is m = tan θ, where θ is its inclination to the positive x-axis (0 ≤ θ < 180°, θ ≠ 90°); through two points, m = (y₂ − y₁)/(x₂ − x₁). Vertical lines have no slope. For two lines of slopes m₁ and m₂, the acute angle θ between them satisfies tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. Lines are parallel if m₁ = m₂ and perpendicular if m₁m₂ = −1 (or one is vertical and the other horizontal).

Worked example

Find the acute angle between lines of slopes 2 and −3.

Try it first, then show the solutionSolution

1. tan θ = |(2 − (−3))/(1 + 2 × (−3))| = |5/(−5)| = 1
2. θ = 45°

Answer: 45°

Standard forms of the equation of a line

Parallel to axes: x = a, y = b. Point-slope: y − y₁ = m(x − x₁). Slope-intercept: y = mx + c. Two-point: y − y₁ = ((y₂ − y₁)/(x₂ − x₁))(x − x₁). Intercept: x/a + y/b = 1 (a, b the x- and y-intercepts, not for lines through the origin). Normal form: x cos α + y sin α = p, where p > 0 is the length of the perpendicular from the origin and α is the angle it makes with the positive x-axis. To change Ax + By + C = 0 into normal form, move C to the right so that it is positive and divide by √(A² + B²).

Worked example

Find the line through (2, 3) and (−1, −3) and write it in intercept form. Also put √3x + y = 8 in normal form.

Try it first, then show the solutionSolution

1. Slope = (−3 − 3)/(−1 − 2) = 2; y − 3 = 2(x − 2) ⇒ 2x − y − 1 = 0
2. Intercept form: 2x − y = 1 ⇒ x/(1/2) + y/(−1) = 1
3. √3x + y = 8: divide by √(3 + 1) = 2: (√3/2)x + (1/2)y = 4, i.e. x cos 30° + y sin 30° = 4

Answer: 2x − y − 1 = 0, intercepts 1/2 and −1; p = 4, α = 30°

General equation; parallel and perpendicular lines

Every line can be written Ax + By + C = 0 (A, B not both zero). Its slope is −A/B, its x-intercept −C/A and its y-intercept −C/B. A line parallel to it is Ax + By + k = 0; a line perpendicular to it is Bx − Ay + k = 0; find k from a given point. Two lines A₁x + B₁y + C₁ = 0 and A₂x + B₂y + C₂ = 0 are parallel if A₁/A₂ = B₁/B₂ and perpendicular if A₁A₂ + B₁B₂ = 0.

Worked example

Find the line through (1, −2) perpendicular to 3x − 4y + 7 = 0.

Try it first, then show the solutionSolution

1. Perpendicular family: 4x + 3y + k = 0
2. Through (1, −2): 4 − 6 + k = 0 ⇒ k = 2

Answer: 4x + 3y + 2 = 0

Distance from a line and between parallel lines

The perpendicular distance of P(x₁, y₁) from Ax + By + C = 0 is d = |Ax₁ + By₁ + C|/√(A² + B²). For parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 (same A and B; rescale first if needed) the distance is |C₁ − C₂|/√(A² + B²). The sign of Ax₁ + By₁ + C tells you which side of the line P is on: points on the same side give the same sign.

Worked example

Find (i) the distance of (−1, 4) from 5x + 12y − 4 = 0 (ii) the distance between 3x − 4y + 7 = 0 and 6x − 8y − 6 = 0.

Try it first, then show the solutionSolution

1. (i) |−5 + 48 − 4|/√(25 + 144) = 39/13 = 3
2. (ii) Rescale: 6x − 8y − 6 = 0 ⇒ 3x − 4y − 3 = 0
3. |7 − (−3)|/5 = 2

Answer: 3; 2

Angle bisectors and families of lines

A point on a bisector of the angles between L₁: A₁x + B₁y + C₁ = 0 and L₂: A₂x + B₂y + C₂ = 0 is equidistant from both, so the bisectors are (A₁x + B₁y + C₁)/√(A₁² + B₁²) = ±(A₂x + B₂y + C₂)/√(A₂² + B₂²). The two bisectors are perpendicular. With both constants C₁, C₂ made positive, the '+' sign gives the bisector of the angle containing the origin. Family: L₁ + λL₂ = 0 represents every line through the intersection of L₁ = 0 and L₂ = 0 (except L₂ itself); one more condition fixes λ.

Worked example

(i) Find the bisectors of the angles between x + 2y − 3 = 0 and 2x + y + 1 = 0. (ii) Find the line through the intersection of x + y − 5 = 0 and 2x − y − 1 = 0 that passes through the origin.

Try it first, then show the solutionSolution

1. (i) Both have √5 in the denominator: x + 2y − 3 = ±(2x + y + 1)
2. '+': x − y + 4 = 0; '−': 3x + 3y − 2 = 0 (slopes 1 and −1: perpendicular ✓)
3. (ii) (x + y − 5) + λ(2x − y − 1) = 0 through (0, 0): −5 − λ = 0, λ = −5
4. −9x + 6y = 0 ⇒ 3x − 2y = 0 (it does pass through the intersection (2, 3) ✓)

Answer: (i) x − y + 4 = 0 and 3x + 3y − 2 = 0 (ii) 3x − 2y = 0

Locus and its equation

A locus is the set of all points (the path) satisfying a given geometric condition. To find its equation: let P(x, y) be any point on it, write the condition using distance, slope or area formulas, and simplify. The result, with no other letters left except constants, is the equation of the locus. For example, points equidistant from two fixed points form their perpendicular bisector.

Worked example

Find the locus of a point that is equidistant from A(1, 2) and B(3, −4).

Try it first, then show the solutionSolution

1. PA² = PB²: (x − 1)² + (y − 2)² = (x − 3)² + (y + 4)²
2. −2x + 1 − 4y + 4 = −6x + 9 + 8y + 16
3. 4x − 12y − 20 = 0

Answer: x − 3y − 5 = 0 (the perpendicular bisector of AB)

Maths in real life: Straight Lines

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Distance from a Line, Angle Bisectors and Locus
• Slope, Angle and Equations of a Line

Straight Lines formula sheet

Section formula

$\left(\frac{m{x}_2\pm n{x}_1}{m\pm n},\frac{m{y}_2\pm n{y}_1}{m\pm n}\right)$

+ internal, − external.

Area of a triangle

$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

Zero ⇔ collinear.

Incentre

$\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

a, b, c opposite A, B, C.

Angle between lines

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Parallel m₁ = m₂; perpendicular m₁m₂ = −1.

Forms of a line

$y-y_1=m(x-x_1);y=mx+c;\frac{x}{a}+\frac{y}{b}=1;x\cos \alpha +y\sin \alpha =p$

General form Ax + By + C = 0, slope −A/B.

Distance of a point from a line

$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

Parallel lines: |C₁ − C₂|/√(A² + B²).

Angle bisectors

$\frac{A_{1}x+B_{1}y+C_1}{\sqrt{A_1^2+B_1^2}}=\pm \frac{A_{2}x+B_{2}y+C_2}{\sqrt{A_2^2+B_2^2}}$

With C₁, C₂ > 0, '+' contains the origin.

Family through an intersection

$L_1+\lambda L_2=0$

Find λ from one extra condition.

Straight Lines: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Straight Lines

The question patterns to expect, and the method that works for each.

Tricks to remember Straight Lines

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Swap and change one sign

  Perpendicular to Ax + By + C = 0: swap A and B and change one sign: Bx − Ay + k = 0.

  Use it when: Perpendicular lines.

• Opposite side weights

  In the incentre, each vertex is weighted by the side opposite it.

  Use it when: Incentre formula.

• Positive p

  Normal form needs p > 0: make the constant positive before dividing.

  Use it when: Normal form.

• Plug in, modulus, root

  Distance = |plug the point in| ÷ √(A² + B²).

  Use it when: Point-to-line distance.

• Lambda glues lines

  L₁ + λL₂ = 0 is every line through the meeting point.

  Use it when: Families of lines.

Straight Lines: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Straight Lines

The slips examiners see most often, and what to write instead.

• ✗Mistake: Dropping the modulus in the angle formula

  Without |…| you may get an obtuse angle's tangent.

  Wrong working

  Slopes 2, −3: tan θ = 5/(−5) = −1 ⇒ θ = 135°.

  Right working

  Acute angle: tan θ = 1 ⇒ 45°.

  ✓Do this: Use the modulus for the acute angle; for lines at a given angle use ±.

  Usually costs: 1 mark.

• ✗Mistake: Negative p in normal form

  Dividing without first making the constant positive.

  Wrong working

  √3x − y + 6 = 0 ⇒ (√3/2)x − (1/2)y = −3.

  Right working

  −√3x + y = 6 ⇒ p = 3, α = 150°.

  ✓Do this: p is a length, so it must be positive.

  Usually costs: 1-2 marks.

• ✗Mistake: Unequal coefficients for parallel-line distance

  Students use |C₁ − C₂| directly for 3x − 4y + 7 = 0 and 6x − 8y − 6 = 0.

  Wrong working

  |7 + 6|/5 = 2.6.

  Right working

  Rescale to 3x − 4y − 3 = 0: |7 + 3|/5 = 2.

  ✓Do this: Make A and B identical first.

  Usually costs: 1-2 marks.

• ✗Mistake: Wrong weights in the incentre

  Students weight each vertex by an adjacent side.

  Wrong working

  Using c for A because c = AB touches A.

  Right working

  A is weighted by a = BC, the side opposite A.

  ✓Do this: Opposite side, always.

  Usually costs: Full marks for the part.

• ✗Mistake: Sign errors in external division

  Students use m + n in the denominator.

  Wrong working

  External 1 : 3 of A(1, 2), B(7, 10): (10/4, 16/4).

  Right working

  ((7 − 3)/(1 − 3), (10 − 6)/(1 − 3)) = (−2, −2).

  ✓Do this: External: minus signs in numerator and denominator.

  Usually costs: 1 mark.

Practice questions on Straight Lines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The slope of the line through (2, 5) and (−1, −1) is:
   (a) 1/2
   (b) 2
   (c) −2
   (d) 3

   Hint

   Rise over run.

   Answer

   (b): (−1 − 5)/(−1 − 2) = 2

2. Q2 · 1 mark · MCQExam level

   The distance of the origin from 3x + 4y = 10 is:
   (a) 2
   (b) 10
   (c) 5/2
   (d) 10/7

   Hint

   |−10|/5.

   Answer

   (a)

3. Q3 · 1 mark · Very short answerExam level

   Write the x-intercept of 4x − 5y = 20.

   Hint

   Put y = 0.

   Answer

   5

4. Q4 · 1 mark · Very short answerExam level

   Find the area of the triangle with vertices (0, 0), (4, 0), (0, 6).

   Hint

   Right angle at the origin.

   Answer

   12 square units

5. Q5 · 2 marks · Short answerExam level

   Write the equations of the lines through (−3, 7) parallel to (i) the y-axis (ii) the x-axis.

   Hint

   x = a, y = b.

   Answer

   (i) x = −3 (ii) y = 7

6. Q6 · 2 marks · Short answerHarder exam level

   Find the point dividing the join of (−1, 7) and (4, −3) internally in the ratio 2 : 3.

   Hint

   (mx₂ + nx₁)/(m + n).

   Answer

   ((8 − 3)/5, (−6 + 21)/5) = (1, 3)

7. Q7 · 2 marks · Short answerHarder exam level

   Find the acute angle between y = 3x + 2 and x + 2y − 4 = 0.

   Hint

   Slopes 3 and −1/2.

   Answer

   tan θ = |(3 + 1/2)/(1 − 3/2)| = 7, θ = tan⁻¹ 7 ≈ 81.9°

8. Q8 · 3 marks · Short answerHarder exam level

   Shift the origin to (2, −3) and find the new form of 2x² + 2y² − 8x + 12y + 1 = 0.

   Hint

   x = X + 2, y = Y − 3.

   Answer

   The first-degree terms vanish: 2X² + 2Y² + (8 + 18 − 16 − 36 + 1) = 0 ⇒ 2X² + 2Y² − 25 = 0

9. Q9 · 3 marks · Short answerHarder exam level

   Find the incentre of the triangle with vertices A(0, 0), B(6, 0), C(0, 8).

   Hint

   a = BC = 10, b = CA = 8, c = AB = 6.

   Answer

   I = ((0 + 8 × 6 + 0)/24, (0 + 0 + 6 × 8)/24) = (2, 2)

10. Q10 · 2 marks · Short answerHarder exam level

    Find the line through (3, −2) perpendicular to the line joining (1, 4) and (2, −1).

    Hint

    Joining slope −5.

    Answer

    Slope 1/5: y + 2 = (x − 3)/5 ⇒ x − 5y − 13 = 0

11. Q11 · 3 marks · Short answerHarder exam level

    Find the foot of the perpendicular from (1, 2) to 3x + 4y − 1 = 0.

    Hint

    The perpendicular through (1, 2) is 4x − 3y + 2 = 0.

    Answer

    Solve 3x + 4y = 1 and 4x − 3y = −2: x = −1/5, y = 2/5. Foot (−1/5, 2/5).

12. Q12 · 3 marks · Short answerHarder exam level

    Find the equation of the locus of a point that is equidistant from (−2, 1) and (4, 5).

    Hint

    PA² = PB².

    Answer

    (x + 2)² + (y − 1)² = (x − 4)² + (y − 5)² ⇒ 4x + 1 − 2y + 4 = −8x + 16 − 10y + 25 ⇒ 12x + 8y − 36 = 0 ⇒ 3x + 2y − 9 = 0

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Find the line through the intersection of x − 2y + 3 = 0 and 3x + y − 5 = 0 parallel to 2x − 3y = 0. [2]
    (b) Find the distance between this line and 2x − 3y + 17 = 0. [2]
    (c) Find the area of the triangle it forms with the coordinate axes. [1]

    Hint

    The lines meet at (1, 2).

    Answer

    (a) 2x − 3y + k = 0 through (1, 2): k = 4, so 2x − 3y + 4 = 0. (b) |4 − 17|/√13 = √13. (c) Intercepts −2 and 4/3: area ½ × 2 × 4/3 = 4/3.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the points on the x-axis whose distance from 4x + 3y = 12 is 4 units.

    Hint

    Let the point be (h, 0).

    Answer

    |4h − 12|/5 = 4 ⇒ 4h − 12 = ±20 ⇒ h = 8 or −2: (8, 0) and (−2, 0)

15. Q15 · 3 marks · Short answerHOTS / competency

    Find the image of the point (2, 9) in the line x + 3y − 7 = 0.

    Hint

    The midpoint lies on the line and the join is perpendicular to it.

    Answer

    Image (2 + t, 9 + 3t) with midpoint (2 + t/2, 9 + 3t/2) on the line: 2 + t/2 + 27 + 9t/2 − 7 = 0 ⇒ 5t = −22 ⇒ t = −4.4. Image (−2.4, −4.2).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.External division formula?Show answer

   ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n)).

2. 2.Condition for collinearity?Show answer

   Area of the triangle formed is zero.

3. 3.Slope of Ax + By + C = 0?Show answer

   −A/B.

4. 4.Normal form?Show answer

   x cos α + y sin α = p, p > 0.

5. 5.Perpendicular distance from a point to a line?Show answer

   |Ax₁ + By₁ + C|/√(A² + B²).

6. 6.Line perpendicular to Ax + By + C = 0?Show answer

   Bx − Ay + k = 0.

7. 7.Lines through the intersection of L₁ = 0 and L₂ = 0?Show answer

   L₁ + λL₂ = 0.

8. 8.Shifting the origin to (h, k)?Show answer

   x = X + h, y = Y + k.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you divide a segment internally and externally in a given ratio?
• ☐Can you find the incentre of a triangle?
• ☐Can you transform an equation by shifting the origin?
• ☐Can you find the angle between two lines and lines at a given angle to a line?
• ☐Can you write a line in intercept form and in normal form?
• ☐Can you find the distance between two parallel lines with different coefficients?
• ☐Can you find the bisector containing the origin?
• ☐Can you use L₁ + λL₂ = 0 with one extra condition?

Straight Lines: frequently asked questions