MathsISC Class 11 · Straight Lines

Slope, Angle between Lines and the Standard Forms of the Equation of a Line

Updated 17 October 2026 · Fact-checked

The slope of a line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁) = tanθ. The standard forms are y − y₁ = m(x − x₁) (point-slope), y = mx + c, x/a + y/b = 1 (intercept) and the two-point form. The angle between lines of slopes m₁ and m₂ satisfies tanθ = |(m₂ − m₁)/(1 + m₁m₂)|.

Slope is the rise divided by the run between any two points, and it equals tanθ, where θ is the angle the line makes with the positive x-axis.

Illustration drawn for QuestionBuddy.

Understand Slope, Angle and Equations of a Line

Slope. For a line inclined at θ to the positive x-axis, $m = tan θ$ . Through two points $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ . Horizontal lines have m = 0, vertical lines have no slope. Three points are collinear if the slopes of AB and BC are equal.

Forms of the equation. *Point-slope:* $y - y_{1} = m (x - x_{1})$ . *Slope-intercept:* $y = m x + c$ . *Two-point:* $y - y_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})$ . *Intercept:* $\frac{x}{a} + \frac{y}{b} = 1$ with x-intercept a and y-intercept b. *Normal form:* $x cos α + y sin α = p$ . Lines parallel to the axes: $y = k$ and $x = h$ .

Angle between two lines. If the slopes are $m_{1}$ and $m_{2}$ , the acute angle θ satisfies $tan θ = ∣ \frac{m _{2} - m _{1}}{1 + m _{1} m _{2}} ∣$ . Parallel lines: $m_{1} = m_{2}$ . Perpendicular lines: $m_{1} m_{2} = - 1$ .

Triangle problems. The median from a vertex passes through the midpoint of the opposite side; the altitude is perpendicular to the opposite side (use $m_{1} m_{2} = - 1$ ). The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ is $\frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ .

Strategy. Find the slope (or use intercepts), choose the form suited to the data, and simplify to $A x + B y + C = 0$ . Check that the given points satisfy the equation.

Slope, Angle and Equations of a Line: formulas to remember

Slope: $m = \tan θ = \frac{y_2 - y_1}{x_2 - x_1}$; Vertical lines have no slope.
Forms: $y - y_1 = m(x - x_1)$ | $y = mx + c$ | $\frac{x}{a} + \frac{y}{b} = 1$; Point-slope, slope-intercept, intercept.
Angle between lines: $\tan θ = |\frac{m_2 - m_1}{1 + m_1 m_2}|$; Perpendicular: $m_{1} m_{2} = - 1$ .
Triangle area: $\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$; Zero if the points are collinear.

Slope, Angle and Equations of a Line in real life

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.

Gradient of a road

Suppose a road rises 20 m over a horizontal distance of 150 m.

Find the slope, the angle of inclination and the equation of the road line y = mx + c taking the start as the origin.

Try it first, then show the solutionSolution

Slope $m = \frac{20}{150} = \frac{2}{15} \approx 0.1333$ .
Angle $θ = tan^{- 1} (0.1333) \approx 7.6°$ .
With c = 0 the equation is $y = \frac{2}{15} x$ .

Answer: m = 2/15; θ ≈ 7.6°; y = 2x/15.

Memory hook: Slope is rise over run.

Tricks to remember Slope, Angle and Equations of a Line

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

m = tan θ
The slope is the tangent of the inclination.
Use it when: Converting angles to slopes.
Perpendicular: negative reciprocal
If m₁m₂ = −1 the lines are at right angles.
Use it when: Altitudes and normals.

Slope, Angle and Equations of a Line: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope between two points · 1 mark · MCQBoard level

The slope of the line through (2, 3) and (4, 7) is:
(a) 1
(b) 2
(c) 3
(d) 1/2

Try it first, then show the solutionSolution

$\frac{7 - 3}{4 - 2} = 2$ .

Answer: (b) 2

Example 2: Equation and intercepts · 3 marks · Short answerBoard level

Find the equation of the line through (1, 2) with slope 3, and the intercepts made by the line $3 x + 4 y = 12$ on the axes.

Try it first, then show the solutionSolution

$y - 2 = 3 (x - 1)$ , i.e. $3 x - y - 1 = 0$ . (1 mark)
$3 x + 4 y = 12$ gives $\frac{x}{4} + \frac{y}{3} = 1$ . (1 mark)
x-intercept 4 and y-intercept 3. (1 mark)

Answer: 3x − y − 1 = 0; intercepts 4 and 3.

Example 3: Angle between two lines · 3 marks · Short answerHarder board level

Find the acute angle between the lines of slopes 2 and 3, and the angle between $y = x$ and $y = 3 x$ .

Try it first, then show the solutionSolution

$tan θ = ∣ \frac{3 - 2}{1 + 6} ∣ = \frac{1}{7}$ , so $θ = tan^{- 1} (\frac{1}{7}) \approx 8.13°$ . (1.5 marks)
For slopes 1 and √3: $tan θ = \frac{3 - 1}{1 + 3} = 2 - 3$ , which is $tan 15°$ . (1.5 marks)

Answer: ≈ 8.13°; 15°.

Example 4: Equal intercepts · 4 marks · Short answerHarder board level

Find the equation of the lines through (3, −2) which make equal intercepts on the axes (including the case of both intercepts zero).

Try it first, then show the solutionSolution

Equal non-zero intercepts: $\frac{x}{k} + \frac{y}{k} = 1$ , i.e. $x + y = k$ . (1 mark)
Through (3, −2): $k = 1$ , so $x + y = 1$ . (1.5 marks)
Both intercepts zero: the line passes through the origin and (3, −2): $y = - \frac{2}{3} x$ , i.e. $2 x + 3 y = 0$ . (1.5 marks)

Answer: x + y = 1 and 2x + 3y = 0.

Example 5: Triangle lines and area · 5 marks · Long answerHOTS / competency

The vertices of a triangle are A(1, 1), B(5, 3) and C(3, 7). Find the equation of the median through A, the altitude through A, and the area of the triangle.

Try it first, then show the solutionSolution

Midpoint of BC $= (4, 5)$ ; slope of median $= \frac{5 - 1}{4 - 1} = \frac{4}{3}$ : $4 x - 3 y - 1 = 0$ . (1.5 marks)
Slope of BC $= \frac{7 - 3}{3 - 5} = - 2$ ; altitude slope $= \frac{1}{2}$ : $y - 1 = \frac{1}{2} (x - 1)$ , i.e. $x - 2 y + 1 = 0$ . (1.5 marks)
Area $= \frac{1}{2} ∣1 (3 - 7) + 5 (7 - 1) + 3 (1 - 3) ∣ = \frac{1}{2} ∣ - 4 + 30 - 6∣ = 10$ . (2 marks)

Answer: Median 4x − 3y − 1 = 0; altitude x − 2y + 1 = 0; area 10.

Where students lose marks in Slope, Angle and Equations of a Line

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

Mistake: Taking slope as Δx/Δy
Inverting the ratio.
Do this: Slope = rise/run = Δy/Δx.
Usually costs: 2 marks.
Mistake: Forgetting the modulus in the angle formula
Getting a negative tan θ for the acute angle.
Do this: Use |…| for the acute angle.
Usually costs: 1 mark.
Mistake: Not simplifying to Ax + By + C = 0
Leaving fractions in the equation.
Do this: Clear fractions and write the general form.
Usually costs: 1 mark.

Exam tips

Sketch the line and points first.
Choose the form best suited to the data.
Check by substituting the given points.
State the final answer as Ax + By + C = 0.

Practice questions on Slope, Angle and Equations of a Line

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

Q1 · 1 mark · MCQBoard level
The slope of a line inclined at 45° to the x-axis is:
(a) 0
(b) 1
(c) √2
(d) undefined
Hint
m = tan 45°.
Answer
(b) 1
Q2 · 2 marks · Short answerBoard level
Find the equation of the line through (−1, 2) and (3, −4).
Hint
Two-point form.
Answer
Slope $= \frac{- 4 - 2}{3 + 1} = - \frac{3}{2}$ ; $y - 2 = - \frac{3}{2} (x + 1)$ , i.e. $3 x + 2 y - 1 = 0$ .
Q3 · 3 marks · Short answerHarder board level
Show that the points (1, 2), (3, 6) and (5, 10) are collinear and find the line.
Hint
Compare slopes.
Answer
Slope of AB $= 2$ , slope of BC $= 2$ : collinear on $y - 2 = 2 (x - 1)$ , i.e. $y = 2 x$ .
Q4 · 3 marks · Short answerHarder board level
Find the angle between the lines $x + y = 3$ and $x - 2 y = 1$ .
Hint
Slopes −1 and 1/2.
Answer
$m_{1} = - 1$ , $m_{2} = \frac{1}{2}$ : $tan θ = ∣ \frac{1/2 + 1}{1 - 1/2} ∣ = 3$ , $θ = tan^{- 1} 3 \approx 71.57°$ .
Q5 · 3 marks · Short answerHarder board level
A line makes an angle of 30° with the positive x-axis and passes through (2, 1). Find its equation.
Hint
m = tan 30°.
Answer
$m = \frac{1}{3}$ ; $y - 1 = \frac{1}{3} (x - 2)$ , i.e. $x - 3 y + 3 - 2 = 0$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a ramp runs from the ground point (0, 0) to a platform at (12, 3) in metres.
(i) Find the slope of the ramp. (1 mark)
(ii) Find the angle of inclination. (1 mark)
(iii) Find its equation. (1 mark)
(iv) Find the height at a horizontal distance of 8 m. (1 mark)
Hint
Rise over run.
Answer
(i) $\frac{3}{12} = \frac{1}{4}$ . (ii) $tan^{- 1} (0.25) \approx 14.04°$ . (iii) $y = \frac{x}{4}$ , i.e. $x - 4 y = 0$ . (iv) $y = 8/4 = 2$ m.

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.Slope in terms of angle?Show answer
tan θ.
2.Perpendicular lines?Show answer
m₁m₂ = −1.
3.Intercept form?Show answer
x/a + y/b = 1.
4.Area of triangle formula?Show answer
½|x₁(y₂ − y₃) + …|.

Slope, Angle and Equations of a Line: frequently asked questions

What is the slope of a vertical line?

It is undefined because Δx = 0.

How do I find the angle between lines when one is vertical?

Use the angle the other line makes with the vertical, or compute through the inclinations.

What does a zero area mean?

The three points are collinear.