Formulas · Mathematics

Straight lines

8 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 11

Slope from two points

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

What each letter means
$m$	the slope (gradient) of the line
$x_{1}$	the x-coordinate of the first point
$y_{1}$	the y-coordinate of the first point
$x_{2}$	the x-coordinate of the second point
$y_{2}$	the y-coordinate of the second point

The slope of a line is how much it rises for each step across: the change in y divided by the change in x between any two of its points.

Why it works

Between two points, the line rises y₂ − y₁ while it moves x₂ − x₁ across. A straight line rises at the same rate everywhere, so this ratio is the same for any two points on it.

When to use it

Finding how steep a line is, testing whether three points are on one line (equal slopes), and telling parallel lines (equal slopes) and perpendicular lines (slopes multiplying to −1).

How to use it

Take the y's away, in the order second minus first.
Take the x's away, in the same order.
Divide. A vertical line (x₂ = x₁) has no slope.

To remember: Rise over run.

Other forms

Rearranged $y_{2} = y_{1} + m (x_{2} - x_{1})$ The second point's height from the first point and the slope.

Worked examples

Story

A road climbs from a point 200 m above sea level to one 260 m above sea level over 1200 m across. What is its slope?

$\frac{260 - 200}{1200 - 0} = \frac{1}{20}$

Answer: 0.05

Picture

Show that (1, 1), (3, 5) and (5, 9) lie on one line.

Slope of the first two: (5 − 1)/(3 − 1) = 2. Slope of the last two: (9 − 5)/(5 − 3) = 2.
$\frac{5 - 1}{3 - 1} = \frac{9 - 5}{5 - 3}$

Answer: Both slopes are 2, so the points lie on one line

Direct

Find the slope of the line through (1, 2) and (4, 11).

$\frac{11 - 2}{4 - 1} = 3$

Answer: 3

Reverse

A line through (2, 3) has slope 4 and passes through a point with x = 5. Find its y.

Try it first, then show the working

y₂ = y₁ + m(x₂ − x₁)
$3 + 4 (5 - 2) = 15$

Answer: 15

Exam

Find k if the line through (k, 3) and (2, 7) has slope 2.

Try it first, then show the working

(7 − 3)/(2 − k) = 2, so 2 − k = 2 and k = 0.
$\frac{7 - 3}{2 - 0} = 2$

Answer: k = 0

Common mistake: Mixing the order: (y₂ − y₁) over (x₁ − x₂) gives the wrong sign. Keep the same point first on the top and the bottom.

Class 11

Slope from the angle with the x-axis

m = tan θ

What each letter means
$m$	the slope of the line
$θ$	the angle the line makes with the positive x-axis, measured anticlockwise (degrees)

A line's slope is the tangent of the angle it makes with the positive x-axis.

Why it works

For a step of 1 across, the line rises m. In the right triangle with that step and that rise, tan θ = rise ÷ run = m ÷ 1.

When to use it

Turning an angle into a slope (a line at 60° has slope √3) and a slope into an angle.

How to use it

Measure θ anticlockwise from the positive x-axis.
The slope is tan θ; a line at more than 90° has a negative slope.

To remember: Slope is the tangent of the tilt.

Other forms

To remember itSlope 1 is 45°, slope √3 is 60°, slope 0 is flat, and a vertical line (90°) has no slope.

Worked examples

Story

A ramp makes 30° with the floor. How much does it rise for each metre across?

$tan \frac{π}{6} = \frac{1}{3}$
About 0.577 m for each metre across.

Answer: 0.5773502691896257

Picture

A line makes 135° with the positive x-axis. Find its slope.

$tan \frac{3 π}{4} = - 1$

Answer: -1

Direct

Find the slope of a line at 60° to the x-axis.

$tan \frac{π}{3} = 3$

Answer: 1.7320508075688772

Reverse

A line has slope 1. What angle does it make with the x-axis?

Try it first, then show the working

tan θ = 1 and θ is between 0° and 180°, so θ = 45°.
$tan \frac{π}{4} = 1$

Answer: 45°

Exam

Find the slope of the line through (3, −2) and (7, −2), and the angle it makes with the x-axis.

Try it first, then show the working

Slope (−2 − (−2))/(7 − 3) = 0: the line is flat.
tan θ = 0, so θ = 0°.
$\frac{- 2 + 2}{7 - 3} = 0$

Answer: Slope 0, angle 0°

Common mistake: Measuring the angle from the y-axis, or clockwise.

Class 11

Slope of a perpendicular line

m_{2} = - \frac{1}{m _{1}}

What each letter means
$m_{2}$	the slope of the perpendicular line
$m_{1}$	the slope of the first line (not 0)

Two lines (neither vertical) are perpendicular exactly when their slopes multiply to −1: each slope is the negative reciprocal of the other.

Why it works

Turning a line through a right angle turns its step (1 across, m up) into (m across, −1 up) or (−m across, 1 up), so the new slope is −1/m.

When to use it

Finding the slope of a perpendicular (an altitude of a triangle, a perpendicular bisector), and testing whether two lines meet at a right angle.

How to use it

Turn the slope upside down.
Change its sign.

To remember: Flip it and flip the sign.

Other forms

Same formula, another form $m_{1} m_{2} = - 1$ The test: perpendicular slopes multiply to −1.

Worked examples

Story

A path has slope ⅓. A fence is built at right angles to it. What is the fence's slope?

$- \frac{1}{\frac{1}{3}} = - 3$

Answer: -3

Picture

Does the triangle with corners (0, 0), (4, 2) and (3, 4) have a right angle at (4, 2)?

Slope from (4, 2) to (0, 0) is ½; from (4, 2) to (3, 4) is −2.
$\frac{1}{2} \times (- 2) = - 1$
The slopes multiply to −1, so the angle at (4, 2) is a right angle.

Answer: Yes, a right angle at (4, 2)

Direct

A line has slope 2. Find the slope of a line perpendicular to it.

$- \frac{1}{2} = - 0.5$

Answer: -0.5

Reverse

A line is perpendicular to one of slope −4/3. Find its slope.

Try it first, then show the working

$- \frac{1}{- \frac{4}{3}} = \frac{3}{4}$

Answer: 0.75

Exam

Find k if the line through (2, 3) and (4, k) is perpendicular to the line y = −2x + 1.

Try it first, then show the working

The given slope is −2, so the perpendicular slope is ½.
(k − 3)/(4 − 2) = ½, so k − 3 = 1 and k = 4.
$\frac{4 - 3}{4 - 2} = \frac{1}{2}$

Answer: k = 4

Common mistake: Only changing the sign (that gives a reflection, not a perpendicular) or only turning it upside down.

Class 11

Point-slope form of a line

y = y_{1} + m (x - x_{1})

What each letter means
$y$	the height of the line at x
$x$	any x-coordinate
$m$	the slope
$x_{1}$	the x-coordinate of the known point
$y_{1}$	the y-coordinate of the known point

A line is fixed by one of its points and its slope. The book writes it y − y₁ = m(x − x₁).

Why it works

For any point (x, y) on the line, the slope from (x₁, y₁) to (x, y) is m: (y − y₁)/(x − x₁) = m. Multiplying out gives the form.

When to use it

Writing the equation of a line when you know one point and the slope (a tangent, an altitude, a line parallel to another).

How to use it

Put the known point and slope into y − y₁ = m(x − x₁).
Multiply out and tidy it into y = mx + c or Ax + By + C = 0 as the question asks.

To remember: From the point, rise m for every step.

Other forms

Same formula, another form $y - y_{1} = m (x - x_{1})$ The form in the book.

Worked examples

Story

A taxi charges ₹60 for 2 km and ₹15 for each further km. Using the line through (2, 60) with slope 15, find the fare for 10 km.

$60 + 15 (10 - 2) = 180$

Answer: 180

Picture

Find the equation of the line through (−2, 3) with slope −4.

y − 3 = −4(x + 2), so y = −4x − 5.
$3 - 4 (0 + 2) = - 5$

Answer: y = −4x − 5

Direct

The line through (2, 3) with slope 4: find y when x = 5.

$3 + 4 (5 - 2) = 15$

Answer: 15

Reverse

On the line through (1, 5) with slope −2, at which x is y = −1?

Try it first, then show the working

−1 = 5 − 2(x − 1), so x − 1 = 3.
$5 - 2 (4 - 1) = - 1$

Answer: 4

Exam

Find the equation of the line through (3, 4) perpendicular to 2x + y = 7.

Try it first, then show the working

The given line has slope −2, so the perpendicular has slope ½.
y − 4 = ½(x − 3), which is x − 2y + 5 = 0.
$4 + \frac{1}{2} (3 - 3) = 4$

Answer: x − 2y + 5 = 0

Common mistake: Putting the point in the wrong places: x₁ goes with x, and y₁ with y.

Class 11

Two-point form of a line

y = y_{1} + \frac{y _{2} - y _{1}}{x _{2} - x _{1}} (x - x_{1})

What each letter means
$y$	the height of the line at x
$x$	any x-coordinate
$x_{1}$	the x-coordinate of the first point
$y_{1}$	the y-coordinate of the first point
$x_{2}$	the x-coordinate of the second point
$y_{2}$	the y-coordinate of the second point

The line through two points: the point-slope form, with the slope worked out from the two points.

Why it works

The slope is (y₂ − y₁)/(x₂ − x₁); putting it into y − y₁ = m(x − x₁) gives the two-point form.

When to use it

Writing the equation of the line through two given points (a side of a triangle, a median).

How to use it

Find the slope from the two points.
Use either point with it in the point-slope form, and tidy.

To remember: Two points give the slope; one point places the line.

Other forms

Same formula, another form $\frac{y - y _{1}}{y _{2} - y _{1}} = \frac{x - x _{1}}{x _{2} - x _{1}}$ The symmetric form.

Worked examples

Story

A candle is 20 cm tall after 1 hour and 14 cm after 4 hours. Using the line through (1, 20) and (4, 14), how tall is it after 6 hours?

$20 + \frac{14 - 20}{4 - 1} (6 - 1) = 10$

Answer: 10

Picture

Find the equation of the side through (−1, 2) and (3, 6) of a triangle.

Slope (6 − 2)/(3 + 1) = 1.
y − 2 = 1(x + 1), so y = x + 3.
$2 + \frac{6 - 2}{3 + 1} (0 + 1) = 3$

Answer: y = x + 3

Direct

On the line through (1, −1) and (3, 5), find y when x = 4.

$- 1 + \frac{5 + 1}{3 - 1} (4 - 1) = 8$

Answer: 8

Reverse

On the line through (0, 2) and (4, 10), at which x is y = 14?

Try it first, then show the working

Slope 2, so 14 = 2 + 2x.
$2 + \frac{10 - 2}{4 - 0} (6 - 0) = 14$

Answer: 6

Exam

Find the equation of the median through A(2, 3) of the triangle with B(4, 1) and C(6, 5).

Try it first, then show the working

The midpoint of BC is (5, 3).
The line through (2, 3) and (5, 3) has slope 0: y = 3.
$\frac{3 - 3}{5 - 2} = 0$

Answer: y = 3

Common mistake: Mixing the order of the points on top and bottom of the slope.

Class 11

Intercept form of a line

y = b (1 - \frac{x}{a})

What each letter means
$y$	the height of the line at x
$x$	any x-coordinate
$a$	where the line cuts the x-axis (the x-intercept)
$b$	where the line cuts the y-axis (the y-intercept)

A line that cuts the x-axis at a and the y-axis at b is x/a + y/b = 1; here it is solved for y.

Why it works

The line passes through (a, 0) and (0, b). Both satisfy x/a + y/b = 1, and the equation is of the first degree, so it is exactly that line.

When to use it

Writing a line from its intercepts, finding the intercepts of a line quickly, and the area of the triangle a line makes with the axes (½ab).

How to use it

Read a and b, or find them (put y = 0 for a, x = 0 for b).
Write x/a + y/b = 1.

To remember: x over its intercept plus y over its intercept makes one.

Other forms

Same formula, another form $\frac{x}{a} + \frac{y}{b} = 1$ The form in the book.

Worked examples

Story

A ladder touches the floor 3 m from a wall and the wall 4 m up. Using the line with intercepts 3 and 4, how high is it 1.5 m from the wall?

$4 (1 - \frac{1.5}{3}) = 2$

Answer: 2

Picture

Find the area of the triangle that 2x + 3y = 12 makes with the axes.

Divide by 12: x/6 + y/4 = 1, so a = 6 and b = 4.
$\frac{1}{2} \times 6 \times 4 = 12$

Answer: 12 square units

Direct

A line cuts the axes at x = 4 and y = 6. Find y when x = 2.

$6 (1 - \frac{2}{4}) = 3$

Answer: 3

Reverse

A line has x-intercept 5 and passes through (2, 6). Find its y-intercept.

Try it first, then show the working

6 = b(1 − 2/5) = 3b/5, so b = 10.
$10 (1 - \frac{2}{5}) = 6$

Answer: 10

Exam

Find the line through (3, 4) whose intercepts are equal.

Try it first, then show the working

x/a + y/a = 1, so x + y = a. Through (3, 4): a = 7.
The line is x + y = 7.
$3 + 4 = 7$

Answer: x + y = 7

Common mistake: Swapping a and b: a goes under x, since it is where the line meets the x-axis.

Class 11

Distance from a point to a line

d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}

What each letter means
$d$	the shortest (perpendicular) distance from the point to the line
$A$	the coefficient of x in Ax + By + C = 0
$B$	the coefficient of y
$C$	the constant term
$x_{1}$	the x-coordinate of the point
$y_{1}$	the y-coordinate of the point

Put the point into the line's equation Ax + By + C, take the size of the result, and divide by √(A² + B²).

Why it works

The point's value Ax₁ + By₁ + C measures how far it is off the line, but stretched by √(A² + B²), the length of the line's normal direction (A, B). Dividing removes the stretch, and the bars make the distance positive.

When to use it

The distance of a point from a line, the height of a triangle from a vertex, and checking whether a line touches a circle (distance = radius).

How to use it

Write the line as Ax + By + C = 0.
Put the point in: Ax₁ + By₁ + C, and drop any minus sign.
Divide by √(A² + B²).

To remember: Put the point in, drop the sign, divide by the root of A² + B².

Other forms

Special case $d = \frac{∣ C ∣}{A ^{2} + B ^{2}}$ When the point is the origin.

Worked examples

Story

A straight road runs along 4x + 3y − 12 = 0 (in km). How far is a house at (6, 4) from the road?

$\frac{∣24 + 12 - 12∣}{16 + 9} = \frac{24}{5}$

Answer: 4.8

Picture

Does the line 3x + 4y = 25 touch the circle of radius 5 centred at the origin?

Distance of (0, 0) from 3x + 4y − 25 = 0:
$\frac{∣0 + 0 - 25∣}{9 + 16} = 5$
The distance equals the radius, so the line touches the circle.

Answer: Yes, it touches the circle

Direct

Find the distance of (3, −5) from the line 3x − 4y − 26 = 0.

$\frac{∣9 + 20 - 26∣}{9 + 16} = \frac{3}{5}$

Answer: 0.6

Reverse

The line 12x + 5y + C = 0 is 2 units from the origin, with C negative. Find C.

Try it first, then show the working

|C|/√(144 + 25) = |C|/13 = 2, so |C| = 26 and C = −26.
$\frac{∣ - 26∣}{144 + 25} = 2$

Answer: C = −26

Exam

Find the length of the altitude from A(1, 2) to the side through B(4, 6) and C(7, 2) of a triangle.

Try it first, then show the working

The line BC: slope −4/3, so 4x + 3y − 34 = 0.
$\frac{∣4 + 6 - 34∣}{16 + 9} = \frac{24}{5}$

Answer: 4.8 units

Common mistake: Forgetting to move everything to one side first (C must be on the left), or forgetting the square root.

Class 11

Distance between two parallel lines

d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}

What each letter means
$d$	the distance between the two parallel lines
$A$	the coefficient of x (the same in both lines)
$B$	the coefficient of y (the same in both lines)
$C_{1}$	the constant of the first line, Ax + By + C₁ = 0
$C_{2}$	the constant of the second line, Ax + By + C₂ = 0

Two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 are |C₁ − C₂|/√(A² + B²) apart.

Why it works

Take any point on the first line and find its distance to the second with the point-to-line formula: its value on the first line is 0, so Ax + By = −C₁, and the distance works out to |C₂ − C₁|/√(A² + B²).

When to use it

The width of a strip between two parallel lines, and the distance between opposite sides of a parallelogram.

How to use it

Write both lines with the same A and B (multiply one through if needed).
Take the constants away, drop the sign, and divide by √(A² + B²).

To remember: Same A and B first; then the gap between the C's over the root.

Other forms

To remember itCheck: if C₁ = C₂ the lines are the same, and the distance is 0.

Worked examples

Story

The two edges of a straight canal lie along 5x + 12y − 13 = 0 and 5x + 12y + 26 = 0 (in metres). How wide is the canal?

$\frac{∣ - 13 - 26∣}{25 + 144} = 3$

Answer: 3

Picture

Opposite sides of a parallelogram lie on 2x + y − 4 = 0 and 2x + y + 6 = 0. How far apart are they?

$\frac{∣ - 4 - 6∣}{4 + 1} = \frac{10}{5}$
That is 2√5, about 4.47.

Answer: 4.47213595499958

Direct

Find the distance between 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.

$\frac{∣7 - 5∣}{9 + 16} = \frac{2}{5}$

Answer: 0.4

Reverse

A line parallel to 4x + 3y + 2 = 0 is 3 units from it, on the side with the larger constant. Find its equation.

Try it first, then show the working

|C − 2|/5 = 3, so C − 2 = 15 and C = 17: the line is 4x + 3y + 17 = 0.
$\frac{∣17 - 2∣}{16 + 9} = 3$

Answer: 4x + 3y + 17 = 0

Exam

Find the distance between the parallel lines l(x + y) + p = 0 and l(x + y) − r = 0.

Try it first, then show the working

Write them as x + y + p/l = 0 and x + y − r/l = 0 (A = B = 1).
d = |p/l + r/l|/√2 = |p + r|/(√2 |l|).
For example, p = 3, r = 1 and l = 2 give:
$\frac{∣ \frac{3}{2} + \frac{1}{2} ∣}{2} = 2$

Answer: |p + r|/(√2 |l|)

Common mistake: Using the constants before making A and B the same in both lines.