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Cambridge IGCSE0606

Straight-line graphs

Mathematics Additional 0606 Chapter Notes

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Straight-line graphs
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In this chapter

  1. 1The equation of a straight line
  2. 2Parallel and perpendicular lines
  3. 3Midpoint and length of a line
  4. 4The perpendicular bisector
  5. 5Points on the perpendicular bisector
  6. 6Straight-line form
  7. 7Finding constants from the straight line

1. The equation of a straight line

A straight line has a gradient m and a y-intercept c. Its equation is y = mx + c.

Given the gradient and one point on the line, use y - y₁ = m(x - x₁). Given two points, find the gradient first.

A line written as ax + by = k must be rearranged to y = mx + c before its gradient is read off.

y = mx + c

m = (y₂ - y₁)/(x₂ - x₁)

y - y₁ = m(x - x₁)

Examiner insight

“Nearly all candidates correctly found the gradient of the line with only a few having the x and y coordinates reversed.”

Worked example 13 marks

Find the equation of the line through the points P(-1, 7) and Q(3, -5). Give your answer in the form y = mx + c.

  1. 1

    gradient m = (-5 - 7)/(3 - (-1)) = (-12)/4 = -3 [1]

  2. 2

    y - 7 = -3(x + 1) [1]

  3. 3

    y = -3x + 4 [1]

2. Parallel and perpendicular lines

Two lines are parallel when their gradients are equal. Two lines are perpendicular when the product of their gradients is -1.

So a line perpendicular to one of gradient m has gradient -1/m: turn the fraction upside down and change the sign.

To show two lines are perpendicular, find both gradients and show their product is -1.

m₁ = m₂ (parallel)

m₁ m₂ = -1 (perpendicular)

Examiner insight

“When attempting to find the gradient of the perpendicular line, some candidates only changed the sign of AB’s gradient instead of using the negative reciprocal.” “The most commonly omitted of the three required elements was the proof of perpendicularity.”

Worked example 12 marks

The point A is the minimum point on the curve y = x² - 6x + 10 and has coordinates (3, 1). The point B has coordinates (6, 7). The curve cuts the y-axis at the point C.

Find the equation of the line through B parallel to the line AC.

  1. 1

    C is (0, 10), where x = 0

  2. 2

    gradient of AC = (10 - 1)/(0 - 3) = -3 [1]

  3. 3

    parallel, so the gradient is -3: y - 7 = -3(x - 6) [1]

3. Midpoint and length of a line

The midpoint of a line joining two points has the mean of the x-coordinates and the mean of the y-coordinates.

The length of the line comes from the difference in x and the difference in y with Pythagoras' theorem. Squaring removes any minus sign.

To compare two lengths, compare their squares first.

midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)

length = √((x₂ - x₁)² + (y₂ - y₁)²)

Examiner insight

“Occasional slips were seen in finding the mid-point as some candidates subtracted the two corresponding coordinates before halving.” “A few candidates made sign errors when finding the mid-point or when forming the equation.”

Worked example 14 marks

Points A and B have coordinates A(-2, 3) and B(8, -3). The point D has coordinates (2, -1). The lengths of AD and BD are such that BD = (√n)/2AD where n is an integer.

Find the value of n.

  1. 1

    AD = √((2+2)² + (-1-3)²) = √32 [1]

  2. 2

    BD = √((8-2)² + (-3+1)²) = √40 [1]

  3. 3

    √40 = (√n)/2√32, so 40 = n/4 × 32 [1]

  4. 4

    n = 5 [1]

4. The perpendicular bisector

The perpendicular bisector of AB cuts AB at right angles at its midpoint.

Find the midpoint of AB. Find the gradient of AB, then the perpendicular gradient. Put the midpoint and that gradient into y - y₁ = m(x - x₁).

A short sketch of the points helps you see which point the line must pass through.

Examiner insight

“The bisector part of the question was often misunderstood or misread and a perpendicular through A or B was given.” “In a few cases the equation of the perpendicular through A was found instead of that through the mid-point of OA and a few other candidates found the equation of the line rather than the perpendicular.”

Worked example 13 marks

Points A and B have coordinates A(-2, 3) and B(8, -3). The perpendicular bisector of AB passes through the point C(0, c).

Find the value of c.

  1. 1

    midpoint of AB = ((-2+8)/2, (3-3)/2) = (3, 0) [1]

  2. 2

    gradient of AB = (-3-3)/(8+2) = -3/5, so the perpendicular gradient is 5/3 [1]

  3. 3

    y - 0 = 5/3(x - 3); at x = 0, c = -5 [1]

5. Points on the perpendicular bisector

When a point on the perpendicular bisector has one unknown coordinate, put the known coordinate into the equation of the bisector to find the other.

The mark scheme also credits an alternative method: a point on the bisector is the same distance from A and from B, so equate the squared distances.

Examiner insight

“The main error was the use of the coordinates of the points A or B, rather than the midpoint to find the equation of the perpendicular bisector.” “Other errors were usually arithmetic, for example, an incorrect midpoint coordinate or gradient.”

Worked example 14 marks

Points A and B have coordinates (1, 3) and (−7, 1). The point P with coordinates (2,a) lies on the perpendicular bisector of AB.

Find the value of a.

  1. 1

    midpoint of AB = (-3, 2) [1]

  2. 2

    gradient of AB = (3-1)/(1+7) = 1/4, so the perpendicular gradient is -4 [1]

  3. 3

    y - 2 = -4(x + 3) [1]

  4. 4

    at x = 2: y = 2 - 4 × 5, so a = -18 [1]

6. Straight-line form

Some relationships are curves. Take logarithms or choose new variables to turn them into Y = mX + c.

For y = Axⁿ, take logs: lg y = lg A + nlg x. Plot lg y against lg x. For y = Ab^(x): ln y = ln A + xln b. Plot ln y against x.

The gradient and the intercept of the straight line hold the unknown constants. Plot the transformed values, not the given ones.

y = Axⁿ → lg y = nlg x + lg A

y = Ab^(x) → ln y = xln b + ln A

y² = Ax³ + B → Y = y², X = x³

Examiner insight

“There were some errors when applying the logarithmic form of the given equation, allocating the values of A and of b incorrectly.” “Inevitably there were candidates who just plotted the given points thus obtaining a curve.”

Worked example 13 marks

The variables x and y are related by y = 3x⁴. State what must be plotted to give a straight line. State the gradient and the intercept of the line.

  1. 1

    lg y = lg 3 + 4lg x [1]

  2. 2

    plot lg y against lg x [1]

  3. 3

    gradient = 4, intercept = lg 3 on the lg y axis [1]

7. Finding constants from the straight line

Find the gradient from two points on the line. Find the intercept by putting one point into Y = mX + c.

Then undo the logarithm to write the original equation. Where Y is ln y, the intercept is ln A, so A = e^(intercept).

When a question says use the graph, use the straight-line graph and its gradient and intercept.

Examiner insight

“It was important that candidates take note of the instruction in the question to use their graph.” “Some candidates realised that they needed to find the gradient, but often did not know what this represented.”

Worked example 16 marks

It is given that x and y are variables. When ln 3y is plotted against x² a straight line is obtained. This straight line passes through the points (4.64, 12.24) and (6.84, 8.94).

Show that y = ke^(ax²) where k and a are exact constants.

  1. 1

    ln 3y = mx² + c [1]

  2. 2

    m = (12.24 - 8.94)/(4.64 - 6.84) = -1.5 [1]

  3. 3

    12.24 = -1.5 × 4.64 + c [1]

  4. 4

    c = 19.2 [1]

  5. 5

    ln 3y = 19.2 - 1.5x², so 3y = e^(19.2 - 1.5x²) [1]

  6. 6

    y = (e^(19.2))/3 e^(-1.5x²) [1]

Recap

  • Straight line: y = mx + c, gradient m = (y₂ - y₁)/(x₂ - x₁).
  • Parallel lines: equal gradients. Perpendicular lines: m₁ m₂ = -1.
  • Midpoint: add the coordinates, then halve.
  • Length: √((x₂ - x₁)² + (y₂ - y₁)²).
  • Perpendicular bisector: through the midpoint, with the perpendicular gradient.
  • To straight-line form: take logs, then read gradient and intercept.
  • y = Axⁿ gives lg y against lg x. y = Ab^(x) gives ln y against x.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State the gradient and the y-intercept of the line y = 4 - 3x.2 marks
  2. Find the equation of the line through (2, −1) and (6, 7). Give your answer in the form y = mx + c.3 marks
  3. Do not use a calculator. Find the equation of the line through (4, −2) that is parallel to the line 3x + 2y = 7.3 marks
  4. Do not use a calculator. The line y = kx + 1 is perpendicular to the line 4x + 3y = 9. Find the value of k.3 marks
  5. Do not use a calculator. The points P and Q have coordinates (−5, 3) and (7, −2). Find the midpoint of PQ and the length of PQ.3 marks
  6. The point M(3, -1) is the midpoint of the line AB. The point A has coordinates (−2, 4). Find the coordinates of B.3 marks
  7. Do not use a calculator. The variables x and y are related by y = Axⁿ. When lg y is plotted against lg x, a straight line with gradient 3 passes through the point (0, 2). Find y in terms of x.3 marks
  8. The points A and B have coordinates (−2, 5) and (4, 1). Find the equation of the perpendicular bisector of AB. Give your answer in the form y = mx + c.5 marks
  9. Do not use a calculator. On a map with units in kilometres, two masts stand at A(-3, 4) and B(5, 0). A straight cable joins A to B. A straight road lies along the perpendicular bisector of AB. A second straight road has equation x + y = 9. The two roads cross at the point C. Find the coordinates of C. Find also the exact shortest distance from C to the cable.7 marks
  10. The number P of bacteria in a dish after t hours is given by P = Ab^(t). When ln P is plotted against t, a straight line passes through (1, 3.0) and (5, 5.0). Find A and b, each to 3 significant figures. Find P when t = 8 to the nearest whole number. Find also the value of t, to 3 significant figures, when P = 5000.8 marks

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