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AI HL · 12.3

Coordinate Geometry

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12.3.2

Straight lines: gradient, parallel and perpendicular

12.3.2

Straight lines: gradient, parallel and perpendicular

Target: I can find the gradient and both intercepts of a line from its equation, and write it as y = mx + c or ax + by + d = 0. (+4 more)

By the end of this section

Key points

Forms of a line

  • \(y = mx + c\): m is the gradient, c is the y-intercept.
  • \(ax + by + d = 0\) has gradient \(-\frac ab\). Put \(y = 0\) to find the x-intercept and \(x = 0\) to find the y-intercept.
  • Through two points: gradient \(m = \frac{y_2-y_1}{x_2-x_1}\), then use one point to find c.

Parallel and perpendicular

  • Parallel lines have equal gradients: \(m_1 = m_2\).
  • Perpendicular lines have gradients that multiply to −1: \(m_1 \times m_2 = -1\).
  • A line parallel to \(ax + by + d = 0\) keeps the same a and b: only d changes. Find d by putting in the point.
  • Three points are on one line if the gradient from the first to the second equals the gradient from the second to the third.

Lines as models

  • In a model \(y = mx + c\), the gradient is the change in y for each 1 unit of x, and c is the value when x is 0 (a fixed charge or a starting amount).
  • Give units with the gradient: euros per day, metres per second.

On the calculator

Solving two equations together

  • Write each line in the form ax + by = c.
  • Use the calculator to solve the pair for x and y.
  • The solution is the point where the two lines cross.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Gradient and intercepts from the general form

A line has equation \(5x + 2y - 20 = 0\). Find its gradient, its y-intercept and its x-intercept.

Working
  1. Make y the subject

    \(2y = -5x + 20\), so \(y = -\frac52 x + 10\)

    Move the x term and the number to the right, then divide by 2.

  2. Read the gradient and y-intercept

    Gradient \(= -\frac52\), y-intercept \(= 10\)

    In \(y = mx + c\), m is the gradient and c is where the line crosses the y-axis.

  3. The x-intercept

    Put \(y = 0\): \(5x - 20 = 0\), so \(x = 4\)

    On the x-axis, y is 0. A line \(ax + by + d = 0\) always has gradient \(-\frac ab\).

Video walkthrough coming
A cost model from two prices

A bike hire shop charges a fixed fee plus the same amount for each day. Hiring for 3 days costs €52 and hiring for 7 days costs €96. Find the cost for each extra day and the fixed fee, then the cost of hiring for 10 days.

Working
  1. Gradient

    \(m = \frac{96-52}{7-3} = 11\)

    The gradient is the change in cost for one more day: €11 per day.

  2. Intercept

    \(52 = 11 \times 3 + c\), so \(c = 19\)

    The intercept is the cost of 0 days: the fixed fee, €19.

  3. The model

    \(C = 11d + 19\)

    A straight-line model: a rate times the number of days, plus a fixed amount.

  4. Ten days

    \(C = 11 \times 10 + 19 = 129\)

    The cost is €129.

Video walkthrough coming
A parallel line through a point

The line L has equation \(3x - 2y = 8\). Find the equation of the line parallel to L that passes through \((4, -5)\). Give your answer in the form \(y = mx + c\).

Working
  1. Gradient of L

    \(2y = 3x - 8\), so \(y = \frac32 x - 4\). Gradient \(= \frac32\)

    Rearrange to find the gradient.

  2. Same gradient

    Parallel line: \(y = \frac32 x + c\)

    Parallel lines never meet, so they have equal gradients: \(m_1 = m_2\).

  3. Use the point

    \(-5 = \frac32 (4) + c\), so \(c = -11\)

    Substitute the point to find c.

  4. The line

    \(y = \frac32 x - 11\)

    In the form \(3x - 2y - 22 = 0\) it has the same x and y coefficients as L: only the number changes.

Video walkthrough coming
Finding k for perpendicular lines

The lines \(6x - ky + 1 = 0\) and \(2x + 3y - 9 = 0\) are perpendicular. Find the value of k.

Working
  1. The two gradients

    First line: \(y = \frac6k x + \frac1k\), gradient \(\frac6k\). Second line: gradient \(-\frac23\)

    Write each line as \(y = mx + c\), or use \(-\frac ab\).

  2. Perpendicular

    \(\frac6k \times \left(-\frac23\right) = -1\), so \(-\frac4k = -1\)

    Perpendicular lines: \(m_1 \times m_2 = -1\).

  3. Solve

    \(k = 4\)

    For parallel lines the equation would be \(\frac6k = -\frac23\), giving \(k = -9\).

Video walkthrough coming
Is it a right angle?

The points \(A(-1, 2)\), \(B(3, 4)\) and \(C(5, 0)\) are the corners of a triangle. Show that angle ABC is a right angle.

ABC
Working
  1. Gradient of AB

    \(\frac{4-2}{3-(-1)} = \frac12\)

    Change in y over change in x.

  2. Gradient of BC

    \(\frac{0-4}{5-3} = -2\)

    Take the points in the same order on top and bottom.

  3. Multiply

    \(\frac12 \times (-2) = -1\)

    The gradients multiply to −1, so AB and BC are perpendicular: angle ABC = 90°. If two gradients were equal, the three points would lie on one line.

Video walkthrough coming