Straight lines: gradient, parallel and perpendicular Problem set B One question at a time

A line has equation \(3x + 4y - 24 = 0\).

1. (a) Write down its gradient and its y-intercept.

   • Make y the subject: \(y = -\frac{3}{4}x + 6\)
   • Gradient \(-\frac{3}{4}\), y-intercept \(6\)

2. (b) Find the x-intercept.

   • Put \(y = 0\): \(3x - 24 = 0\)
   • \(x = 8\)

3. (c) Write the equation in the form \(y = mx + c\).

   • \(y = -\frac{3}{4}x + 6\)

Find the equation of the line parallel to \(3x - 4y = 12\) that passes through \((-4, 2)\). Give it in the form \(ax + by + d = 0\) with whole-number coefficients.

• Gradient of the given line: \(\frac{3}{4}\)
• Parallel line, same gradient: \(y - (2) = \frac{3}{4}(x - (-4))\)
• \(y = \frac{3}{4}x + 5\)
• \(3x - 4y + 20 = 0\)

For each pair of lines, say whether they are parallel, perpendicular or neither.

1. (a) \(y = -2x + 7\) and \(4x + 2y - 3 = 0\)

   • Gradients: \(-2\) and \(-2\)
   • Equal, so parallel

2. (b) \(5x + 2y = 10\) and \(2x - 5y = 3\)

   • Gradients: \(-\frac{5}{2}\) and \(\frac{2}{5}\)
   • Product: \(-1\), so perpendicular

3. (c) \(y = 2x + 1\) and \(y = -x + 5\)

   • Gradients: \(2\) and \(-1\)
   • Not equal, and the product is \(-2\), not −1: neither

The lines \(kx + 3y = 7\) and \(5x - 2y = 3\) have gradients \(m_1\) and \(m_2\).

1. (a) Find the value of k for which the lines are parallel.

   • \(m_1 = -\frac k3, \quad m_2 = \frac{5}{2}\)
   • Parallel: \(m_1 = m_2\), so \(-\frac k3 = \frac{5}{2}\)
   • \(k = -\frac{15}{2}\)

2. (b) Find the value of k for which the lines are perpendicular.

   • Perpendicular: \(m_1 m_2 = -1\), so \(m_1 = -\frac{2}{5}\)
   • \(-\frac k3 = -\frac{2}{5}\)
   • \(k = \frac{6}{5}\)

The points \(P(-4, -3)\), \(Q(0, 3)\) and \(R(2, 6)\) are given.

1. (a) Show that P, Q and R lie on a straight line.

   • \(PQ: \frac{6}{4} = \frac{3}{2}\)
   • \(QR: \frac{3}{2} = \frac{3}{2}\)
   • The gradients are equal and Q is common, so the three points are on one line.

2. (b) The point \(S(10, t)\) is on the same line. Find t.

   • Gradient from Q to S: \(\frac{t - (3)}{10 - (0)} = \frac{3}{2}\)
   • \(t = 18\)

A plumber charges a call-out fee plus the same amount for each hour of work. A job of 3 hours costs €135 and a job of 6 hours costs €240. Treat the cost as a linear function of the number of hours.

1. (a) Find the gradient of the line through these two points and say what it means.

   • \(m = \frac{240 - 135}{6 - 3} = 35\)
   • Each extra hour adds €35 to the cost.

2. (b) Find the intercept and say what it means.

   • \(135 = 35 \times 3 + c\), so \(c = 30\)
   • €30 is the cost with 0 hours: a fixed charge.

3. (c) Find the cost after 8 hours.

   • Cost = 35 × 8 + 30
   • €310

4. (d) The cost is €380. Find the number of hours.

   • 35 × n + 30 = 380
   • 35n = 350, so n = 10

On a plan a straight cycle path passes through \(M(-2, 4)\) and \(N(6, 0)\). A new straight path is to be built through \(T(3, 10)\), parallel to MN. Distances are in metres.

1. (a) Find the gradient of the path MN.

   • \(\frac{0 - 4}{6 - (-2)} = -\frac{1}{2}\)

2. (b) Find the equation of the new path. Give it in the form \(y = mx + c\).

   • Parallel lines have the same gradient: \(y = -\frac{1}{2}x + c\)
   • Substitute \((3, 10)\): \(10 = -\frac{1}{2}(3) + c\)
   • \(y = -\frac{1}{2}x + \frac{23}{2}\)

3. (c) Find where the new path meets the y-axis.

   • At \(x = 0\): \(y = \frac{23}{2}\)
   • It meets the y-axis at \((0, \frac{23}{2})\)

A triangular sail has corners \(P(1, 2)\), \(Q(5, 5)\) and \(R(11, -3)\), with coordinates in centimetres.

1. (a) Find the gradients of PQ and QR.

   • \(PQ: \frac{3}{4} = \frac{3}{4}\)
   • \(QR: \frac{-8}{6} = -\frac{4}{3}\)

2. (b) Show that angle PQR is a right angle.

   • \(\frac{3}{4} \times (-\frac{4}{3}) = -1\)
   • The sides are perpendicular, so the angle is 90°.

3. (c) Find the area of the sail.

   • \(PQ = \sqrt{25}\), \(QR = \sqrt{100}\)
   • Area \(= \frac12 \times PQ \times QR\) = 25

The lines \((k+2)x + 3y = 5\) and \(4x + (k-1)y = 9\) have gradients \(m_1\) and \(m_2\).

1. (a) Write \(m_1\) and \(m_2\) in terms of k.

   • Gradient of \(ax + by = c\) is \(-\frac ab\)
   • \(m_1 = -\frac{k+2}{3}\), \(m_2 = -\frac{4}{k-1}\)

2. (b) Find the values of k for which the lines are parallel. Give your answers to 3 significant figures.

   • Parallel: \(m_1 = m_2\), so \(\frac{k+2}{3} = \frac{4}{k-1}\)
   • \((k+2)(k-1) = 12\), so \(k^2 + k - 14 = 0\)
   • Solve on the calculator: \(k = 3.27\) or \(k = -4.27\)

3. (c) Find the value of k for which the lines are perpendicular.

   • Perpendicular: \(m_1 m_2 = -1\), so \(\frac{4(k+2)}{3(k-1)} = -1\)
   • \(4k + 8 = -3k + 3\), so \(k = -\frac57\)

ABCD is a parallelogram. The points \(A(-2, 0)\), \(B(3, -1)\) and \(C(7, 3)\) are three of its vertices, taken in order round the shape.

1. (a) Find the gradients of AB and BC.

   • \(AB: \frac{-1}{5} = -\frac{1}{5}\)
   • \(BC: \frac{4}{4} = 1\)

2. (b) Find the equation of the line through A parallel to BC, and the equation of the line through C parallel to AB. Give each in the form \(ax + by + d = 0\).

   • Opposite sides are parallel, so AD has gradient \(1\) and CD has gradient \(-\frac{1}{5}\)
   • AD: \(y = x + 2\), which is \(x - y + 2 = 0\)
   • CD: \(y = -\frac{1}{5}x + \frac{22}{5}\), which is \(x + 5y - 22 = 0\)

3. (c) Solve the two equations together to find the coordinates of D.

   • Solve the pair on the calculator
   • \(D = (2, 4)\)
   • Check: the midpoint of AC is \((2.5, 1.5)\), the same as the midpoint of BD