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

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

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

   • Make y the subject: \(y = \frac{4}{3}x + 8\)
   • Gradient \(\frac{4}{3}\), y-intercept \(8\)

2. (b) Find the x-intercept.

   • Put \(y = 0\): \(4x + 24 = 0\)
   • \(x = -6\)

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

   • \(y = \frac{4}{3}x + 8\)

Find the equation of the line parallel to \(2x + 5y = 10\) that passes through \((5, -3)\). Give it in the form \(ax + by + d = 0\) with whole-number coefficients.

• Gradient of the given line: \(-\frac{2}{5}\)
• Parallel line, same gradient: \(y - (-3) = -\frac{2}{5}(x - (5))\)
• \(y = -\frac{2}{5}x - 1\)
• \(2x + 5y + 5 = 0\)

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

1. (a) \(y = 3x - 1\) and \(6x - 2y + 5 = 0\)

   • Gradients: \(3\) and \(3\)
   • Equal, so parallel

2. (b) \(2x + 3y = 6\) and \(3x - 2y = 1\)

   • Gradients: \(-\frac{2}{3}\) and \(\frac{3}{2}\)
   • Product: \(-1\), so perpendicular

3. (c) \(x + 2y = 4\) and \(y = \frac12 x + 1\)

   • Gradients: \(-\frac{1}{2}\) and \(\frac{1}{2}\)
   • Not equal, and the product is \(-\frac{1}{4}\), not −1: neither

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

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

   • \(m_1 = -\frac k4, \quad m_2 = \frac{2}{3}\)
   • Parallel: \(m_1 = m_2\), so \(-\frac k4 = \frac{2}{3}\)
   • \(k = -\frac{8}{3}\)

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

   • Perpendicular: \(m_1 m_2 = -1\), so \(m_1 = -\frac{3}{2}\)
   • \(-\frac k4 = -\frac{3}{2}\)
   • \(k = 6\)

The points \(P(-2, 1)\), \(Q(1, 7)\) and \(R(4, 13)\) are given.

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

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

2. (b) The point \(S(5, t)\) is on the same line. Find t.

   • Gradient from Q to S: \(\frac{t - (7)}{5 - (1)} = 2\)
   • \(t = 15\)

A gym charges a fixed joining fee plus the same amount each month. A membership of 4 months costs €140 and a membership of 10 months costs €290. Treat the cost as a linear function of the number of months.

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

   • \(m = \frac{290 - 140}{10 - 4} = 25\)
   • Each extra month adds €25 to the cost.

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

   • \(140 = 25 \times 4 + c\), so \(c = 40\)
   • €40 is the cost with 0 months: a fixed charge.

3. (c) Find the cost after 14 months.

   • Cost = 25 × 14 + 40
   • €390

4. (d) The cost is €365. Find the number of months.

   • 25 × n + 40 = 365
   • 25n = 325, so n = 13

On a map a straight road passes through \(P(1, 2)\) and \(Q(7, 5)\). A new straight road is to be built through \(T(2, 9)\), parallel to PQ. Distances are in kilometres.

1. (a) Find the gradient of the road PQ.

   • \(\frac{5 - 2}{7 - (1)} = \frac{1}{2}\)

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

   • Parallel lines have the same gradient: \(y = \frac{1}{2}x + c\)
   • Substitute \((2, 9)\): \(9 = \frac{1}{2}(2) + c\)
   • \(y = \frac{1}{2}x + 8\)

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

   • At \(x = 0\): \(y = 8\)
   • It meets the y-axis at \((0, 8)\)

A triangular plot of land has corners \(P(-2, 1)\), \(Q(2, 3)\) and \(R(3, 1)\), with coordinates in metres.

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

   • \(PQ: \frac{2}{4} = \frac{1}{2}\)
   • \(QR: \frac{-2}{1} = -2\)

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

   • \(\frac{1}{2} \times (-2) = -1\)
   • The sides are perpendicular, so the angle is 90°.

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

   • \(PQ = \sqrt{20}\), \(QR = \sqrt{5}\)
   • Area \(= \frac12 \times PQ \times QR\) = 5

The lines \((k+1)x + 2y = 7\) and \(3x + (k-2)y = 4\) 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 \(-\frac ab\)
   • \(m_1 = -\frac{k+1}{2}\), \(m_2 = -\frac{3}{k-2}\)

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 \(\frac{k+1}{2} = \frac{3}{k-2}\)
   • \((k+1)(k-2) = 6\), so \(k^2 - k - 8 = 0\)
   • Solve on the calculator: \(k = 3.37\) or \(k = -2.37\)

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

   • Perpendicular: \(m_1 m_2 = -1\), so \(\frac{3(k+1)}{2(k-2)} = -1\)
   • \(3k + 3 = -2k + 4\), so \(k = \frac15\)

ABCD is a parallelogram. The points \(A(1, 1)\), \(B(6, 3)\) and \(C(9, 8)\) are three of its vertices, taken in order round the shape.

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

   • \(AB: \frac{2}{5} = \frac{2}{5}\)
   • \(BC: \frac{5}{3} = \frac{5}{3}\)

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 \(\frac{5}{3}\) and CD has gradient \(\frac{2}{5}\)
   • AD: \(y = \frac{5}{3}x - \frac{2}{3}\), which is \(5x - 3y - 2 = 0\)
   • CD: \(y = \frac{2}{5}x + \frac{22}{5}\), which is \(2x - 5y + 22 = 0\)

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

   • Solve the pair on the calculator
   • \(D = (4, 6)\)
   • Check: the midpoint of AC is \((5, 4.5)\), the same as the midpoint of BD