Unit 7 Review Exercises
Linear Functions
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed.
1 Paper 1 \(A\) is \((-1, 2)\) and \(B\) is \((5, 10)\). Find
(a) the length of \(AB\),
(b) the midpoint of \(AB\),
(c) the gradient of \(AB\).
2 Paper 1 The points \(P(x - 2, 1)\), \(Q(2x - 1, 4)\) and \(R(x + 2, 2x + 1)\) are collinear. Find the possible values of \(x\).
\(m_{PQ} = \dfrac{4 - 1}{(2x - 1) - (x - 2)} = \dfrac{3}{x + 1}\) and \(m_{PR} = \dfrac{(2x + 1) - 1}{(x + 2) - (x - 2)} = \dfrac{2x}{4} = \dfrac{x}{2}\).
\[ \frac{3}{x + 1} = \frac{x}{2} \;\Longrightarrow\; x^2 + x - 6 = 0 \;\Longrightarrow\; (x + 3)(x - 2) = 0 . \]
So \(x = -3\) or \(x = 2\). See 7.1.
3 Paper 1 The line \(l\) has equation \(4x - 3y + 6 = 0\).
(a) Find the gradient of \(l\) and its intercepts with the axes.
(b) Find the equation of the line through \((2, -1)\) perpendicular to \(l\), in the form \(ax + by + d = 0\).
(a) \(y = \dfrac{4}{3}x + 2\): gradient \(\dfrac{4}{3}\), \(y\)-intercept \(2\). Put \(y = 0\): \(x = -\dfrac{3}{2}\).
(b) Gradient \(-\dfrac{3}{4}\): \(\;y + 1 = -\dfrac{3}{4}(x - 2) \Rightarrow 4y + 4 = -3x + 6 \Rightarrow 3x + 4y - 2 = 0\).
See 7.2.
4 Paper 1 The lines \(l_1\): \(kx - 2y + 1 = 0\) and \(l_2\): \(3x + (k - 1)y - 4 = 0\) are given, with \(k \ne 1\).
(a) Find the value of \(k\) for which \(l_1\) and \(l_2\) are perpendicular.
(b) Show that \(l_1\) and \(l_2\) can never be parallel.
Gradients: \(m_1 = \dfrac{k}{2}\) and \(m_2 = -\dfrac{3}{k - 1}\).
(a) \(\dfrac{k}{2} \times \left(-\dfrac{3}{k - 1}\right) = -1 \Rightarrow 3k = 2(k - 1) \Rightarrow k = -2\).
(b) \(\dfrac{k}{2} = -\dfrac{3}{k - 1} \Rightarrow k^2 - k + 6 = 0\). The discriminant is \(1 - 24 = -23 < 0\), so there is no real \(k\).
5 Paper 1 Two adjacent vertices of a square \(ABCD\) are \(A(1, 2)\) and \(B(5, 5)\).
(a) Find the two possible pairs of coordinates of \(C\) and \(D\).
(b) Find the centre of the square in each case.
(a) From \(A\) to \(B\): \(4\) right and \(3\) up. A perpendicular step of the same length is \(3\) left and \(4\) up, or \(3\) right and \(4\) down (gradients \(\dfrac{3}{4}\) and \(-\dfrac{4}{3}\) multiply to \(-1\)).
\(C = B + (-3, 4) = (2, 9)\) and \(D = A + (-3, 4) = (-2, 6)\); \(\;\) or \(\;C = (8, 1)\) and \(D = (4, -2)\).
(b) The centre is the midpoint of \(AC\): \(\left(\dfrac{3}{2}, \dfrac{11}{2}\right)\) or \(\left(\dfrac{9}{2}, \dfrac{3}{2}\right)\).
6 Paper 1 A triangle has vertices \(A(7, 1)\), \(B(3, 5)\) and \(C(1, -1)\). \(M\) is the midpoint of \(AB\).
(a) Find the length of \(CM\).
(b) Find the equation of \(AB\) and the foot of the perpendicular from \(C\) to \(AB\).
(c) What do (a) and (b) tell you about triangle \(ABC\)?
(a) \(M = (5, 3)\). \(\;CM = \sqrt{4^2 + 4^2} = 4\sqrt{2}\).
(b) \(m_{AB} = \dfrac{5 - 1}{3 - 7} = -1\), so \(AB\): \(y = -x + 8\). The perpendicular through \(C\) has gradient \(1\): \(y = x - 2\). Then \(x - 2 = -x + 8\), so \(x = 5\): the foot is \((5, 3)\).
(c) The foot of the perpendicular is \(M\), so \(CM\) is the perpendicular bisector of \(AB\). So \(CA = CB\) (\(= \sqrt{40}\)) and the triangle is isosceles.
7 Paper 1 \(A\) is \((-3, 4)\) and \(B\) is \((5, 0)\). The perpendicular bisector of \(AB\) meets the line \(y = 6\) at \(C\). Find the coordinates of \(C\) and the area of triangle \(ABC\).
\(M = (1, 2)\), \(m_{AB} = \dfrac{-4}{8} = -\dfrac{1}{2}\), so the perpendicular gradient is \(2\): \(\;y - 2 = 2(x - 1)\), that is \(y = 2x\).
At \(y = 6\): \(x = 3\), so \(C\) is \((3, 6)\).
\(AB = \sqrt{64 + 16} = 4\sqrt{5}\) and \(CM = \sqrt{4 + 16} = 2\sqrt{5}\). Area \(= \dfrac{1}{2} \times 4\sqrt{5} \times 2\sqrt{5} = 20\).
See 7.3.
8 Paper 2 A straight mountain road rises \(84\) m over a horizontal distance of \(1.2\) km.
(a) Find the gradient of the road and its angle of inclination.
(b) A second road has angle of inclination \(6^\circ\). Find its gradient. Which road is steeper?
(a) \(m = \dfrac{84}{1200} = 0.07\), and \(\theta = \tan^{-1} 0.07 = 4.00^\circ\) (3 s.f.).
(b) \(m = \tan 6^\circ = 0.105\) (3 s.f.). The second road is steeper.
See 7.1.
9 Paper 2 The lines \(l_1\): \(y = 3x - 2\) and \(l_2\): \(x + 2y = 10\) meet at the point \(P\).
(a) Find the coordinates of \(P\).
(b) Find the angle of inclination of each line.
(c) Hence find the acute angle between \(l_1\) and \(l_2\).
(a) \(x + 2(3x - 2) = 10 \Rightarrow 7x = 14 \Rightarrow x = 2\), \(y = 4\). \(P\) is \((2, 4)\).
(b) \(l_1\): \(\theta_1 = \tan^{-1} 3 = 71.6^\circ\). \(\quad l_2\): \(m = -\dfrac{1}{2}\), so \(\theta_2 = \tan^{-1}\left(-\dfrac{1}{2}\right) + 180^\circ = -26.6^\circ + 180^\circ = 153.4^\circ\).
(c) \(153.43^\circ - 71.57^\circ = 81.9^\circ\) (3 s.f.).
10 Paper 1 The line \(y = x + 2\) meets the curve \(y = x^2 - 2x - 2\) at \(A\) and \(B\).
(a) Find the coordinates of \(A\) and \(B\).
(b) The perpendicular bisector of \(AB\) meets the \(x\)-axis at \(P\). Find the coordinates of \(P\), and show that \(PA = PB\).
(a) \(x^2 - 2x - 2 = x + 2 \Rightarrow x^2 - 3x - 4 = 0 \Rightarrow (x - 4)(x + 1) = 0\). So \(A(4, 6)\) and \(B(-1, 1)\).
(b) \(M = \left(\dfrac{3}{2}, \dfrac{7}{2}\right)\), \(m_{AB} = 1\), so the bisector has gradient \(-1\): \(\;y - \dfrac{7}{2} = -\left(x - \dfrac{3}{2}\right)\), that is \(y = -x + 5\).
At \(y = 0\): \(P\) is \((5, 0)\). \(\;PA = \sqrt{1 + 36} = \sqrt{37}\) and \(PB = \sqrt{36 + 1} = \sqrt{37}\), so \(PA = PB\).
See 7.3.