Edexcel A level Maths (9MA0) · Pure mathematics › Coordinate geometry
Practise Straight lines. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
The points \(A(-1, -4)\) and \(B(7, 2)\) are the ends of the line segment \(AB\).
(a) Find an equation of the perpendicular bisector of \(AB\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers.[4]
(b) The perpendicular bisector of \(AB\) meets the \(x\)-axis at the point \(C\). Find the exact area of triangle \(ABC\).[3]
Show the answer and mark scheme
(a)Answer: \(4x + 3y - 9 = 0\)
B1 for the midpoint of \(AB\), \((3, -1)\)
M1 for the gradient of \(AB\) \(= \frac{3}{4}\) and use of \(m_1 m_2 = -1\) to obtain \(-\frac{4}{3}\)
dM1 for \(y + 1 = -\frac{4}{3}(x - 3)\) or equivalent, using their midpoint and their perpendicular gradient
A1 for \(4x + 3y - 9 = 0\) or any integer multiple
Worked solution: Midpoint of \(AB\) \(= \left(\frac{-1 + 7}{2}, \frac{-4 + 2}{2}\right) = (3, -1)\) Gradient of \(AB\) \(= \frac{2 - (-4)}{7 - (-1)} = \frac{3}{4}\), so the perpendicular gradient is \(-\frac{4}{3}\). \(y + 1 = -\frac{4}{3}(x - 3)\) which rearranges to \(4x + 3y - 9 = 0\).
(b)Answer: \(\frac{25}{4}\)
M1 for substituting \(y = 0\) into their perpendicular bisector to find \(C\left(\frac{9}{4}, 0\right)\)
M1 for a complete method for the area, e.g. \(\frac{1}{2} \times AB \times MC\) where \(M\) is the midpoint of \(AB\), or the shoelace formula
A1 for \(\frac{25}{4}\) cao
Worked solution: At \(C\), \(y = 0\): \(4x - 9 = 0\) so \(C\left(\frac{9}{4}, 0\right)\). \(AB = \sqrt{8^2 + 6^2} = 10\) and, with \(M(3, -1)\), \(MC = \sqrt{\left(-\frac{3}{4}\right)^2 + (-1)^2} = \frac{5}{4}\). \(C\) is on the perpendicular bisector, so \(CM\) is perpendicular to \(AB\): \(\text{Area} = \frac{1}{2} \times 10 \times \frac{5}{4} = \frac{25}{4}\)
Question 3Hard7 marks
The points \(A\) and \(B\) have coordinates \((-2, -5)\) and \((6, 1)\) respectively. The line \(l\) has equation \(y = -x + 7\). The point \(C\) lies on \(l\) such that the area of triangle \(ABC\) is \(7\).
Find the two possible sets of coordinates of \(C\).[7]
Show the answer and mark scheme
Answer: \(C(7, 0)\) or \(C(5, 2)\)
M1 for a general point on \(l\), e.g. \(C\left(t, -t + 7\right)\)
M1 for a correct method for the area of triangle \(ABC\) in terms of \(t\), e.g. the shoelace formula or \(\frac{1}{2} \times \text{base} \times \text{height}\)
A1 for area \(= \frac{1}{2}\left|-14t + 84\right|\) oe
M1 for setting their area equal to \(7\) and considering both signs (or squaring)
A1 for \(t = 7\) and \(t = 5\)
M1 for substituting their values of \(t\) into the equation of \(l\)
A1 for \((7, 0)\) and \((5, 2)\)
Worked solution: Let \(C = \left(t, -t + 7\right)\). Area \(= \frac{1}{2}\left|(x_B - x_A)(y_C - y_A) - (x_C - x_A)(y_B - y_A)\right|\) \(= \frac{1}{2}\left|8\left(-t + 12\right) - 6\left(t + 2\right)\right| = \frac{1}{2}\left|-14t + 84\right|\) Setting this equal to \(7\): \(-14t + 84 = \pm 14\), so \(t = 5\) or \(t = 7\). Hence \(C(7, 0)\) or \(C(5, 2)\).