Edexcel IAL P1 Coordinate Geometry Questions
Straight lines, gradient, midpoint, perpendicular/parallel, equation of a line, distance formula.
- P1 · Pure Mathematics 1
- Non-calculator
- 68 practice questions
What the questions test
The 68 P1 coordinate geometry practice questions cover:
Worked example 1
Three points define the vertices of a triangle: \(A(-2, -1)\), \(B(4, 7)\), and \(C(8, 4)\).
By calculating the gradients of \([AB]\) and \([BC]\), prove that triangle \(ABC\) is a right-angled triangle.
Calculate the exact lengths of \([AB]\) and \([BC]\).
Hence, find the exact area of the triangle \(ABC\).
Mark scheme
Gradient \(m_{AB} = \frac{7 - (-1)}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}\). A1
Gradient \(m_{BC} = \frac{4 - 7}{8 - 4} = \frac{-3}{4}\). A1
Since \(m_{AB} \times m_{BC} = \frac{4}{3} \times -\frac{3}{4} = -1\), \([AB]\) and \([BC]\) are perpendicular. M1
Thus, triangle \(ABC\) has a right angle at \(B\). B1Length \(AB = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{36 + 64} = \sqrt{100} = 10\). A1
Length \(BC = \sqrt{(8 - 4)^2 + (4 - 7)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\). A1Since the triangle is right-angled at \(B\), \(AB\) and \(BC\) act as the base and height. M1
Area \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 5 = 25\) sq units. A1
Worked example 2
The line \(L_1\) has equation \(y = x + 2\) and the line \(L_2\) has equation \(y = -2x + 14\). The lines intersect at point \(A\).
Find the coordinates of \(A\).
Point \(B\) lies on \(L_1\) such that its \(x\)-coordinate is less than the \(x\)-coordinate of \(A\). The distance \(AB\) is exactly \(\sqrt{18}\) units. Find the coordinates of \(B\).
The line \(L_3\) passes through point \(B\) and is perpendicular to \(L_1\). Find the equation of \(L_3\) and determine where it intersects \(L_2\).
Mark scheme
Solve simultaneously: \(x + 2 = -2x + 14 \implies 3x = 12 \implies x = 4\). M1
\(y = 4 + 2 = 6\). \(A(4, 6)\). A1Let \(B\) be \((x, x+2)\). Distance \(AB = \sqrt{18}\).
\((x - 4)^2 + (x+2 - 6)^2 = 18 \implies (x - 4)^2 + (x - 4)^2 = 18\). M1
\(2(x - 4)^2 = 18 \implies (x - 4)^2 = 9\). A1
\(x - 4 = \pm 3 \implies x = 7\) or \(x = 1\). A1
Since \(x\) for \(B\) is less than \(x\) for \(A\) (which is \(4\)), \(x = 1\).
\(y = 1 + 2 = 3\). \(B(1, 3)\). A1\(L_3\) is perpendicular to \(L_1\) (\(m = 1\)), so \(m_3 = -1\). M1
Passes through \(B(1, 3)\): \(y - 3 = -1(x - 1) \implies y = -x + 4\).
Intersection with \(L_2\) (\(y = -2x + 14\)):
\(-x + 4 = -2x + 14 \implies x = 10\). \(y = -10 + 4 = -6\).
Intersection point is \((10, -6)\). A1
FAQ
What coordinate geometry is in Edexcel IAL P1?
Straight lines, gradient, midpoint, perpendicular/parallel, equation of a line, distance formula.
Is P1 a calculator paper?
No. P1 is non-calculator, so every coordinate geometry question has to be done by hand, with exact values where the question asks for them.
How are P1 coordinate geometry questions marked?
With M marks for a correct method, A marks for accurate answers that depend on the M mark before them, and B marks for independent results, as in the worked examples on this page.