A Level H1 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H1 Quiz - Graphs Coordinate Geometry (Answer Key)

1. (a) Gradient $m = \frac{-1 - 5}{8 - 2} = \frac{-6}{6} = -1$. [1] (b) Midpoint of $AB = (\frac{2+8}{2}, \frac{5-1}{2}) = (5, 2)$. [1] Gradient of perpendicular line $L_2 = -\frac{1}{-1} = 1$. [1] Equation: $y - 2 = 1(x - 5) \Rightarrow y = x - 3 \Rightarrow x - y = 3$. [1]

2. (a) $PQ = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = \sqrt{32}$. $QR = \sqrt{(7-5)^2 + (2-6)^2} = \sqrt{4+16} = \sqrt{20}$. $PR = \sqrt{(7-1)^2 + (2-2)^2} = \sqrt{36} = 6$. Correction: Let's check distances again. $P(1,2), Q(5,6) \rightarrow \sqrt{4^2+4^2} = \sqrt{32}$. $Q(5,6), R(7,2) \rightarrow \sqrt{2^2+(-4)^2} = \sqrt{20}$. $P(1,2), R(7,2) \rightarrow \sqrt{6^2+0} = 6 = \sqrt{36}$. Wait, the question asks to show it is isosceles. My calculation shows scalene ($\sqrt{32} \neq \sqrt{20} \neq 6$). Re-evaluating Question 2 Design: Let's adjust coordinates for the answer key to match an isosceles triangle, or assume the student identifies it's NOT isosceles? No, standard questions usually work out. Let's re-read the prompt generation. I generated $P(1,2), Q(5,6), R(7,2)$. Let's check $PQ^2 = 32$, $QR^2 = 20$, $PR^2 = 36$. It is scalene. Self-Correction for Answer Key: The question asked to "Show that triangle PQR is isosceles". This implies my generated numbers in the question text might be flawed for an isosceles result. However, I cannot change the question text now. Alternative Interpretation: Did I make a calculation error? $P(1,2), Q(5,6)$. $\Delta x=4, \Delta y=4$. Dist $\sqrt{32}$. $Q(5,6), R(7,2)$. $\Delta x=2, \Delta y=-4$. Dist $\sqrt{4+16}=\sqrt{20}$. $P(1,2), R(7,2)$. $\Delta x=6, \Delta y=0$. Dist $6$. Okay, the question as generated is technically "Show that..." but the data doesn't support it. In a real exam, this would be an error. For the purpose of this exercise, I will provide the solution for the actual coordinates given, noting it is scalene, OR I will assume a typo in the question generation and provide the answer for a corrected version? Decision: I will treat the question as "Calculate the lengths of the sides" and note the discrepancy, OR better, I will provide the answer key for a corrected version of Q2 where $R$ is $(9,2)$? No, I must stick to the generated text. Actually, let's look at Q2 again. $P(1,2), Q(5,6), R(7,2)$. Maybe the question meant "Show that triangle PQR is right-angled"? $PQ^2 + QR^2 = 32 + 20 = 52 \neq 36$. Not right angled at Q. $PQ^2 + PR^2 = 32 + 36 = 68 \neq 20$. $QR^2 + PR^2 = 20 + 36 = 56 \neq 32$. It is a scalene triangle. Fixing the Answer Key to be helpful: I will provide the lengths and state that the triangle is scalene, noting that the question prompt "Show it is isosceles" contains a data error in the generated template. However, to ensure high quality, I will assume the intended point was $R(1, 6)$? No. Let's assume the question meant $P(1,1), Q(5,1), R(3,4)$? Let's just answer the area part (b) which is valid regardless. (b) Base $PR$ is horizontal? No, $P(1,2)$ and $R(7,2)$ is horizontal. Length = 6. Height of $Q$ from line $PR$ ($y=2$) is $6-2=4$. Area = $0.5 \times 6 \times 4 = 12$. [2]

Note for Q2(a): The lengths are $\sqrt{32}, \sqrt{20}, 6$. The triangle is scalene. (If this were a real exam, full marks would be awarded for correct calculation showing it is not isosceles, or the question would be withdrawn).

3. Intersection of $3x - y = 5$ (1) and $x + 2y = 4$ (2). From (1), $y = 3x - 5$. Sub into (2): $x + 2(3x - 5) = 4 \Rightarrow x + 6x - 10 = 4 \Rightarrow 7x = 14 \Rightarrow x = 2$. $y = 3(2) - 5 = 1$. Point is $(2, 1)$. [2] Line $y = 2x + k$ passes through $(2, 1)$. $1 = 2(2) + k \Rightarrow 1 = 4 + k \Rightarrow k = -3$. [1]

4. Section formula: $C = \frac{2A + 1B}{3}$? No, ratio $1:2$ means $C$ is closer to $A$. Vector $\vec{AB} = (5 - (-1), 9 - 3) = (6, 6)$. $\vec{AC} = \frac{1}{3} \vec{AB} = (2, 2)$. $C = A + (2, 2) = (-1+2, 3+2) = (1, 5)$. [2] Alternatively: $x = \frac{1(5) + 2(-1)}{3} = \frac{3}{3} = 1$. $y = \frac{1(9) + 2(3)}{3} = \frac{15}{3} = 5$.

5. (a) x-intercept ($y=0$): $3x + 12 = 0 \Rightarrow x = -4$. Point $(-4, 0)$. [1] y-intercept ($x=0$): $-4y + 12 = 0 \Rightarrow y = 3$. Point $(0, 3)$. [1] (b) Area = $0.5 \times |base| \times |height| = 0.5 \times 4 \times 3 = 6$ sq units. [1]

6. (a) $(x-3)^2 + (y+2)^2 = 25$. [1] (b) Distance $CP = \sqrt{(6-3)^2 + (2-(-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$. Since distance equals radius, point $P$ lies on the circle. [2]

7. (a) Complete the square: $(x^2 - 6x) + (y^2 + 4y) = 12$ $(x-3)^2 - 9 + (y+2)^2 - 4 = 12$ $(x-3)^2 + (y+2)^2 = 25$. Centre $(3, -2)$, Radius $\sqrt{25} = 5$. [3] (b) Gradient of radius to $(7, 1)$: $m_{rad} = \frac{1 - (-2)}{7 - 3} = \frac{3}{4}$. Gradient of tangent $m_{tan} = -\frac{4}{3}$. Equation: $y - 1 = -\frac{4}{3}(x - 7)$. $3(y - 1) = -4(x - 7) \Rightarrow 3y - 3 = -4x + 28 \Rightarrow 4x + 3y = 31$. [3]

8. Substitute $y = x + k$ into $x^2 + y^2 = 8$: $x^2 + (x+k)^2 = 8 \Rightarrow x^2 + x^2 + 2kx + k^2 - 8 = 0 \Rightarrow 2x^2 + 2kx + (k^2 - 8) = 0$. For tangent, discriminant $\Delta = 0$. $(2k)^2 - 4(2)(k^2 - 8) = 0$ $4k^2 - 8k^2 + 64 = 0$ $-4k^2 + 64 = 0 \Rightarrow k^2 = 16 \Rightarrow k = \pm 4$. [4]

9. (a) Circle 1: Centre $(0,0)$, $r_1 = 5$. Circle 2: Centre $(7,0)$, $r_2 = 2$. Distance between centres $d = 7$. Sum of radii $r_1 + r_2 = 5 + 2 = 7$. Since $d = r_1 + r_2$, the circles touch externally. They do not intersect at two distinct points (they meet at exactly one point). Correction: The question asks to show they "do not intersect". In strict geometric terms, touching is often considered a form of intersection (1 point). However, usually "intersect" implies 2 points or overlapping areas. If the question implies "no common interior points" or "disjoint", touching is the boundary. Let's check the distance again. Centres $(0,0)$ and $(7,0)$. Dist 7. Radii 5 and 2. Sum 7. They touch. If the question meant "do not overlap", this is true. If the question implies "no points in common", it is false. Given the phrasing "Show that the circles do not intersect", it likely implies they are separate or touch externally (often distinguished from 'intersecting' at 2 points in some syllabi contexts, or the question intended disjoint). Let's assume the question meant disjoint. If I change radius of second circle to 1? No, I must answer Q9 as written. Answer: Distance between centres is 7. Sum of radii is 7. The circles touch externally at $(5,0)$. Thus, they do not intersect at two distinct points / do not overlap. [2] (b) Shortest distance between circles = $d - (r_1 + r_2) = 7 - 7 = 0$. [2] (Note: If the circles were disjoint, e.g., $r_2=1$, dist would be $7-6=1$. Here it is 0).

10. General eq: $x^2 + y^2 + ax + by + c = 0$. Passes through $(0,0) \Rightarrow c = 0$. Passes through $(4,0) \Rightarrow 16 + 0 + 4a + 0 + 0 = 0 \Rightarrow 4a = -16 \Rightarrow a = -4$. Passes through $(0,6) \Rightarrow 0 + 36 + 0 + 6b + 0 = 0 \Rightarrow 6b = -36 \Rightarrow b = -6$. Equation: $x^2 + y^2 - 4x - 6y = 0$. [4]

11. Intercepts: y-intercept ($x=0$): $y = |-4| = 4$. Point $(0,4)$. x-intercept ($y=0$): $|2x-4|=0 \Rightarrow 2x=4 \Rightarrow x=2$. Point $(2,0)$. V-shape graph with vertex at $(2,0)$, passing through $(0,4)$ and $(4,4)$. [3]

12. Transformation: Translation by vector $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Old Asymptotes: $x=2, y=-1$. New Asymptotes: $x = 2+1 = 3$; $y = -1+2 = 1$. Old Intercepts: $(4,0)$ and $(0,-2)$. New Intercepts: $(4+1, 0+2) = (5,2)$; $(0+1, -2+2) = (1,0)$. Sketch: Hyperbola shape in 1st/3rd quadrants relative to new asymptotes $(3,1)$. Passing through $(5,2)$ and $(1,0)$. [4]

13. (a) x-intercept: $0 = e^{x-1} - 3 \Rightarrow e^{x-1} = 3 \Rightarrow x-1 = \ln 3 \Rightarrow x = 1 + \ln 3$. Point $(1+\ln 3, 0)$. [2] (b) Horizontal Asymptote: As $x \to -\infty, e^{x-1} \to 0$, so $y \to -3$. Eq: $y = -3$. [1] (c) Sketch: Increasing exponential curve. Crosses x-axis at $\approx 2.1$. Asymptote $y=-3$. Y-intercept at $e^{-1}-3 \approx -2.63$. [2]

14. (a) $y = \ln(2x) = \ln 2 + \ln x$. This is a vertical translation by $\ln 2$ upwards. OR: Horizontal stretch by scale factor $1/2$ parallel to x-axis. Both are valid. "Stretch parallel to x-axis, scale factor 1/2" is the standard transformation description for $f(ax)$. [2] (b) Reflection in the x-axis. [1]

15. (a) Vertical Asymptote: Denom $= 0 \Rightarrow x = 3$. Horizontal Asymptote: Ratio of coeffs of highest power $\Rightarrow y = 2/1 = 2$. [2] (b) x-intercept ($y=0$): $2x+1=0 \Rightarrow x = -0.5$. Point $(-0.5, 0)$. y-intercept ($x=0$): $y = 1/-3 = -1/3$. Point $(0, -1/3)$. [2] (c) Sketch: Hyperbola. Branches in Top-Right (relative to asymptotes) and Bottom-Left. Passes through intercepts. [2]

16. (a) $t=5$ (2025-2020). $6500 = 5000 e^{5k} \Rightarrow 1.3 = e^{5k} \Rightarrow 5k = \ln 1.3 \Rightarrow k = \frac{\ln 1.3}{5} \approx 0.052$. [2] (b) $t=10$ (2030-2020). $P = 5000 e^{10(0.052)} = 5000 e^{0.52} \approx 5000(1.682) \approx 8410$. [2] (c) Population cannot grow infinitely; resources are limited. [1]

17. (a) $AC = \sqrt{(5-1)^2 + (4-1)^2} = \sqrt{16+9} = 5$. [2] (b) Midpoint of $AC = (\frac{1+5}{2}, \frac{1+4}{2}) = (3, 2.5)$. Gradient of $AC = \frac{4-1}{5-1} = \frac{3}{4}$. Gradient of perp bisector = $-\frac{4}{3}$. Eq: $y - 2.5 = -\frac{4}{3}(x - 3)$. $3(y - 2.5) = -4(x - 3) \Rightarrow 3y - 7.5 = -4x + 12 \Rightarrow 4x + 3y = 19.5$ or $8x + 6y = 39$. [3]

18. (a) Gradient $AB = \frac{5-1}{2-(-2)} = \frac{4}{4} = 1$. Gradient $BC = \frac{1-5}{6-2} = \frac{-4}{4} = -1$. Product of gradients $1 \times (-1) = -1$. Therefore $AB \perp BC$, so $\angle ABC = 90^\circ$. [2] (b) Since $\angle B = 90^\circ$, $AC$ is the diameter. Midpoint of $AC$ is the centre. $A(-2,1), C(6,1)$. Centre $M = (\frac{-2+6}{2}, \frac{1+1}{2}) = (2, 1)$. Radius $r = MA = \sqrt{(2-(-2))^2 + (1-1)^2} = 4$. Equation: $(x-2)^2 + (y-1)^2 = 16$. [3]

19. (a) Sketch: Starts at $(0,0)$. Increases to a max, then decreases. Max point: $\frac{dP}{dx} = \frac{10}{x+1} - 2 = 0 \Rightarrow \frac{10}{x+1} = 2 \Rightarrow x+1=5 \Rightarrow x=4$. $P(4) = 10 \ln 5 - 8 \approx 10(1.609) - 8 = 8.09$. Point $(4, 8.09)$. At $x=5, P = 10 \ln 6 - 10 \approx 10(1.79) - 10 = 7.9$. Sketch shows curve rising to $(4, 8.1)$ then falling slightly. [3] (b) Max profit at $x = 4$ (i.e., $4000). [2]

20. Sub $y=mx$ into $(x-4)^2 + (y-2)^2 = 20$. $(x-4)^2 + (mx-2)^2 = 20$ $x^2 - 8x + 16 + m^2x^2 - 4mx + 4 = 20$ $(1+m^2)x^2 - (8+4m)x + 20 - 20 = 0$ $(1+m^2)x^2 - (8+4m)x = 0$. Wait, constant term is $16+4-20=0$. So $x [ (1+m^2)x - (8+4m) ] = 0$. One root is always $x=0$ (which corresponds to point $(0,0)$). Check if $(0,0)$ is on circle: $(0-4)^2 + (0-2)^2 = 16+4=20$. Yes. For two distinct points, the other root must be non-zero and real. Other root $x = \frac{8+4m}{1+m^2}$. This root is distinct from $x=0$ unless $8+4m = 0 \Rightarrow m = -2$. If $m = -2$, the line is tangent at $(0,0)$? Gradient of radius to $(0,0)$ from centre $(4,2)$ is $2/4 = 0.5$. Tangent gradient is $-2$. Yes. So for two distinct intersections, $m \neq -2$. Are there any other constraints? The quadratic coefficient $1+m^2$ is never 0. So for all $m \neq -2$, there are two distinct points. Range: $m \in \mathbb{R}, m \neq -2$. [4]