Secondary 4 Additional Mathematics Graphs Coordinate Geometry Quiz

Secondary 4 Additional Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

1. (a) Midpoint of $AB = \left(\frac{-2+4}{2}, \frac{5+(-1)}{2}\right) = (1, 2)$. [1] Gradient of $AB = \frac{-1-5}{4-(-2)} = \frac{-6}{6} = -1$. Gradient of perpendicular bisector $= -\frac{1}{-1} = 1$. [1] Equation: $y - 2 = 1(x - 1) \Rightarrow y = x + 1$. [1]

(b) At x-axis, $y=0$. $0 = x + 1 \Rightarrow x = -1$. Coordinates of $C$ are $(-1, 0)$. [1]

2. (a) $3x - 4y + 12 = 0 \Rightarrow 4y = 3x + 12 \Rightarrow y = \frac{3}{4}x + 3$. Gradient $m = \frac{3}{4}$. [1]

(b) Gradient of $L_2 = \frac{3}{4}$. Passes through $(2, -3)$. $y - (-3) = \frac{3}{4}(x - 2)$ $4(y + 3) = 3(x - 2)$ $4y + 12 = 3x - 6$ $3x - 4y - 18 = 0$. [2]

3. (a) Gradient $PQ = \frac{6-2}{5-1} = \frac{4}{4} = 1$. Gradient $QR = \frac{2-6}{9-5} = \frac{-4}{4} = -1$. Product of gradients $m_{PQ} \times m_{QR} = 1 \times (-1) = -1$. Therefore, $PQ \perp QR$ and $\angle PQR = 90^\circ$. Triangle is right-angled. [2]

(b) Length $PQ = \sqrt{(5-1)^2 + (6-2)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$. Length $QR = \sqrt{(9-5)^2 + (2-6)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$. Area $= \frac{1}{2} \times PQ \times QR = \frac{1}{2} \times 4\sqrt{2} \times 4\sqrt{2} = \frac{1}{2} \times 32 = 16$. [2]

4. Section formula: $P = \frac{2A + 1B}{1+2} = \frac{2(3,4) + 1(9,10)}{3}$. $x = \frac{6+9}{3} = \frac{15}{3} = 5$. $y = \frac{8+10}{3} = \frac{18}{3} = 6$. $P(5, 6)$. [2]

5. $2x + 1 = -x + 7$ $3x = 6 \Rightarrow x = 2$. $y = 2(2) + 1 = 5$. $K(2, 5)$. [2]

6. (a) Complete the square: $(x^2 - 6x) + (y^2 + 8y) = 11$ $(x-3)^2 - 9 + (y+4)^2 - 16 = 11$ $(x-3)^2 + (y+4)^2 = 36$. Centre $(3, -4)$. [2] (b) $r^2 = 36 \Rightarrow r = 6$. [1]

7. Radius squared $r^2 = (5-2)^2 + (1-(-3))^2 = 3^2 + 4^2 = 9 + 16 = 25$. Equation: $(x-2)^2 + (y+3)^2 = 25$. [3]

8. Substitute $y = x+k$ into $x^2 + y^2 = 18$: $x^2 + (x+k)^2 = 18$ $x^2 + x^2 + 2kx + k^2 - 18 = 0$ $2x^2 + 2kx + (k^2 - 18) = 0$. For tangent, discriminant $\Delta = 0$. $(2k)^2 - 4(2)(k^2 - 18) = 0$ $4k^2 - 8k^2 + 144 = 0$ $-4k^2 + 144 = 0 \Rightarrow k^2 = 36$. $k = \pm 6$. [4]

9. (a) Centre is midpoint of $AB$: $(\frac{1+5}{2}, \frac{2+6}{2}) = (3, 4)$. Radius squared $r^2 = (5-3)^2 + (6-4)^2 = 4 + 4 = 8$. Equation: $(x-3)^2 + (y-4)^2 = 8$. [3] (b) Distance squared of $D(6,3)$ from centre $(3,4)$: $(6-3)^2 + (3-4)^2 = 3^2 + (-1)^2 = 9 + 1 = 10$. Since $10 > 8$ (radius squared), $D$ lies outside the circle. [2]

10. (a) $C_1$: Centre $(0,0)$, $r_1=5$. $C_2$: $(x-5)^2 + (y-5)^2 = -25 + 25 + 25 = 25$? $x^2 - 10x + 25 + y^2 - 10y + 25 = 25 \Rightarrow (x-5)^2 + (y-5)^2 = 25$. Centre $(5,5)$, $r_2=5$. Distance between centres $d = \sqrt{5^2+5^2} = \sqrt{50} = 5\sqrt{2} \approx 7.07$. Sum of radii $= 5+5=10$. Difference $= 0$. Since $0 < 7.07 < 10$, they intersect. [2] (b) Subtract equations: $(x^2 + y^2 - 10x - 10y + 25) - (x^2 + y^2 - 25) = 0$ $-10x - 10y + 50 = 0 \Rightarrow x + y = 5 \Rightarrow y = 5-x$. Sub into $C_1$: $x^2 + (5-x)^2 = 25$ $x^2 + 25 - 10x + x^2 = 25$ $2x^2 - 10x = 0 \Rightarrow 2x(x-5)=0$. $x=0 \Rightarrow y=5$. Point $(0,5)$. $x=5 \Rightarrow y=0$. Point $(5,0)$. Intersections: $(0,5)$ and $(5,0)$. [3]

11. Touches y-axis at $(0,4) \Rightarrow$ Centre has y-coordinate $4$. Let Centre be $(a, 4)$. Radius $r = |a|$ (distance to y-axis). Equation: $(x-a)^2 + (y-4)^2 = a^2$. Passes through $(2,6)$: $(2-a)^2 + (6-4)^2 = a^2$ $4 - 4a + a^2 + 4 = a^2$ $8 - 4a = 0 \Rightarrow 4a = 8 \Rightarrow a = 2$. Centre $(2,4)$, $r=2$. Equation: $(x-2)^2 + (y-4)^2 = 4$ or $x^2 + y^2 - 4x - 8y + 16 = 0$. [4]

12. (a) Gradient of $OM$ (Centre to Midpoint) $= \frac{4-0}{3-0} = \frac{4}{3}$. Chord is perpendicular to radius, so gradient of chord $= -\frac{3}{4}$. Equation: $y - 4 = -\frac{3}{4}(x - 3)$ $4y - 16 = -3x + 9$ $3x + 4y = 25$. [2] (b) Distance $OM = \sqrt{3^2+4^2} = 5$. Radius $R = \sqrt{50} = 5\sqrt{2}$. Half-chord length $= \sqrt{R^2 - OM^2} = \sqrt{50 - 25} = \sqrt{25} = 5$. Total length $AB = 10$. [2]

13. (a) Vertical axis: $y$. Horizontal axis: $x^2$. [1] (b) Equation of line: $Y = mX + c \Rightarrow y = -2(x^2) + 5$. Comparing to $y = ax^2 + b$: $a = -2$, $b = 5$. [2]

14. (a) Diagonals of a rhombus bisect each other. Midpoint of $AC = (\frac{-1+7}{2}, \frac{3+3}{2}) = (3, 3)$. Let $D = (x,y)$. Midpoint of $BD = (\frac{3+x}{2}, \frac{7+y}{2})$. $\frac{3+x}{2} = 3 \Rightarrow 3+x=6 \Rightarrow x=3$. $\frac{7+y}{2} = 3 \Rightarrow 7+y=6 \Rightarrow y=-1$. $D(3, -1)$. [2] (b) Diagonal $AC$ length $= 7 - (-1) = 8$ (horizontal). Diagonal $BD$ length $= 7 - (-1) = 8$ (vertical). Area $= \frac{1}{2} d_1 d_2 = \frac{1}{2} \times 8 \times 8 = 32$. [2]

15. $PA = 2 PB \Rightarrow PA^2 = 4 PB^2$. $(x-2)^2 + y^2 = 4 [ (x-8)^2 + y^2 ]$ $x^2 - 4x + 4 + y^2 = 4 [ x^2 - 16x + 64 + y^2 ]$ $x^2 - 4x + 4 + y^2 = 4x^2 - 64x + 256 + 4y^2$ $3x^2 - 60x + 3y^2 + 252 = 0$ Divide by 3: $x^2 - 20x + y^2 + 84 = 0$. [3]

16. Substitute $y=mx$ into $(x-4)^2 + (y-2)^2 = 4$: $(x-4)^2 + (mx-2)^2 = 4$ $x^2 - 8x + 16 + m^2x^2 - 4mx + 4 = 4$ $(1+m^2)x^2 - (8+4m)x + 16 = 0$. For 2 distinct points, $\Delta > 0$: $(8+4m)^2 - 4(1+m^2)(16) > 0$ $64 + 64m + 16m^2 - 64(1+m^2) > 0$ Divide by 16: $4 + 4m + m^2 - 4(1+m^2) > 0$ $4 + 4m + m^2 - 4 - 4m^2 > 0$ $-3m^2 + 4m > 0$ $m(4 - 3m) > 0$. Critical values $m=0, m=4/3$. Since coefficient of $m^2$ is negative, range is between roots: $0 < m < \frac{4}{3}$. [4]

17. (a) Gradient $m = \frac{1-5}{8-2} = \frac{-4}{6} = -\frac{2}{3}$. Equation: $y - 5 = -\frac{2}{3}(x - 2)$ $3(y - 5) = -2(x - 2)$ $3y - 15 = -2x + 4$ $2x + 3y = 19$. [2]

(b) Perpendicular distance from $(0,0)$ to $2x + 3y - 19 = 0$: $d = \frac{|2(0) + 3(0) - 19|}{\sqrt{2^2 + 3^2}} = \frac{19}{\sqrt{13}}$. $d = \frac{19\sqrt{13}}{13}$ or approx $5.27$. [2]

18. (a) $(x-3)^2 + (y+2)^2 = 25$. [1]

(b) Centre $C(3, -2)$, Radius $r=5$. Distance from centre to line $3x + 4y - 10 = 0$: $d = \frac{|3(3) + 4(-2) - 10|}{\sqrt{3^2 + 4^2}}$ $d = \frac{|9 - 8 - 10|}{\sqrt{25}} = \frac{|-9|}{5} = \frac{9}{5} = 1.8$. Since $1.8 \neq 5$, the line is NOT a tangent. Correction for Question Validity: Let's re-evaluate the line equation or circle. If the line was $3x + 4y = 1$, distance is $\frac{|9-8-1|}{5} = 0$ (secant through centre). If the line was $3x + 4y = 25$? Distance $\frac{|9-8-25|}{5} = \frac{24}{5} = 4.8$. Let's adjust the question line to be a tangent. Tangent at $(6, 2)$? Gradient radius to $(6,2)$ from $(3,-2)$ is $\frac{4}{3}$. Tangent gradient $-\frac{3}{4}$. Eq: $y-2 = -\frac{3}{4}(x-6) \Rightarrow 4y-8 = -3x+18 \Rightarrow 3x+4y=26$. Distance: $\frac{|9-8-26|}{5} = \frac{25}{5} = 5$. Yes. Assuming the question intended a valid tangent, e.g., $3x + 4y = 26$: Distance $= 5$. Since distance equals radius, it is a tangent. [3] (Note: Based on the provided question text $3x+4y=10$, the answer is it is NOT a tangent. However, in exam keys, usually the question is correct. If forced to answer the provided text: Distance is 1.8, which is less than 5, so it is a secant.) Standard Key Answer for "Show that... is tangent": Calculate distance from centre to line. If distance = radius, it is a tangent. Here, $d = 1.8 \neq 5$. The statement in the question is false for the given numbers. For the purpose of this key, assuming a typo in the question constant to make it a tangent (e.g. RHS=26 or similar), the method is:

1. Find distance from centre to line.
2. Compare with radius.
3. Conclude.

19. (a) Centroid $G = (\frac{1+4+7}{3}, \frac{1+5+1}{3}) = (\frac{12}{3}, \frac{7}{3}) = (4, \frac{7}{3})$. [2]

(b) Midpoint of $PR$: $M_{PR} = (\frac{1+7}{2}, \frac{1+1}{2}) = (4, 1)$. Median passes through $Q(4,5)$ and $M_{PR}(4,1)$. Since x-coordinates are same, the line is vertical. Equation: $x = 4$. [2]

20. (a) Vertical axis: $y$. Horizontal axis: $\frac{1}{x}$. [1]

(b) Equation: $y = k(\frac{1}{x}) + h$. Points: $(\frac{1}{x}, y) \rightarrow (0.5, 7)$ and $(2, 4)$? No, $x$ values are $0.5$ and $2$? If $x=0.5, 1/x = 2$. Point $(2, 7)$. If $x=2, 1/x = 0.5$. Point $(0.5, 4)$. Gradient $k = \frac{7-4}{2-0.5} = \frac{3}{1.5} = 2$. $y = 2(\frac{1}{x}) + h$. Using $(2, 7)$: $7 = 2(2) + h \Rightarrow 7 = 4 + h \Rightarrow h = 3$. $k = 2, h = 3$. [3]