Secondary 3 Additional Mathematics Graphs Coordinate Geometry Quiz

Secondary 3 Additional Mathematics Quiz - Graphs Coordinate Geometry

Answer Key

Section A

1. [2 marks]
Using the section formula:
$\left(\frac{2(7) + 1(1)}{2+1}, \frac{2(-2) + 1(4)}{2+1}\right) = \left(\frac{14 + 1}{3}, \frac{-4 + 4}{3}\right) = \left(\frac{15}{3}, \frac{0}{3}\right) = (5, 0)$
Answer: $(5, 0)$

2. [2 marks]
Gradient formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$
$-4 = \frac{-5 - 3}{2 - k} = \frac{-8}{2 - k}$
$-4(2 - k) = -8$
$-8 + 4k = -8$
$4k = 0 \implies k = 0$
Answer: $k = 0$

3. [3 marks]
Rewrite $2x - 4y + 7 = 0$ in gradient form:
$4y = 2x + 7 \implies y = \frac{1}{2}x + \frac{7}{4}$
Gradient of given line $= \frac{1}{2}$, so gradient of parallel line $= \frac{1}{2}$.
Using point $(3, -1)$:
$y - (-1) = \frac{1}{2}(x - 3)$
$y + 1 = \frac{1}{2}x - \frac{3}{2}$
$y = \frac{1}{2}x - \frac{3}{2} - 1 = \frac{1}{2}x - \frac{5}{2}$
Multiply by 2: $2y = x - 5$, so $x - 2y - 5 = 0$.
Answer: $x - 2y - 5 = 0$ (or $y = \frac{1}{2}x - \frac{5}{2}$)

4. [2 marks]
Line 1: $3x + 2y = 8 \implies y = -\frac{3}{2}x + 4$, so $m_1 = -\frac{3}{2}$.
Line 2: $2x - 3y = 6 \implies y = \frac{2}{3}x - 2$, so $m_2 = \frac{2}{3}$.
Check: $m_1 \times m_2 = -\frac{3}{2} \times \frac{2}{3} = -1$.
Since the product of the gradients is $-1$, the lines are perpendicular.
Answer: Yes, the lines are perpendicular because $m_1 \times m_2 = -1$.

5. [2 marks]
Midpoint $= \left(\frac{-3 + 7}{2}, \frac{5 + (-1)}{2}\right) = \left(\frac{4}{2}, \frac{4}{2}\right) = (2, 2)$
Answer: $(2, 2)$

6. [3 marks]
Complete the square:
$x^2 - 6x + y^2 + 4y = 12$
$(x - 3)^2 - 9 + (y + 2)^2 - 4 = 12$
$(x - 3)^2 + (y + 2)^2 = 12 + 9 + 4 = 25$
Centre $= (3, -2)$, radius $= \sqrt{25} = 5$.
Answer: Centre $(3, -2)$, radius $= 5$

7. [2 marks]
$AB = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$
Answer: 10 units

8. [2 marks]
Substitute $(4, 7)$ and $m = 3$ into $y = mx + c$:
$7 = 3(4) + c$
$7 = 12 + c \implies c = -5$
Answer: $c = -5$

9. [3 marks]
Set the two equations equal:
$2x + 1 = -x + 7$
$3x = 6 \implies x = 2$
Substitute $x = 2$ into $y = 2x + 1$:
$y = 2(2) + 1 = 5$
Answer: $(2, 5)$

10. [3 marks]
Centre $= (2, -3)$, so $a = 2$, $b = -3$.
$r = \sqrt{(5 - 2)^2 + (1 - (-3))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
Equation: $(x - 2)^2 + (y + 3)^2 = 25$.
Answer: $(x - 2)^2 + (y + 3)^2 = 25$

Section B

11.
(a) [1 mark]
$m_{AB} = \frac{4 - 2}{5 - 1} = \frac{2}{4} = \frac{1}{2}$
Answer: $\frac{1}{2}$

(b) [2 marks]
Gradient of line perpendicular to $AB = -\frac{1}{m_{AB}} = -2$.
Line through $C(3, 8)$ with gradient $-2$:
$y - 8 = -2(x - 3)$
$y - 8 = -2x + 6$
$y = -2x + 14$
Answer: $y = -2x + 14$ (or $2x + y - 14 = 0$)

(c) [2 marks]
Equation of $AB$: using point $A(1, 2)$ and gradient $\frac{1}{2}$:
$y - 2 = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x + \frac{3}{2}$
Set equal to perpendicular line:
$\frac{1}{2}x + \frac{3}{2} = -2x + 14$
$\frac{1}{2}x + 2x = 14 - \frac{3}{2}$
$\frac{5}{2}x = \frac{25}{2} \implies x = 5$
$y = \frac{1}{2}(5) + \frac{3}{2} = \frac{5}{2} + \frac{3}{2} = 4$
Answer: $(5, 4)$

12.
(a) [3 marks]
$x^2 + 4x + y^2 - 8y + 11 = 0$
$(x + 2)^2 - 4 + (y - 4)^2 - 16 + 11 = 0$
$(x + 2)^2 + (y - 4)^2 = 9$
Answer: $(x + 2)^2 + (y - 4)^2 = 9$

(b) [1 mark]
Centre $= (-2, 4)$, radius $= \sqrt{9} = 3$.
Answer: Centre $(-2, 4)$, radius $= 3$

(c) [2 marks]
Distance from $(-1, 4)$ to centre $(-2, 4)$:
$d = \sqrt{(-1 - (-2))^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = 1$
Since $d = 1 < r = 3$, the point lies inside the circle.
Answer: Inside the circle, because the distance from the point to the centre (1 unit) is less than the radius (3 units).

13.
(a) [1 mark]
$3x + 4y = 12 \implies y = -\frac{3}{4}x + 3$
Answer: Gradient of $l_1 = -\frac{3}{4}$

(b) [3 marks]
Gradient of $l_2 = -\frac{1}{-\frac{3}{4}} = \frac{4}{3}$ (negative reciprocal).
Line through $(6, -2)$ with gradient $\frac{4}{3}$:
$y - (-2) = \frac{4}{3}(x - 6)$
$y + 2 = \frac{4}{3}x - 8$
$y = \frac{4}{3}x - 10$
Multiply by 3: $3y = 4x - 30$, so $4x - 3y - 30 = 0$.
Answer: $4x - 3y - 30 = 0$ (or $y = \frac{4}{3}x - 10$)

(c) [2 marks]
From $l_1$: $3x + 4y = 12$.
From $l_2$: $4x - 3y = 30$.
Multiply first equation by 3: $9x + 12y = 36$.
Multiply second equation by 4: $16x - 12y = 120$.
Add: $25x = 156 \implies x = \frac{156}{25} = 6.24$.
Substitute into $3x + 4y = 12$:
$3\left(\frac{156}{25}\right) + 4y = 12$
$\frac{468}{25} + 4y = 12 = \frac{300}{25}$
$4y = \frac{300 - 468}{25} = \frac{-168}{25}$
$y = \frac{-42}{25} = -1.68$
Answer: $\left(\frac{156}{25}, -\frac{42}{25}\right)$ or $(6.24, -1.68)$

14.
(a) [3 marks]
$PQ = \sqrt{(8 - 2)^2 + (3 - 1)^2} = \sqrt{36 + 4} = \sqrt{40}$
$QR = \sqrt{(5 - 8)^2 + (7 - 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
$PR = \sqrt{(5 - 2)^2 + (7 - 1)^2} = \sqrt{9 + 36} = \sqrt{45}$
Since $PQ = \sqrt{40} \approx 6.32$, $QR = 5$, $PR = \sqrt{45} \approx 6.71$, none of the sides are equal.
Rechecking:
$PQ = \sqrt{(8-2)^2 + (3-1)^2} = \sqrt{36 + 4} = \sqrt{40}$
$PR = \sqrt{(5-2)^2 + (7-1)^2} = \sqrt{9 + 36} = \sqrt{45}$
$$QR = \sqrt{(5-8)^2 + (7-3)^2} = \sqrt{9 + 16} = 5$None are equal — let me re-examine. Actually,$PQ = \sqrt{40}$,$QR = 5 = \sqrt{25}$,$PR = \sqrt{45}$. These are not equal. *Correction:* The triangle is not isosceles with these coordinates. Let me verify the question is correctly set. **Note:** With the given coordinates,$PQ = \sqrt{40}$,$QR = 5$,$PR = \sqrt{45}$. Since no two sides are equal, the triangle is scalene. However, if the question asks to "show it is isosceles," the coordinates may need adjustment. For this answer key, we proceed with the calculation as shown. **Answer:**$PQ = \sqrt{40}$,$QR = 5$,$PR = \sqrt{45}$. Since$PQ \neq QR \neq PR$, the triangle is not isosceles with these coordinates. (If the question requires showing it is isosceles, the coordinates in the question should be adjusted so that two sides are equal.)

(b) [2 marks]
Using the shoelace formula:
$\text{Area} = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
$= \frac{1}{2}|2(3 - 7) + 8(7 - 1) + 5(1 - 3)|$
$= \frac{1}{2}|2(-4) + 8(6) + 5(-2)|$
$= \frac{1}{2}|-8 + 48 - 10| = \frac{1}{2}|30| = 15$
Answer: 15 square units

15.
(a) [1 mark]
Centre $= (-1, 3)$, radius $= \sqrt{25} = 5$.
Answer: Centre $(-1, 3)$, radius $= 5$

(b) [4 marks]
Verify $(3, 6)$ lies on the circle: $(3 + 1)^2 + (6 - 3)^2 = 16 + 9 = 25$ ✓
Gradient of radius from centre $(-1, 3)$ to $(3, 6)$:
$m_r = \frac{6 - 3}{3 - (-1)} = \frac{3}{4}$
Gradient of tangent $= -\frac{1}{m_r} = -\frac{4}{3}$.
Equation of tangent at $(3, 6)$:
$y - 6 = -\frac{4}{3}(x - 3)$
$y - 6 = -\frac{4}{3}x + 4$
$y = -\frac{4}{3}x + 10$
Multiply by 3: $3y = -4x + 30$, so $4x + 3y - 30 = 0$.
Answer: $4x + 3y - 30 = 0$ (or $y = -\frac{4}{3}x + 10$)

16.
(a) [3 marks]
Substitute $y = 3x - 5$ into the circle equation:
$x^2 + (3x - 5)^2 - 4x + 2(3x - 5) - 20 = 0$
$x^2 + 9x^2 - 30x + 25 - 4x + 6x - 10 - 20 = 0$
$10x^2 - 28x - 5 = 0$
Note: The question states $10x^2 - 34x + 5 = 0$. Let me recheck:
$x^2 + (3x - 5)^2 - 4x + 2(3x - 5) - 20 = 0$
$x^2 + 9x^2 - 30x + 25 - 4x + 6x - 10 - 20 = 0$
$10x^2 - 28x - 5 = 0$
The derived equation is $10x^2 - 28x - 5 = 0$, not $10x^2 - 34x + 5 = 0$. The question may contain a typo. Proceeding with the correct derivation:
Answer: Substituting and simplifying gives $10x^2 - 28x - 5 = 0$.

(b) [3 marks]
Using $10x^2 - 28x - 5 = 0$:
$x = \frac{28 \pm \sqrt{784 + 200}}{20} = \frac{28 \pm \sqrt{984}}{20} = \frac{28 \pm 2\sqrt{246}}{20} = \frac{14 \pm \sqrt{246}}{10}$
$x_1 = \frac{14 + \sqrt{246}}{10} \approx \frac{14 + 15.68}{10} \approx 2.968$
$x_2 = \frac{14 - \sqrt{246}}{10} \approx \frac{14 - 15.68}{10} \approx -0.168$
Corresponding $y$ values:
$y_1 = 3(2.968) - 5 \approx 8.904 - 5 = 3.904$
$y_2 = 3(-0.168) - 5 \approx -0.504 - 5 = -5.504$
Answer: $A \approx (2.97, 3.90)$ and $B \approx (-0.168, -5.50)$ (to 3 s.f.)

17.
(a) [2 marks]
Using section formula (ratio $3:1$):
$C = \left(\frac{3(6) + 1(-2)}{3 + 1}, \frac{3(-5) + 1(3)}{3 + 1}\right) = \left(\frac{18 - 2}{4}, \frac{-15 + 3}{4}\right) = \left(\frac{16}{4}, \frac{-12}{4}\right) = (4, -3)$
Answer: $(4, -3)$

(b) [3 marks]
Line $x + 2y = 6 \implies y = -\frac{1}{2}x + 3$, so gradient $= -\frac{1}{2}$.
Parallel line through $C(4, -3)$ with gradient $-\frac{1}{2}$:
$y - (-3) = -\frac{1}{2}(x - 4)$
$y + 3 = -\frac{1}{2}x + 2$
$y = -\frac{1}{2}x - 1$
Multiply by 2: $2y = -x - 2$, so $x + 2y + 2 = 0$.
Answer: $x + 2y + 2 = 0$ (or $y = -\frac{1}{2}x - 1$)

Section C

18.
(a) [2 marks]
$AC = \sqrt{(12 - 0)^2 + (8 - 0)^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}$
Answer: $4\sqrt{13}$ coordinate units (or $\approx 14.42$ units)

(b) [4 marks]
Midpoint of $AC = \left(\frac{0 + 12}{2}, \frac{0 + 8}{2}\right) = (6, 4)$.
Gradient of $AC = \frac{8 - 0}{12 - 0} = \frac{2}{3}$.
Gradient of perpendicular bisector $= -\frac{3}{2}$.
Equation through $(6, 4)$:
$y - 4 = -\frac{3}{2}(x - 6)$
$y - 4 = -\frac{3}{2}x + 9$
$y = -\frac{3}{2}x + 13$
Multiply by 2: $2y = -3x + 26$, so $3x + 2y - 26 = 0$.
Answer: $3x + 2y - 26 = 0$ (or $y = -\frac{3}{2}x + 13$)

(c) [3 marks]
The point on the perpendicular bisector closest to the origin is the foot of the perpendicular from the origin to the line $3x + 2y - 26 = 0$.
The line through the origin perpendicular to $3x + 2y - 26 = 0$ has gradient $\frac{2}{3}$ (negative reciprocal of $-\frac{3}{2}$).
Equation: $y = \frac{2}{3}x$.
Substitute into $3x + 2y = 26$:
$3x + 2\left(\frac{2}{3}x\right) = 26$
$3x + \frac{4}{3}x = 26$
$\frac{9x + 4x}{3} = 26 \implies \frac{13x}{3} = 26 \implies x = 6$
$y = \frac{2}{3}(6) = 4$
Answer: $(6, 4)$ — this is the midpoint of $AC$, which lies on the perpendicular bisector.

19.
(a) [2 marks]
$C_1$: centre $(0, 0)$, radius $= \sqrt{16} = 4$.
$C_2$: centre $(6, 0)$, radius $= \sqrt{4} = 2$.
Answer: $C_1$: centre $(0, 0)$, radius $4$; $C_2$: centre $(6, 0)$, radius $2$

(b) [3 marks]
Distance between centres:
$d = \sqrt{(6 - 0)^2 + (0 - 0)^2} = \sqrt{36} = 6$
Sum of radii $= 4 + 2 = 6$.
Since the distance between centres equals the sum of the radii, the circles touch externally.
Answer: The circles touch externally because the distance between centres (6) equals the sum of the radii (6).

(c) [3 marks]
The point of contact divides the line joining centres in the ratio of the radii $4:2 = 2:1$.
Point of contact from $C_1(0,0)$ towards $C_2(6,0)$:
$= \left(\frac{2(6) + 1(0)}{2 + 1}, \frac{2(0) + 1(0)}{2 + 1}\right) = \left(\frac{12}{3}, 0\right) = (4, 0)$
The common tangent at the point of contact is perpendicular to the line joining the centres (which is horizontal).
Therefore, the tangent is vertical: $x = 4$.
Answer: $x = 4$

20.
(a) [4 marks]
Set $x^2 - 4x + 7 = x + 3$:
$x^2 - 4x + 7 - x - 3 = 0$
$x^2 - 5x + 4 = 0$
$(x - 1)(x - 4) = 0$
$x = 1 \text{ or } x = 4$
When $x = 1$: $y = 1 + 3 = 4$.
When $x = 4$: $y = 4 + 3 = 7$.
Answer: $P(1, 4)$ and $Q(4, 7)$

(b) [4 marks]
Midpoint of $PQ = \left(\frac{1 + 4}{2}, \frac{4 + 7}{2}\right) = \left(\frac{5}{2}, \frac{11}{2}\right)$.
Gradient of $PQ = \frac{7 - 4}{4 - 1} = \frac{3}{3} = 1$.
Gradient of perpendicular bisector $= -1$.
Equation through $\left(\frac{5}{2}, \frac{11}{2}\right)$:
$y - \frac{11}{2} = -1\left(x - \frac{5}{2}\right)$
$y - \frac{11}{2} = -x + \frac{5}{2}$
$y = -x + \frac{5}{2} + \frac{11}{2} = -x + 8$
Answer: $y = -x + 8$ (or $x + y - 8 = 0$)

(c) [2 marks]
The axis of symmetry of the parabola $y = x^2 - 4x + 7$ is:
$x = -\frac{-4}{2(1)} = 2$
The perpendicular bisector of $PQ$ is $y = -x + 8$, which is a line with gradient $-1$, not a vertical line.
The axis of symmetry $x = 2$ is a vertical line.
The perpendicular bisector of $PQ$ passes through the midpoint of $PQ$, which is $\left(\frac{5}{2}, \frac{11}{2}\right)$. The axis of symmetry $x = 2$ passes through $x = 2$.
Note: The axis of symmetry of the parabola passes through the midpoint of $PQ$ only if $P$ and $Q$ are symmetric about the axis. Here, $P(1, 4)$ and $Q(4, 7)$ — the midpoint has $x = 2.5$, not $x = 2$.
Answer: The perpendicular bisector of $PQ$ ($y = -x + 8$) is not the same as the axis of symmetry of the parabola ($x = 2$). They are different lines. The axis of symmetry is vertical while the perpendicular bisector has gradient $-1$.