Secondary 2 Mathematics Graphs Coordinate Geometry Quiz

Secondary 2 Mathematics Quiz - Graphs Coordinate Geometry (Answer Key)

Total Marks: 40

Section A

1. Length of $AB$
Answer: $10$ units
Working:
Use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$AB = \sqrt{(8 - 2)^2 + (13 - 5)^2}$
$AB = \sqrt{6^2 + 8^2}$
$AB = \sqrt{36 + 64} = \sqrt{100} = 10$
Marks: [2] (1 for substitution, 1 for correct answer)

2. Midpoint of $PQ$
Answer: $(1, 3)$
Working:
Midpoint formula: $\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
$x = \frac{-3 + 5}{2} = \frac{2}{2} = 1$
$y = \frac{7 + (-1)}{2} = \frac{6}{2} = 3$
Marks: [2] (1 for x-coord, 1 for y-coord)

3. Gradient of line
Answer: $2$
Working:
Gradient $m = \frac{y_2 - y_1}{x_2 - x_1}$
$m = \frac{10 - 4}{4 - 1} = \frac{6}{3} = 2$
Marks: [1]

4. Equation of line
Answer: $y = -3x + 5$
Working:
The equation of a line is $y = mx + c$.
Given gradient $m = -3$ and $y$-intercept $c = 5$ (since it passes through $(0,5)$).
Marks: [1]

5. Gradient of $2y = 6x - 8$
Answer: $3$
Working:
Rearrange into $y = mx + c$:
$2y = 6x - 8$
$y = \frac{6x}{2} - \frac{8}{2}$
$y = 3x - 4$
Gradient $m = 3$.
Marks: [1]

Section B

6. $y$-intercept of $3x + 4y = 12$
Answer: $3$ (or coordinate $(0, 3)$)
Working:
At $y$-intercept, $x = 0$.
$3(0) + 4y = 12$
$4y = 12$
$y = 3$
Marks: [1]

7. Gradient of parallel line
Answer: $2$
Working:
Parallel lines have equal gradients.
$L_1: y = 2x + 1 \implies m = 2$.
Therefore, gradient of $L_2$ is also $2$.
Marks: [1]

8. Gradient of perpendicular line
Answer: $-\frac{1}{4}$
Working:
Product of gradients of perpendicular lines is $-1$.
$m_1 \times m_2 = -1$
$4 \times m_2 = -1$
$m_2 = -\frac{1}{4}$
Marks: [1]

9. Value of $k$
Answer: $3$
Working:
Substitute $x = k$ and $y = 6$ into $y = 3x - 3$:
$6 = 3(k) - 3$
$9 = 3k$
$k = 3$
Marks: [2] (1 for substitution, 1 for answer)

10. Equation of line
Answer: $y = 2x + 1$
Working:
$y = mx + c$. Given $m = 2$, so $y = 2x + c$.
Passes through $(1, 3)$:
$3 = 2(1) + c$
$3 = 2 + c \implies c = 1$
Equation: $y = 2x + 1$
Marks: [2] (1 for finding c, 1 for equation)

Section C

11. Triangle $ABC$
Visual Context: $A(1,1)$, $B(5,1)$, $C(3,5)$.

(a) Length of $AC$
Answer: $\sqrt{20}$ or $2\sqrt{5}$ or approx $4.47$
Working:
$AC = \sqrt{(3 - 1)^2 + (5 - 1)^2}$
$AC = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20}$
Marks: [2]

(b) Area of $\triangle ABC$
Answer: $8$ units$^2$
Working:
Base $AB$ is horizontal. Length $AB = 5 - 1 = 4$ units.
Height is vertical distance from $C$ to line $AB$. $y_C - y_A = 5 - 1 = 4$ units.
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times 4 \times 4 = 8$
Marks: [2] (1 for base/height identification, 1 for calculation)

12. Equation through $A(2, 5)$ and $B(6, 17)$
Answer: $y = 3x - 1$
Working:
Step 1: Find gradient $m$.
$m = \frac{17 - 5}{6 - 2} = \frac{12}{4} = 3$
Step 2: Use $y = mx + c$.
$y = 3x + c$
Substitute $(2, 5)$: $5 = 3(2) + c \implies 5 = 6 + c \implies c = -1$
Equation: $y = 3x - 1$
Marks: [3] (1 for gradient, 1 for intercept, 1 for equation)

13. Line $4x + 2y = 10$
(a) Gradient
Answer: $-2$
Working:
$2y = -4x + 10$
$y = -2x + 5$
Gradient $m = -2$
Marks: [1]

(b) $x$-intercept
Answer: $(2.5, 0)$ or $x = 2.5$
Working:
At $x$-intercept, $y = 0$.
$4x + 2(0) = 10$
$4x = 10$
$x = \frac{10}{4} = 2.5$
Marks: [2] (1 for setting y=0, 1 for answer)

Section D

14. Points $A(1, 2)$ and $B(7, 8)$
(a) Midpoint $M$
Answer: $(4, 5)$
Working:
$x = \frac{1+7}{2} = 4$
$y = \frac{2+8}{2} = 5$
Marks: [1]

(b) Gradient of perpendicular
Answer: $-1$
Working:
Gradient of $AB = \frac{8-2}{7-1} = \frac{6}{6} = 1$.
Gradient of perpendicular $= -\frac{1}{1} = -1$.
Marks: [2] (1 for grad AB, 1 for perp grad)

15. Quadrilateral $ABCD$ with $A(1, 1)$, $B(5, 3)$, $C(7, 7)$, $D(3, 5)$
(a) Midpoints of diagonals
Answer: Both midpoints are $(4, 4)$.
Working:
Midpoint of $AC$: $\left(\frac{1+7}{2}, \frac{1+7}{2}\right) = (4, 4)$
Midpoint of $BD$: $\left(\frac{5+3}{2}, \frac{3+5}{2}\right) = (4, 4)$
Since midpoints are identical, diagonals bisect each other.
Marks: [2]

(b) Geometric Name
Answer: Rhombus (Parallelogram is also acceptable but Rhombus is more specific).
Note: To be a rhombus, adjacent sides must be equal. $AB = \sqrt{16+4}=\sqrt{20}$, $BC=\sqrt{4+16}=\sqrt{20}$. It is a Rhombus.
Marks: [2] (1 for Parallelogram, 1 for Rhombus if checked, otherwise 1 for correct classification based on property shown). Award full marks for Rhombus.

16. Lines $L_1$ and $L_2$
(a) Equation of $L_1$ through $(0, 4)$ and $(2, 0)$
Answer: $y = -2x + 4$
Working:
Gradient $m = \frac{0 - 4}{2 - 0} = \frac{-4}{2} = -2$.
$y$-intercept is given as $4$ (point $(0,4)$).
Equation: $y = -2x + 4$.
Marks: [2]

(b) Equation of $L_2$ perpendicular to $L_1$ through $(2, 0)$
Answer: $y = \frac{1}{2}x - 1$
Working:
Gradient of $L_1$ is $-2$.
Gradient of $L_2$ ($m_{\perp}$) $= -\frac{1}{-2} = \frac{1}{2}$.
Equation: $y = \frac{1}{2}x + c$.
Passes through $(2, 0)$:
$0 = \frac{1}{2}(2) + c$
$0 = 1 + c \implies c = -1$.
Equation: $y = \frac{1}{2}x - 1$.
Marks: [2] (1 for correct gradient, 1 for correct equation)

17. Triangle $PQR$ with $P(2, 3)$, $Q(6, 3)$, $R(4, 7)$
(a) Show Isosceles
Answer: $PR = QR = \sqrt{20}$
Working:
$PR = \sqrt{(4-2)^2 + (7-3)^2} = \sqrt{2^2 + 4^2} = \sqrt{4+16} = \sqrt{20}$
$QR = \sqrt{(4-6)^2 + (7-3)^2} = \sqrt{(-2)^2 + 4^2} = \sqrt{4+16} = \sqrt{20}$
Since $PR = QR$, the triangle is isosceles.
Marks: [2]

(b) Area of $\triangle PQR$
Answer: $8$ units$^2$
Working:
Base $PQ$ is horizontal. Length $PQ = 6 - 2 = 4$.
Height is vertical distance from $R$ to $PQ$. $y_R - y_P = 7 - 3 = 4$.
Area $= \frac{1}{2} \times 4 \times 4 = 8$.
Marks: [2]

18. Intersection of $y = 2x + 1$ and $y = -x + 7$
(a) Coordinates of $K$
Answer: $(2, 5)$
Working:
$2x + 1 = -x + 7$
$3x = 6 \implies x = 2$
$y = 2(2) + 1 = 5$
$K(2, 5)$
Marks: [2]

(b) Distance $OK$
Answer: $\sqrt{29}$
Working:
$O(0,0)$, $K(2,5)$
$OK = \sqrt{(2-0)^2 + (5-0)^2} = \sqrt{4 + 25} = \sqrt{29}$
Marks: [2]

19. Line through $A(-2, 1)$ and $B(4, 5)$
(a) Equation in $ax + by = c$
Answer: $2x - 3y = -7$ (or equivalent e.g., $-2x + 3y = 7$)
Working:
Gradient $m = \frac{5-1}{4-(-2)} = \frac{4}{6} = \frac{2}{3}$
$y - 1 = \frac{2}{3}(x + 2)$
$3(y - 1) = 2(x + 2)$
$3y - 3 = 2x + 4$
$2x - 3y = -7$
Marks: [2]

(b) Check point $C(10, 9)$
Answer: Yes, it lies on the line.
Working:
LHS: $2(10) - 3(9) = 20 - 27 = -7$
RHS: $-7$
LHS = RHS, so $C$ lies on the line.
Marks: [2]

20. Midpoint of $A(3, k)$ and $B(7, 5)$ is $M(5, 2)$
(a) Value of $k$
Answer: $-1$
Working:
$y$-coord of midpoint: $\frac{k + 5}{2} = 2$
$k + 5 = 4$
$k = -1$
Marks: [2]

(b) Length of $AB$
Answer: $\sqrt{52}$ or $2\sqrt{13}$
Working:
$A(3, -1)$, $B(7, 5)$
$AB = \sqrt{(7-3)^2 + (5-(-1))^2}$
$AB = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$
Marks: [2]