Level 1 — RecognitionCoordinate Geometry

Coordinate Geometry

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Level: 1 (Recognition) Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Choose the single correct option.

Q1. The point $(-3, 5)$ lies in which quadrant? (a) I (b) II (c) III (d) IV

Q2. The distance between the origin and the point $(3, 4)$ is: (a) $5$ (b) $7$ (c) $\sqrt{7}$ (d) $12$

Q3. The midpoint of the segment joining $(2, -4)$ and $(6, 8)$ is: (a) $(4, 2)$ (b) $(8, 4)$ (c) $(4, 4)$ (d) $(2, 2)$

Q4. The slope of the line through $(1, 2)$ and $(4, 8)$ is: (a) $2$ (b) $\tfrac{1}{2}$ (c) $-2$ (d) $6$

Q5. The line $y = -3x + 7$ has y-intercept: (a) $-3$ (b) $7$ (c) $3$ (d) $-7$

Q6. Two lines are perpendicular if the product of their slopes equals: (a) $1$ (b) $0$ (c) $-1$ (d) $\infty$

Q7. The x-intercept of $2x + 5y = 10$ is: (a) $(0, 2)$ (b) $(5, 0)$ (c) $(2, 0)$ (d) $(0, 5)$

Q8. The centre of the circle $(x - 2)^2 + (y + 3)^2 = 16$ is: (a) $(2, 3)$ (b) $(-2, 3)$ (c) $(2, -3)$ (d) $(-2, -3)$

Q9. The radius of the circle in Q8 is: (a) $16$ (b) $8$ (c) $4$ (d) $256$

Q10. A line parallel to $y = 4x - 1$ has slope: (a) $-\tfrac{1}{4}$ (b) $4$ (c) $-4$ (d) $1$

Section B — Matching (1 mark each, 5 marks total)

Q11. Match each formula in Column X to its description in Column Y.

Column X		Column Y
(i) $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$		(P) slope
(ii) $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$		(Q) area of triangle
(iii) $\frac{y_2-y_1}{x_2-x_1}$		(R) distance
(iv) $\frac{	ax_0+by_0+c	}{\sqrt{a^2+b^2}}$
(v) $\frac{1}{2}\lvert x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\rvert$		(T) distance point-to-line

Write the correct pairing for (i)–(v).

Section C — True/False with Justification (3 marks each)

State True or False (1 mark) and give a brief justification (2 marks).

Q12. The three points $(0,0)$ , $(2,2)$ and $(5,5)$ are collinear.

Q13. The lines $y = 2x + 1$ and $2x - y + 5 = 0$ are parallel.

Q14. The point that divides the segment from $A(0,0)$ to $B(6,9)$ internally in the ratio $2:1$ is $(4,6)$ .

Q15. The equation $x^2 + y^2 - 6x + 8y + 9 = 0$ represents a circle of radius $4$ .

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (b) II — $x<0$ , $y>0$ ⇒ second quadrant. (1)

Q2. (a) $5$ — $\sqrt{3^2+4^2}=\sqrt{25}=5$ . (1)

Q3. (c) $(4,2)$ — $\left(\frac{2+6}{2},\frac{-4+8}{2}\right)=(4,2)$ . (1) (Note: correct answer is $(4,2)$ ; option (a).) Correct option: (a) $(4,2)$ .

Q4. (a) $2$ — $\frac{8-2}{4-1}=\frac{6}{3}=2$ . (1)

Q5. (b) $7$ — In $y=mx+c$ , $c=7$ . (1)

Q6. (c) $-1$ — Perpendicular condition $m_1 m_2 = -1$ . (1)

Q7. (b) $(5,0)$ — Set $y=0$ : $2x=10\Rightarrow x=5$ . (1)

Q8. (c) $(2,-3)$ — Centre $(h,k)=(2,-3)$ . (1)

Q9. (c) $4$ — $r^2=16\Rightarrow r=4$ . (1)

Q10. (b) $4$ — Parallel lines have equal slopes. (1)

Section B

Q11. (i)–R, (ii)–S, (iii)–P, (iv)–T, (v)–Q. (1 each, 5 total)

Section C

Q12. True. (1) Slope $(0,0)\to(2,2)$ is $1$ ; slope $(0,0)\to(5,5)$ is $1$ . Equal slopes through common point ⇒ collinear (all lie on $y=x$ ). (2)

Q13. False. (1) Line 1 slope $=2$ . Line 2: $2x-y+5=0\Rightarrow y=2x+5$ , slope $=2$ . Slopes are equal but they are two distinct parallel lines — however since the statement claims they are parallel and they are indeed parallel...

Correction: Both have slope $2$ , so they are parallel ⇒ True. (1) Justification: rewriting the second as $y=2x+5$ gives slope $2$ , matching line 1's slope $2$ ; different intercepts confirm distinct parallel lines. (2)

Q14. True. (1) Section formula: $\left(\frac{2\cdot6+1\cdot0}{3},\frac{2\cdot9+1\cdot0}{3}\right)=\left(\frac{12}{3},\frac{18}{3}\right)=(4,6)$ . (2)

Q15. True. (1) Complete squares: $(x-3)^2+(y+4)^2 = 9+16-9 = 16$ , so $r=\sqrt{16}=4$ . (2)

[
  {"claim":"Distance origin to (3,4) is 5","code":"result = sqrt(3**2+4**2)==5"},
  {"claim":"Midpoint of (2,-4),(6,8) is (4,2)","code":"result = (Rational(2+6,2), Rational(-4+8,2))==(4,2)"},
  {"claim":"Slope through (1,2),(4,8) is 2","code":"result = Rational(8-2,4-1)==2"},
  {"claim":"Section 2:1 of (0,0),(6,9) is (4,6)","code":"result = (Rational(2*6+1*0,3), Rational(2*9+1*0,3))==(4,6)"},
  {"claim":"Circle x2+y2-6x+8y+9=0 has radius 4","code":"result = sqrt(9+16-9)==4"}
]