Noesis
Level 2 — RecallCoordinate Geometry

Coordinate Geometry

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 (Recall — definitions, standard problems, short derivations) Time Limit: 30 minutes Total Marks: 40

Q1. State the quadrant in which each of the following points lies: (−3,5)(-3, 5), (2,−7)(2, -7), (−4,−1)(-4, -1). (3 marks)

Q2. Find the distance between the points A(1,2)A(1, 2) and B(4,6)B(4, 6). (3 marks)

Q3. Find the coordinates of the midpoint of the segment joining P(−2,3)P(-2, 3) and Q(6,−5)Q(6, -5). (3 marks)

Q4. A point RR divides the join of A(2,1)A(2, 1) and B(8,7)B(8, 7) internally in the ratio 2:12:1. Find the coordinates of RR using the section formula. (4 marks)

Q5. Find the slope of the line passing through the points (−1,4)(-1, 4) and (3,−2)(3, -2), and state whether the line is increasing or decreasing. (4 marks)

Q6. Write the equation of the line with slope 22 passing through the point (1,−3)(1, -3). Then find its xx-intercept and yy-intercept. (5 marks)

Q7. Determine whether the lines 2x−3y+5=02x - 3y + 5 = 0 and 4x−6y−7=04x - 6y - 7 = 0 are parallel, perpendicular, or neither. Justify. (4 marks)

Q8. Find the perpendicular distance from the point (2,−1)(2, -1) to the line 3x+4y−12=03x + 4y - 12 = 0. (4 marks)

Q9. Find the area of the triangle whose vertices are A(1,1)A(1, 1), B(4,2)B(4, 2), and C(2,5)C(2, 5). Hence state whether the points are collinear. (5 marks)

Q10. Find the centre and radius of the circle x2+y2−6x+4y−12=0x^2 + y^2 - 6x + 4y - 12 = 0. (5 marks)

End of paper.

Answer keyMark scheme & solutions

Q1. (3 marks)

  • (−3,5)(-3, 5): x<0,y>0x<0, y>0 → Quadrant II (1)
  • (2,−7)(2, -7): x>0,y<0x>0, y<0 → Quadrant IV (1)
  • (−4,−1)(-4, -1): x<0,y<0x<0, y<0 → Quadrant III (1)

Why: Sign combination of coordinates fixes the quadrant.

Q2. (3 marks) d=(4−1)2+(6−2)2d = \sqrt{(4-1)^2 + (6-2)^2} (setup, 1) =32+42=9+16=25= \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} (1) =5= \mathbf{5} units (1)

Why: Distance formula from Pythagoras.

Q3. (3 marks) M=(−2+62,3+(−5)2)M = \left(\dfrac{-2+6}{2}, \dfrac{3+(-5)}{2}\right) (setup, 1) =(42,−22)= \left(\dfrac{4}{2}, \dfrac{-2}{2}\right) (1) =(2,−1)= \mathbf{(2, -1)} (1)

Why: Midpoint = average of coordinates.

Q4. (4 marks) Internal division, ratio m:n=2:1m:n = 2:1, A(2,1)A(2,1), B(8,7)B(8,7). R=(mx2+nx1m+n,my2+ny1m+n)R = \left(\dfrac{mx_2 + nx_1}{m+n}, \dfrac{my_2 + ny_1}{m+n}\right) (formula, 1) x=2(8)+1(2)3=183=6x = \dfrac{2(8)+1(2)}{3} = \dfrac{18}{3} = 6 (1) y=2(7)+1(1)3=153=5y = \dfrac{2(7)+1(1)}{3} = \dfrac{15}{3} = 5 (1) R=(6,5)R = \mathbf{(6, 5)} (1)

Q5. (4 marks) m=−2−43−(−1)m = \dfrac{-2 - 4}{3-(-1)} (formula, 1) =−64= \dfrac{-6}{4} (1) =−32= -\dfrac{3}{2} (1) Since m<0m<0, the line is decreasing. (1)

Q6. (5 marks) Point-slope: y−(−3)=2(x−1)y - (-3) = 2(x-1) (1) y+3=2x−2⇒y=2x−5y + 3 = 2x - 2 \Rightarrow y = 2x - 5 (1) y-intercept: set x=0x=0: y=−5y = -5 → (0,−5)(0,-5) (1) x-intercept: set y=0y=0: 0=2x−5⇒x=520 = 2x-5 \Rightarrow x = \dfrac{5}{2} → (52,0)\left(\tfrac{5}{2},0\right) (1) Equation correct and both intercepts (1)

Q7. (4 marks) Slope of line 1: −2−3=23-\dfrac{2}{-3} = \dfrac{2}{3} (1) Slope of line 2: −4−6=23-\dfrac{4}{-6} = \dfrac{2}{3} (1) Slopes equal ⇒\Rightarrow lines are parallel (1) (They are not the same line since intercepts differ.) (justification, 1)

Q8. (4 marks) d=∣3x0+4y0−12∣32+42d = \dfrac{|3x_0 + 4y_0 - 12|}{\sqrt{3^2+4^2}} (formula, 1) =∣3(2)+4(−1)−12∣25= \dfrac{|3(2)+4(-1)-12|}{\sqrt{25}} (1) =∣6−4−12∣5=∣−10∣5= \dfrac{|6-4-12|}{5} = \dfrac{|-10|}{5} (1) =105=2= \dfrac{10}{5} = \mathbf{2} units (1)

Q9. (5 marks) Area =12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣= \dfrac12 |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| (formula, 1) =12∣1(2−5)+4(5−1)+2(1−2)∣= \dfrac12 |1(2-5) + 4(5-1) + 2(1-2)| (1) =12∣−3+16−2∣= \dfrac12 |{-3} + 16 - 2| (1) =12(11)=112=5.5= \dfrac12 (11) = \dfrac{11}{2} = \mathbf{5.5} sq. units (1) Area ≠0\neq 0 → points are not collinear. (1)

Q10. (5 marks) General form: x2+y2+2gx+2fy+c=0x^2+y^2+2gx+2fy+c=0 with 2g=−6,2f=4,c=−122g=-6, 2f=4, c=-12 (1) g=−3, f=2g=-3,\ f=2 → Centre (−g,−f)=(3,−2)(-g,-f) = \mathbf{(3, -2)} (2) Radius =g2+f2−c=9+4+12=25= \sqrt{g^2+f^2-c} = \sqrt{9+4+12} = \sqrt{25} (1) =5= \mathbf{5} units (1)

[
  {"claim":"Q2 distance AB = 5","code":"result = sqrt((4-1)**2+(6-2)**2)==5"},
  {"claim":"Q4 section point = (6,5)","code":"x=(2*8+1*2)/3; y=(2*7+1*1)/3; result = (x==6) and (y==5)"},
  {"claim":"Q8 distance from point to line = 2","code":"result = Rational(abs(3*2+4*(-1)-12), sqrt(9+16))==2"},
  {"claim":"Q9 triangle area = 11/2","code":"result = Rational(1,2)*abs(1*(2-5)+4*(5-1)+2*(1-2))==Rational(11,2)"},
  {"claim":"Q10 centre (3,-2) radius 5","code":"g=-3; f=2; c=-12; result = (-g==3) and (-f==-2) and (sqrt(g**2+f**2-c)==5)"}
]