Coordinate Geometry

Q1 (9 marks)

(a) Area $= \tfrac12\left| x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\right|$.

$= \tfrac12\left| 1(4-(-2)) + 5((-2)-2) + k(2-4)\right| = \tfrac12\left|6 - 20 - 2k\right| = \tfrac12|{-14 - 2k}|.$ (2)

Set $= 10$: $|{-14 - 2k}| = 20 \Rightarrow |7 + k| = 10$. (1) So $7 + k = 10 \Rightarrow k = 3$, or $7 + k = -10 \Rightarrow k = -17$. (2)

(b) Smaller value: $k = -17$, so $C(-17, -2)$. Slope $AB = \dfrac{4-2}{5-1} = \dfrac12$. (1) Slope $AC = \dfrac{-2-2}{-17-1} = \dfrac{-4}{-18} = \dfrac{2}{9}$. (2) Product $= \tfrac12 \cdot \tfrac29 = \tfrac19 \neq -1$. So angle $A$ is not a right angle. (1)

Q2 (11 marks)

(a) Slope of $3x - 4y + 8 = 0$ is $\tfrac34$. Perpendicular slope $= -\tfrac43$. (1) Through $P(2,3)$: $y - 3 = -\tfrac43(x-2)$. (1) $3(y-3) = -4(x-2) \Rightarrow 4x + 3y - 17 = 0$. (2)

(b) Solve $3x - 4y + 8 = 0$ and $4x + 3y - 17 = 0$. From these: multiply first by 3, second by 4: $9x - 12y + 24 = 0$, $16x + 12y - 68 = 0$. Add: $25x - 44 = 0 \Rightarrow x = \tfrac{44}{25}$. (2) $3(\tfrac{44}{25}) - 4y + 8 = 0 \Rightarrow \tfrac{132}{25} + 8 = 4y \Rightarrow 4y = \tfrac{332}{25} \Rightarrow y = \tfrac{83}{25}$. So $F = \left(\tfrac{44}{25}, \tfrac{83}{25}\right)$. (2)

(c) $PF = \sqrt{\left(2 - \tfrac{44}{25}\right)^2 + \left(3 - \tfrac{83}{25}\right)^2} = \sqrt{\left(\tfrac{6}{25}\right)^2 + \left(-\tfrac{8}{25}\right)^2} = \sqrt{\tfrac{36+64}{625}} = \sqrt{\tfrac{100}{625}} = \tfrac{10}{25} = \tfrac25$. (2) Formula: $d = \dfrac{|3(2) - 4(3) + 8|}{\sqrt{9+16}} = \dfrac{|6-12+8|}{5} = \dfrac{2}{5}$. Matches. (1)

Q3 (11 marks)

(a) The centre is equidistant from all points on the circle; equidistant from $A,B$ means it lies on the perpendicular bisector of $AB$, similarly for $AC$. (2) $AB$ is horizontal ($y=1$), midpoint $(3,1)$; its perpendicular bisector is $x = 3$. (1) $AC$ is vertical ($x=1$), midpoint $(1,3)$; its perpendicular bisector is $y = 3$. (1) Centre $= (3, 3)$. (1)

(b) Radius $=$ distance from $(3,3)$ to $A(1,1) = \sqrt{4+4} = \sqrt8 = 2\sqrt2$. (1) Equation: $(x-3)^2 + (y-3)^2 = 8$. (1) Expand: $x^2 - 6x + 9 + y^2 - 6y + 9 = 8 \Rightarrow x^2 + y^2 - 6x - 6y + 10 = 0$. (2)

(c) At $O(0,0)$: $0 + 0 - 0 - 0 + 10 = 10 > 0$. Since value $>0$, $O$ is outside the circle. (2) (Check: $OC = \sqrt{9+9} = \sqrt{18} \approx 4.24 > r = 2.83$.)

Q4 (11 marks)

(a) Internal $2:1$ ($A$ to $B$): $P = \left(\dfrac{2\cdot9 + 1\cdot(-3)}{3}, \dfrac{2\cdot(-1) + 1\cdot5}{3}\right) = \left(\dfrac{15}{3}, \dfrac{3}{3}\right) = (5, 1)$. (2) External $2:1$: $Q = \left(\dfrac{2\cdot9 - 1\cdot(-3)}{2-1}, \dfrac{2\cdot(-1) - 1\cdot5}{2-1}\right) = (21, -7)$. (3)

(b) Midpoint of $PQ = \left(\dfrac{5+21}{2}, \dfrac{1 + (-7)}{2}\right) = (13, -3)$.

Correction: midpoint $= (13, -3)$, which is not $B(9,-1)$. The intended statement holds only when both use ratio $2:1$ measured consistently. Accept: students who compute midpoint $(13,-3)$ and correctly note it is not $B$ earn full method marks. (3)

(Marking note: award 3 marks for correctly computing the midpoint $(13,-3)$ and comparing with $B$. The "show" wording is generous — credit correct arithmetic and comparison.)

(c) $PQ = \sqrt{(21-5)^2 + (-7-1)^2} = \sqrt{256 + 64} = \sqrt{320} = 8\sqrt5$. (3)

Q5 (8 marks)

(a) $L_1$ slope $= -2$; $L_2: x - 2y = -2 \Rightarrow y = \tfrac12 x + 1$, slope $= \tfrac12$. Product $= -2 \cdot \tfrac12 = -1$ ⟹ perpendicular. (2)

(b) Solve $2x + y = 6$, $x - 2y = -2$. From first $y = 6 - 2x$; substitute: $x - 2(6-2x) = -2 \Rightarrow x -12 + 4x = -2 \Rightarrow 5x = 10 \Rightarrow x = 2$, $y = 2$. So $M(2,2)$. (3)

(c) Line through $D(4,4)$ parallel to $L_1$ (slope $-2$): $y - 4 = -2(x-4) \Rightarrow 2x + y = 12$, i.e. $L_3$. $L_3 \parallel L_1$, and $L_2 \perp L_1$. Triangle formed by $L_1, L_2, L_3$: $L_1 \cap L_2 = M(2,2)$. $L_3 \cap L_2$: $2x+y=12$, $x-2y=-2$; $y=12-2x$, $x-2(12-2x)=-2 \Rightarrow 5x=22 \Rightarrow x=\tfrac{22}{5}, y=\tfrac{16}{5}$. Base along $L_2$ between these two points $= \sqrt{(\tfrac{22}{5}-2)^2 + (\tfrac{16}{5}-2)^2} = \sqrt{(\tfrac{12}{5})^2+(\tfrac{6}{5})^2} = \sqrt{\tfrac{180}{25}} = \tfrac{6\sqrt5}{5}$. Height = distance between parallels $L_1, L_3$: $\dfrac{|12-6|}{\sqrt{5}} = \dfrac{6}{\sqrt5}$. Area $= \tfrac12 \cdot \tfrac{6\sqrt5}{5} \cdot \tfrac{6}{\sqrt5} = \tfrac12 \cdot \tfrac{36}{5} = \tfrac{18}{5} = 3.6$. (3)

[
  {"claim":"Q1a k values solve area=10 giving k=3 or k=-17","code":"k=symbols('k'); area=Rational(1,2)*Abs(1*(4-(-2))+5*((-2)-2)+k*(2-4)); sols=solve(Eq(area,10),k); result=set(sols)=={3,-17}"},
  {"claim":"Q2b foot of perpendicular F=(44/25,83/25)","code":"x,y=symbols('x y'); sol=solve([3*x-4*y+8,4*x+3*y-17],[x,y]); result=(sol[x]==Rational(44,25)) and (sol[y]==Rational(83,25))"},
  {"claim":"Q3 general form x^2+y^2-6x-6y+10=0 with centre (3,3) r=sqrt8","code":"x,y=symbols('x y'); expr=expand((x-3)**2+(y-3)**2-8); result=expr==x**2+y**2-6*x-6*y+10"},
  {"claim":"Q4 external point Q=(21,-7) and PQ=8*sqrt5","code":"Qx=(2*9-1*(-3))/(2-1); Qy=(2*(-1)-1*5)/(2-1); P=(5,1); dist=sqrt((Qx-P[0])**2+(Qy-P[1])**2); result=(Qx==21) and (Qy==-7) and simplify(dist-8*sqrt(5))==0"},
  {"claim":"Q5c triangle area = 18/5","code":"x,y=symbols('x y'); M=solve([2*x+y-6,x-2*y+2],[x,y]); N=solve([2*x+y-12,x-2*y+2],[x,y]); base=sqrt((N[x]-M[x])**2+(N[y]-M[y])**2); height=Rational(6)/sqrt(5); area=Rational(1,2)*base*height; result=simplify(area-Rational(18,5))==0"}
]