Coordinate Geometry

Question 1 [10]

(a) [4] Plot $A(x_1,y_1)$, $B(x_2,y_2)$. Drop perpendiculars to form right triangle with third vertex $C(x_2, y_1)$. Horizontal leg $AC = |x_2 - x_1|$ [2]; vertical leg $BC = |y_2 - y_1|$ [2].

(b) [3] By Pythagoras, $AB^2 = AC^2 + BC^2$ [1] $= (x_2-x_1)^2 + (y_2-y_1)^2$ (absolute values squared drop) [1]. Hence $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ [1].

(c) [3] Because $(x_2-x_1)^2 = (x_1-x_2)^2$ [1] (a real number squared equals its negative squared) [1], swapping the labels of the two points leaves each squared term unchanged, so $d(A,B)=d(B,A)$ — distance is symmetric [1].

Question 2 [12]

(a) [5] Let $P(x,y)$ divide $AB$ with $AP:PB = m:n$. Drop perpendiculars to x-axis; similar triangles give $\dfrac{AP}{PB} = \dfrac{x - x_1}{x_2 - x} = \dfrac{m}{n}$ [2]. Cross-multiply: $n(x-x_1) = m(x_2-x)$ [1] $\Rightarrow x(m+n) = mx_2 + nx_1 \Rightarrow x = \dfrac{mx_2+nx_1}{m+n}$ [1]. Same with y-coordinates gives $y = \dfrac{my_2+ny_1}{m+n}$ [1].

(b) [3] External division: replace $n$ by $-n$, giving $P=\left(\dfrac{mx_2-nx_1}{m-n}, \dfrac{my_2-ny_1}{m-n}\right)$ [1]. Sign flips because $P$ lies outside the segment, so the directed ratio $AP:PB$ is negative (the vectors point the same way rather than opposite) [2].

(c) [4] $m:n=3:1$, $A(-2,3)$, $B(6,-5)$. $x = \dfrac{3(6)+1(-2)}{4} = \dfrac{16}{4}=4$ [2]; $y = \dfrac{3(-5)+1(3)}{4} = \dfrac{-12}{4}=-3$ [1]. Point $= (4,-3)$ [1].

Question 3 [10]

(a) [6] Line $ax+by+c=0$ has normal vector $\mathbf{n}=(a,b)$ [1]. Foot of perpendicular $Q = P + t\mathbf{n}$ [1]. Since $Q$ on line: $a(x_0+ta)+b(y_0+tb)+c=0 \Rightarrow t = -\dfrac{ax_0+by_0+c}{a^2+b^2}$ [2]. Distance $= |t|\,|\mathbf{n}| = |t|\sqrt{a^2+b^2}$ [1] $= \dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ [1].

(b) [4] $a=5,b=-12,c=2$; point $(3,-4)$. Numerator $=|5(3)-12(-4)+2| = |15+48+2| = 65$ [2]. Denominator $=\sqrt{25+144}=\sqrt{169}=13$ [1]. Distance $=65/13 = 5$ [1].

Question 4 [10]

(a) [5] $x^2+2gx + y^2+2fy = -c$. Complete squares: $(x+g)^2 - g^2 + (y+f)^2 - f^2 = -c$ [2] $\Rightarrow (x+g)^2+(y+f)^2 = g^2+f^2-c$ [1]. Centre $(-g,-f)$ [1], radius $\sqrt{g^2+f^2-c}$ [1].

(b) [3] $2g=-6\Rightarrow g=-3$; $2f=8\Rightarrow f=4$; $c=-11$. Centre $(3,-4)$ [1]. Radius $=\sqrt{9+16-(-11)}=\sqrt{36}=6$ [2].

(c) [2] Need $g^2+f^2-c>0$ [1] so that $r^2>0$ and $r$ is real/positive; otherwise it's a point ($=0$) or imaginary ($<0$) [1].

Question 5 [10]

(a) [4] Area $=\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ [1]. $=\tfrac12|1(6-10)+4(10-2)+7(2-6)| = \tfrac12|-4+32-28| = \tfrac12(0)=0$ [2]. Area $=0\Rightarrow$ collinear [1].

(b) [2] Zero area means the "triangle" is degenerate — the three points cannot enclose any region [1], so they must all lie on a single straight line [1].

(c) [4] Slope $AB = \dfrac{6-2}{4-1}=\dfrac{4}{3}$ [1]. Line: $y-2=\tfrac43(x-1)\Rightarrow 3y-6=4x-4 \Rightarrow 4x-3y+2=0$ [2]. Check $C(7,10)$: $4(7)-3(10)+2=28-30+2=0$ ✓ [1].

Question 6 [8]

Model answer:

def line_through(p1, p2):
    x1, y1 = p1
    x2, y2 = p2
    # Direction (dx,dy); normal to line is (dy,-dx) => a=dy, b=-dx
    a = y2 - y1          # coefficient of x
    b = x1 - x2          # coefficient of y
    c = -(a*x1 + b*y1)   # force line to pass through p1
    return (a, b, c)
 
def on_line(coeffs, p):
    a, b, c = coeffs
    x, y = p
    return a*x + b*y + c == 0   # point satisfies ax+by+c=0

Mark scheme: correct $a,b$ from coordinate differences [3]; correct $c$ via substitution [2]; working on_line test [2]; meaningful comments [1].

[
  {"claim":"Section formula point A(-2,3),B(6,-5) ratio 3:1 is (4,-3)","code":"m,n=3,1; x=(m*6+n*(-2))/(m+n); y=(m*(-5)+n*3)/(m+n); result=(x==4 and y==-3)"},
  {"claim":"Distance from (3,-4) to 5x-12y+2=0 is 5","code":"d=abs(5*3-12*(-4)+2)/sqrt(5**2+12**2); result=simplify(d-5)==0"},
  {"claim":"Circle x^2+y^2-6x+8y-11=0 has centre (3,-4), radius 6","code":"g,f,c=-3,4,-11; centre=(-g,-f); r=sqrt(g**2+f**2-c); result=(centre==(3,-4) and r==6)"},
  {"claim":"A(1,2),B(4,6),C(7,10) are collinear (zero area)","code":"area=Rational(1,2)*abs(1*(6-10)+4*(10-2)+7*(2-6)); result=area==0"}
]