Level 5 — MasteryCoordinate Geometry

Coordinate Geometry

⏱ 75 minutes50 marksprintable — key stays hidden on paper

Time limit: 75 minutes
Total marks: 50
Instructions: Answer all three questions. Show full reasoning; unsupported answers earn no method marks. Use $...$ for mathematics. Pseudocode/Python is acceptable where a program is requested.

Question 1 — Geometry of a triangle and its circumscribed circle (20 marks)

A triangle has vertices $A(1,1)$ , $B(7,3)$ , $C(3,7)$ .

(a) Compute the three side lengths and the area of $\triangle ABC$ using the coordinate area formula. Verify the area independently using Heron's formula. (6)

(b) The circumcircle passes through $A$ , $B$ , $C$ . Starting from the general form $x^2+y^2+Dx+Ey+F=0$ , set up and solve the linear system to find $D$ , $E$ , $F$ . Hence state the centre and radius. (7)

(c) Prove that the circumcentre found in (b) is equidistant from all three vertices, and prove from first principles that the circumcentre lies on the perpendicular bisector of $AB$ (i.e. show the perpendicular-bisector condition holds). (4)

(d) Determine whether the point $P(5,5)$ lies inside, on, or outside the circumcircle. Justify. (3)

Question 2 — Physics: projectile trajectory meets a straight ramp (18 marks)

A ball is launched from the origin with speed $u=20\ \mathrm{m\,s^{-1}}$ at angle $45^\circ$ above the horizontal. Take $g=10\ \mathrm{m\,s^{-2}}$ . Its trajectory is the parabola $y = x - \frac{g\,x^2}{2u^2\cos^2 45^\circ}.$

(a) Show that the trajectory simplifies to $y = x - \dfrac{x^2}{40}$ , and find the horizontal range (the non-zero $x$ -intercept). (4)

(b) A straight ramp is described by the line $L:\ x - 2y + 4 = 0$ . Find its slope and its $x$ - and $y$ -intercepts. (4)

(c) The ball strikes the ramp. Find the coordinates of the impact point(s) by solving the trajectory against $L$ . (5)

(d) At the moment before impact the ball's velocity direction is tangent to the parabola. Find the slope of the trajectory at the impact point of largest $x$ , and hence the angle between the ball's path and the ramp at impact. (5)

Question 3 — Build & prove: a coordinate-geometry mini-library (12 marks)

(a) Write a function (pseudocode or Python) collinear(P, Q, R) that returns True iff three points are collinear, using the triangle-area test. Explain why area $=0$ is the correct criterion. (4)

(b) Prove the general result: the perpendicular distance from point $(x_0,y_0)$ to the line $ax+by+c=0$ equals $\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ . Give a clean vector or foot-of-perpendicular derivation. (5)

(c) Using your formula from (b), compute the distance from $(3,7)$ to the line $x-2y+4=0$ , and use collinear reasoning to confirm whether $(3,7)$ , $(1,1)$ , $(7,3)$ are collinear. (3)

Answer keyMark scheme & solutions

Question 1

(a) Side lengths:

$AB=\sqrt{(7-1)^2+(3-1)^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}$
$BC=\sqrt{(3-7)^2+(7-3)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt2$
$CA=\sqrt{(1-3)^2+(1-7)^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$ (3)

Coordinate area: $\text{Area}=\tfrac12|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)|$ $=\tfrac12|1(3-7)+7(7-1)+3(1-3)|=\tfrac12|-4+42-6|=\tfrac12(32)=16.$ (2)

Heron check: $s=\tfrac12(2\sqrt{10}+4\sqrt2+2\sqrt{10})=2\sqrt{10}+2\sqrt2$ . Rather than messy surds, note $AB=CA$ (isosceles). Base $BC=4\sqrt2$ , height from $A$ to line $BC$ . Line $BC$ : through $(7,3),(3,7)$ , slope $-1$ , equation $x+y-10=0$ ; distance from $A(1,1)=|1+1-10|/\sqrt2=8/\sqrt2=4\sqrt2$ . Area $=\tfrac12(4\sqrt2)(4\sqrt2)=\tfrac12(32)=16.$ ✓ (1)

(b) Substitute each vertex into $x^2+y^2+Dx+Ey+F=0$ :

$A$ : $1+1+D+E+F=0 \Rightarrow D+E+F=-2$
$B$ : $49+9+7D+3E+F=0 \Rightarrow 7D+3E+F=-58$
$C$ : $9+49+3D+7E+F=0 \Rightarrow 3D+7E+F=-58$ (2)

Subtract eq(B)−eq(C): $4D-4E=0\Rightarrow D=E$ . Subtract eq(B)−eq(A): $6D+2E=-56$ ; with $D=E$ : $8D=-56\Rightarrow D=-7=E$ . Then $F=-2-D-E=-2+14=12.$ (3)

So $x^2+y^2-7x-7y+12=0$ . Centre $\left(-\tfrac{D}{2},-\tfrac{E}{2}\right)=(3.5,3.5)$ . Radius $r=\sqrt{(D/2)^2+(E/2)^2-F}=\sqrt{12.25+12.25-12}=\sqrt{12.5}=\tfrac{5}{\sqrt2}=\tfrac{5\sqrt2}{2}.$ (2)

(c) Distances from centre $O(3.5,3.5)$ :

$OA=\sqrt{2.5^2+2.5^2}=\sqrt{12.5}$
$OB=\sqrt{3.5^2+0.5^2}=\sqrt{12.25+0.25}=\sqrt{12.5}$
$OC=\sqrt{0.5^2+3.5^2}=\sqrt{12.5}$ — all equal ✓ (2)

Perp-bisector of $AB$ : midpoint $M_{AB}=(4,2)$ ; slope $AB=\tfrac{2}{6}=\tfrac13$ , so bisector slope $=-3$ . Check $O$ satisfies it: slope $OM_{AB}=\tfrac{3.5-2}{3.5-4}=\tfrac{1.5}{-0.5}=-3$ ✓, and $-3\times\tfrac13=-1$ confirms perpendicularity. Hence $O$ lies on the perpendicular bisector of $AB$ . (2)

(d) $P(5,5)$ : $OP^2=(5-3.5)^2+(5-3.5)^2=1.5^2+1.5^2=4.5$ . Compare $r^2=12.5$ . Since $4.5<12.5$ , $P$ is inside the circle. (3)

Question 2

(a) $\cos^2 45^\circ=\tfrac12$ , so $2u^2\cos^2 45^\circ=2(400)(0.5)=400$ . Thus $y=x-\dfrac{10x^2}{400}=x-\dfrac{x^2}{40}$ . ✓ (2) Range: $y=0\Rightarrow x(1-\tfrac{x}{40})=0\Rightarrow x=40$ m. (2)

(b) $L: x-2y+4=0\Rightarrow y=\tfrac12 x+2$ . Slope $=\tfrac12$ . (2) $y$ -intercept: $x=0\Rightarrow y=2$ , point $(0,2)$ . $x$ -intercept: $y=0\Rightarrow x=-4$ , point $(-4,0)$ . (2)

(c) Set $y=x-\tfrac{x^2}{40}$ into $x-2y+4=0$ : $x-2\left(x-\tfrac{x^2}{40}\right)+4=0 \Rightarrow x-2x+\tfrac{x^2}{20}+4=0 \Rightarrow \tfrac{x^2}{20}-x+4=0.$ Multiply by 20: $x^2-20x+80=0\Rightarrow x=\dfrac{20\pm\sqrt{400-320}}{2}=10\pm\sqrt{20}=10\pm2\sqrt5.$ (3) $x_1=10-2\sqrt5\approx5.53$ , $x_2=10+2\sqrt5\approx14.47$ . Corresponding $y=\tfrac12 x+2$ : $y_1\approx4.76$ , $y_2\approx9.24$ . Impact points $(10\pm2\sqrt5,\ 7\pm\sqrt5)$ . (2)

(d) $\dfrac{dy}{dx}=1-\dfrac{x}{20}$ . At $x_2=10+2\sqrt5$ : slope $=1-\dfrac{10+2\sqrt5}{20}=\dfrac{10-2\sqrt5}{20}=\dfrac{5-\sqrt5}{10}\approx0.276.$ (2) Ramp slope $m_L=\tfrac12$ . Angle between: $\tan\theta=\left|\frac{m_L-m_t}{1+m_L m_t}\right|=\left|\frac{0.5-0.276}{1+0.5(0.276)}\right|=\frac{0.224}{1.138}\approx0.1968.$ $\theta\approx11.1^\circ.$ (3)

Question 3

(a)

def collinear(P, Q, R):
    area2 = P[0]*(Q[1]-R[1]) + Q[0]*(R[1]-P[1]) + R[0]*(P[1]-Q[1])
    return area2 == 0

Area of a triangle is zero exactly when the three points fail to enclose any region, i.e. they lie on one straight line; the signed-area expression (without the $\tfrac12$ ) vanishes iff collinear. (4)

(b) Line $ax+by+c=0$ has normal vector $\mathbf n=(a,b)$ . Let $F$ be the foot of perpendicular from $P_0=(x_0,y_0)$ ; then $\vec{P_0F}=t\,\mathbf n$ for some scalar $t$ , and $F=P_0+t(a,b)$ lies on the line: $a(x_0+ta)+b(y_0+tb)+c=0 \Rightarrow ax_0+by_0+c+t(a^2+b^2)=0 \Rightarrow t=-\frac{ax_0+by_0+c}{a^2+b^2}.$ Distance $=|\vec{P_0F}|=|t|\,|\mathbf n|=\dfrac{|ax_0+by_0+c|}{a^2+b^2}\cdot\sqrt{a^2+b^2}=\dfrac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}.$ ∎ (5)

(c) $(3,7)$ to $x-2y+4=0$ : $\dfrac{|3-14+4|}{\sqrt{1+4}}=\dfrac{7}{\sqrt5}=\dfrac{7\sqrt5}{5}\approx3.13.$ (2) collinear((1,1),(7,3),(3,7)): signed area $\times2 = 1(3-7)+7(7-1)+3(1-3)=-4+42-6=32\neq0$ → not collinear (consistent with Q1 area $16\neq0$ ). (1)

[
 {"claim":"Triangle ABC coordinate area is 16","code":"xa,ya,xb,yb,xc,yc=1,1,7,3,3,7\nresult=(Rational(1,2)*abs(xa*(yb-yc)+xb*(yc-ya)+xc*(ya-yb))==16)"},
 {"claim":"Circumcircle D=E=-7, F=12","code":"D,E,F=symbols('D E F')\nsol=solve([D+E+F+2, 7*D+3*E+F+58, 3*D+7*E+F+58],[D,E,F])\nresult=(sol[D]==-7 and sol[E]==-7 and sol[F]==12)"},
 {"claim":"Radius squared = 12.5 and P(5,5) inside","code":"cx,cy=Rational(7,2),Rational(7,2)\nr2=(cx-1)**2+(cy-1)**2\nOP2=(5-cx)**2+(5-cy)**2\nresult=(r2==Rational(25,2) and OP2<r2)"},
 {"claim":"Projectile-ramp impact x-values are 10 +/- 2*sqrt5","code":"x=symbols('x')\nsol=solve(Eq(x**2/20 - x + 4,0),x)\nresult=(set(sol)=={10-2*sqrt(5),10+2*sqrt(5)})"},
 {"claim":"Distance from (3,7) to x-2y+4=0 is 7/sqrt5","code":"result=(Abs(3-2*7+4)/sqrt(1**2+(-2)**2)==7/sqrt(5))"}
]