Sec 3 A Math: Coordinate Geometry, practice questions & worked solutions

Coordinate Geometry practice questions selected from 2025 Singapore school A Math prelim papers, each with a full worked solution.

About this topic & key methods

These questions practise Sec 3 topics; the source papers are Sec 4 prelims. Questions requiring later Sec 4 methods are excluded.

Attempt each question on paper before opening its worked solution. Keep the source question number when checking against the original paper.

Key methods

Find gradients, lengths, midpoints and equations of straight lines.
Use perpendicular and parallel line conditions.
Find intersections by solving simultaneous equations.
Calculate the area of a triangle or a rectilinear figure.

For guided practice, see our Sec 3 A Maths tuition programme.

Questions & worked solutions

Unless the question specifies otherwise, give numerical answers to 3 significant figures and angles in degrees to 1 decimal place. Angles in radians are stated explicitly.

Q1Jurongville Secondary School2025Prelim P1 Q4●●○ Standard5 marks

Coordinate Geometry

The line $y - x = 5$ intersects the curve $x^2 + xy + y^2 = 91$ at two points. Find the coordinates of these two points.

Show worked solution▾

$\begin{aligned} y &= x + 5\\ x^2 + x(x + 5) + (x + 5)^2 &= 91\\ x^2 + x^2 + 5x + x^2 + 10x + 25 - 91 &= 0\\ 3x^2 + 15x - 66 &= 0\\ x^2 + 5x - 22 &= 0 \end{aligned}$ Using Quadratic Formula, $\begin{aligned} x &= 2.82 \ \text{or}\ -7.82\\ (2.82,\ 7.82) \ &\text{or}\ (-7.82,\ -2.82) \end{aligned}$

Q2Singapore Chinese Girls’ School2025Prelim P1 Q2●●○ Standard5 marks

Coordinate Geometry

The straight line $2x + y = 1$ intersects the curve $x^2 - xy - y^2 + 5 = 0$ at the points $A$ and $B$ . Show that the length $AB$ is $5\sqrt{5}$ units.

Show worked solution▾

$\begin{aligned} y &= 1 - 2x \text{-----(1)}\\ x^2 - xy - y^2 + 5 &= 0 \text{-----(2)} \end{aligned}$ Sub (1) into (2) $\begin{aligned} x^2 - x(1 - 2x) - (1 - 2x)^2 + 5 &= 0\\ x^2 - x + 2x^2 - 1 + 4x - 4x^2 + 5 &= 0\\ -x^2 + 3x + 4 &= 0\\ x^2 - 3x - 4 &= 0\\ (x - 4)(x + 1) &= 0\\ x = 4 \ &\text{or} \ -1\\ y = -7 \ &\text{or} \ 3 \end{aligned}$ $\begin{aligned} \text{Length} &= \sqrt{(4 + 1)^2 + (-7 - 3)^2}\\ &= \sqrt{125}\\ &= 5\sqrt{5}\text{ units} \end{aligned}$

Q3Dunman Secondary School2025Prelim P1 Q1●●○ Standard6 marks

Coordinate Geometry / Surds

The line $y = x - 4$ and the curve $y = \dfrac{2}{x-3}$ intersect at the points $A$ and $B$ .

Find the length of $AB$ .

Show worked solution▾

$\begin{aligned} x-4 &= \frac{2}{x-3}\\ (x-4)(x-3) &= 2\\ x^2-7x+12 &= 2\\ x^2-7x+10 &= 0\\ (x-2)(x-5) &= 0\\ x=2 \;&\text{or}\; 5 \end{aligned}$ When $x=2$ , $y=-2$ . When $x=5$ , $y=1$ . $\begin{aligned} \text{Length} &= \sqrt{(2-5)^2+(-2-1)^2}\\ &= 4.24\ \text{unit} \end{aligned}$

Q4Methodist Girls’ School2025Prelim P1 Q3●●○ Standard6 marks

Coordinate Geometry

Solutions to this question by accurate drawing will not be accepted.

Methodist Girls’ School 2025 P1 Q3: question diagram

The diagram shows a quadrilateral $OABC$ where $O$ is the origin. The point $A$ is $(-1, 5)$ and the point $B$ is $(3, 7)$ . The perpendicular bisector of $AB$ meets the $x$ -axis at $C$ .

Find the coordinates of $C$ .
Find the area of the quadrilateral $OABC$ .

Show worked solution▾

(a)

$\begin{aligned} \text{Gradient of } AB &= \frac{7 - 5}{3 + 1} = \frac{1}{2} \end{aligned}$ Gradient of perpendicular to $AB = -2$ $\begin{aligned} \text{Midpoint of } AB &= \left(\frac{-1 + 3}{2},\ \frac{5 + 7}{2}\right) = (1, 6) \end{aligned}$ Equation of perpendicular bisector of $AB$ : $\begin{aligned} y - 6 &= -2(x - 1)\\ y &= -2x + 8 \end{aligned}$ At $x$ -axis, $y = 0$ $\begin{aligned} 0 &= -2x + 8\\ x &= 4 \end{aligned}$ $\text{Coordinates of } C = (4, 0)$

(b)

$\begin{aligned} \text{Area} &= \frac{1}{2}\left| \begin{matrix} 0 & 4 & 3 & -1 & 0\\ 0 & 0 & 7 & 5 & 0 \end{matrix} \right|\\ &= \frac{1}{2}\left[(0 + 28 + 15 + 0) - (0 - 7 + 0 + 0)\right]\\ &= 25 \text{ units}^2 \end{aligned}$

Q5School of Science and Technology2025Prelim P1 Q5●●○ Standard6 marks

Coordinate Geometry

The line $4y = 5x - 6$ cuts the curve $x + 4xy - 3y^2 = 7$ at the points $P$ and $Q$ . Find the length of the line $PQ$ in the form $a\sqrt{b}$ , where $a$ and $b$ are integers.

Show worked solution▾

$\begin{aligned} 5x &= 4y + 6\\ x &= \frac{4y}{5} + \frac{6}{5} \text{----- (1)}\\ x + 4xy - 3y^2 &= 7 \text{----- (2)} \end{aligned}$ Substitute (1) into (2): $\begin{aligned} \left(\frac{4y}{5} + \frac{6}{5}\right)(1 + 4y) - 3y^2 &= 7\\ (4y + 6)(1 + 4y) - 15y^2 &= 35\\ 16y^2 + 28y + 6 - 15y^2 - 35 &= 0\\ y^2 + 28y - 29 &= 0\\ (y - 1)(y + 29) &= 0\\ y = 1 \ &\text{or} \ -29\\ x = 2 \ &\text{or} \ -22 \end{aligned}$ Point $P$ is $(2, 1)$ and Point $Q$ is $(-22, -29)$ .

Length of line $PQ$ $\begin{aligned} &= \sqrt{\left(2 - (-22)\right)^2 + (1 + 29)^2}\\ &= \sqrt{1476}\\ &= 6\sqrt{41}\ \text{units} \end{aligned}$

Q6Bedok View Secondary School2025Prelim P1 Q10●●○ Standard7 marks

Coordinate Geometry

An isosceles triangle $PQR$ has vertices $P(1, 0)$ , $Q(0, 2)$ and $R(2, 3)$ . $PQ = QR$ .

Find the area of the triangle $PQR$ .
If $S$ is a point such that $PQRS$ is a kite and it also lies on the line $x - y = 5$ , find the coordinates of $S$ .

Show worked solution▾

(a)

$\begin{aligned} \text{Area of } \triangle PQR &= \frac{1}{2}\left| \begin{matrix} 1 & 2 & 0 & 1\\ 0 & 3 & 2 & 0 \end{matrix} \right|\\ &= \frac{1}{2}[3 + 4 - 2]\\ &= 2.5\ \text{units}^2 \end{aligned}$

(b)

$\begin{aligned} m_{PR} &= \frac{3 - 0}{2 - 1}\\ &= 3\\ m_{QS} &= -\frac{1}{3} \end{aligned}$ Equation of $QS$ : $\begin{aligned} y - 2 &= -\frac{1}{3}(x - 0)\\ y &= -\frac{1}{3}x + 2 \text{---(1)}\\ x - y &= 5 \text{---(2)} \end{aligned}$ Sub (1) into (2): $\begin{aligned} x + \frac{1}{3}x - 2 &= 5\\ x &= \frac{21}{4}\\ y &= \frac{1}{4} \end{aligned}$ $\text{coordinates of } S = \left(\frac{21}{4},\ \frac{1}{4}\right)$

Q7Tanjong Katong Secondary School2025Prelim P1 Q8●●○ Standard7 marks

Coordinate Geometry

The equations of the lines $AB$ , $BC$ and $CA$ are $x - y = 0$ , $x + 3y = 20$ and $5x - y = 4$ respectively. The coordinates of $C$ are $(2, 6)$ .

Find the coordinates of $A$ and of $B$ .
Given that $ABCD$ is a parallelogram, find the coordinates of $D$ .
Find the area of the parallelogram $ABCD$ .

Show worked solution▾

(a)

$\begin{aligned} x - y &= 0, x + 3y = 20\\ x + 3x &= 20\\ x &= 5,\\ y &= 5\\ \text{Coord } B &= (5, 5) \end{aligned}$ $\begin{aligned} 5x - y &= 4, x - y = 0,\\ 5x - x &= 4\\ x &= 1\\ y &= 1\\ \text{Coord } A &= (1, 1) \end{aligned}$

(b)

$\begin{aligned} \text{Coord } D &= \left(1 - (5 - 2),\ 1 + 1\right)\\ &= (-2, 2) \end{aligned}$

(c)

$\begin{aligned} \text{Area} &= 2 \times \frac{1}{2}\left| \begin{matrix} 1 & 5 & 2 & 1\\ 1 & 5 & 6 & 1 \end{matrix} \right|\\ &= 5 + 30 + 2 - 5 - 10 - 6\\ &= 16\ \text{units}^2 \end{aligned}$

Q8Ahmad Ibrahim Secondary School2025Prelim P1 Q7●●● Challenging8 marks

Coordinate Geometry

Ahmad Ibrahim Secondary School 2025 P1 Q7: question diagram

The diagram shows an isosceles triangle $ABC$ in which $A(-3, 0)$ , $B\left(\dfrac{1}{2}, 3\right)$ and $C(5, 2)$ . $M$ is the foot of perpendicular from $B$ to $AC$ .

Find the coordinates of $M$ .
Find the equation of the perpendicular bisector of $AC$ .
Given that $ABCD$ is a kite with $BM = \dfrac{2}{7}BD$ , find the coordinates of $D$ .
Find the area of the kite $ABCD$ .

Show worked solution▾

(a)

$M = \left(\frac{-3 + 5}{2},\ \frac{0 + 2}{2}\right) = (1, 1)$

(b)

$\begin{aligned} \text{Gradient of } AC &= \frac{2 - 0}{5 - (-3)} = \frac{1}{4} \end{aligned}$ Gradient of perpendicular $BM = -4$ $\text{or } m_{BM} = \frac{3 - 1}{\frac{1}{2} - 1} = -4$ Equation of perpendicular bisector: $\begin{aligned} y - 1 &= -4(x - 1)\\ y &= -4x + 5 \end{aligned}$

(c)

Correct ratio: $2 : 5$ $\frac{1}{2} : 1\frac{1}{4}$ Coordinates of $D$ $\begin{aligned} &= \left(1 + \frac{1}{4},\ 1 - 5\right)\\ &= \left(2\frac{1}{4}, -4\right) \end{aligned}$

Alternative: using vectors

$\begin{aligned} \overrightarrow{BM} &= \frac{2}{7}\overrightarrow{BD}\\ \begin{pmatrix} 0.5\\ -2 \end{pmatrix} &= \frac{2}{7}\left(\overrightarrow{OD} - \overrightarrow{OB}\right)\\ \overrightarrow{OD} &= 3.5\begin{pmatrix} 0.5\\ -2 \end{pmatrix} + \begin{pmatrix} 0.5\\ 3 \end{pmatrix} = \begin{pmatrix} 2.25\\ -4 \end{pmatrix}\\ D &= \left(2\frac{1}{4}, -4\right) \end{aligned}$

(d)

$\begin{aligned} \text{Area of } ABCD &= \frac{1}{2}\left| \begin{matrix} -3 & 2\frac{1}{4} & 5 & \frac{1}{2} & -3\\ 0 & -4 & 2 & 3 & 0 \end{matrix} \right|\\ &= \frac{1}{2}(31.5 + 28) (\text{anti-clockwise})\\ &= 29.75\ \text{units}^2 \left(\text{or } \frac{119}{4}\right) \end{aligned}$

Q9Maris Stella High School2025Prelim P1 Q11●●● Challenging8 marks

Coordinate Geometry

Solutions to this equation by accurate drawing will not be accepted.

$ABC$ is a triangle with vertices $A(0,\, h)$ , $B(2h,\, -2)$ and $C(7,\, 5)$ , where $h$ is a positive integer. The area of the triangle is $20$ units $^2$ .

Show that $h = 2$ .
Find the equation of the perpendicular bisector of $AB$ .
Explain why triangle $ABC$ is isosceles.

Show worked solution▾

(a)

$\begin{aligned} \text{Area of triangle } \frac{1}{2}\left|\begin{array}{cccc} 0 & 2h & 7 & 0\\ h & -2 & 5 & h \end{array}\right|\\ 10h+7h-2h^2+14 &= 40\\ 2h^2-17h+26 &= 0\\ (h-2)(2h-13) &= 0\\ h=2,\ h&=\frac{13}{2}\ (\text{rejected}) \end{aligned}$

(b)

$\begin{aligned} \text{Midpoint of } AB &= \left(\frac{0+4}{2},\,\frac{2-2}{2}\right)=(2,0)\\ \text{gradient of } AB &= \frac{-2-2}{4-0}=-1 \end{aligned}$ Equation of perpendicular bisector of $AB$ : $y=x-2$ .

(c)

$\begin{aligned} AC &= \sqrt{(0-7)^2+(2-5)^2}=\sqrt{58}\\ BC &= \sqrt{(4-7)^2+(-2-5)^2}=\sqrt{58} \end{aligned}$ since $AC=BC$ , triangle $ABC$ is isosceles.

Or from part (b), when $x=7$ , $y=5$ ; vertex $C$ lies on the perpendicular bisector of $AB$ , $AC=BC$ ; triangle $ABC$ is isosceles.

Q10St Joseph’s Institution2025Prelim P1 Q10●●● Challenging8 marks

Coordinate Geometry

St Joseph’s Institution 2025 P1 Q10: question diagram

The diagram shows a trapezium with vertices $A(-2, 4)$ , $B(k, 8)$ , $C(6, 4)$ and $D$ . The sides $AD$ and $BC$ are parallel and angle $BAD = 90^\circ$ .

Show that $k = 2$ .
$BDEF$ is a quadrilateral. Point $A$ is the midpoint of both $BE$ and $DF$ . Given that the area of $BDEF$ is 96 square units, find the coordinates of $D$ .

Show worked solution▾

(a)

$\begin{aligned} \frac{8 - 4}{k + 2} \times \frac{8 - 4}{k - 6} &= -1\\ \frac{4}{k + 2} \times \frac{4}{k - 6} &= -1\\ (k + 2)(k - 6) &= -16\\ k^2 - 4k - 12 + 16 &= 0\\ k^2 - 4k + 4 &= 0\\ (k - 2)^2 &= 0\\ k &= 2 \end{aligned}$

(b)

Let the coordinates of $E$ be $(E_x, E_y)$ . $\left(\frac{2 + E_x}{2},\ \frac{8 + E_y}{2}\right) = (-2, 4)$ $(E_x, E_y) = (-6, 0)$ $2 \times \frac{1}{2}\left| \begin{matrix} 2 & -6 & 2a & 2\\ 8 & 0 & -a & 8 \end{matrix} \right| = 96$ $\begin{aligned} \left[6a + 16a - (-48 - 2a)\right] &= 96\\ 24a &= 48\\ a &= 2 \end{aligned}$ $D\,(4, -2)$

$\begin{aligned} \frac{-a - 4}{2a + 2} &= \frac{8 - 4}{2 - 6}\\ \frac{-a - 4}{2a + 2} &= -1\\ a + 4 &= 2a + 2\\ a &= 2 \end{aligned}$ $D(4, -2)$

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Frequently asked questions

Do these A Math questions include worked solutions?▾

Yes. Every question includes the worked solution for all the parts shown.

Which year are the questions from?▾

All the selected questions are from 2025 Singapore school Preliminary Examination papers.

How should I use this topic page?▾

Attempt a question fully on paper, then open the worked solution and compare your method and answer.

Sec 3 A Math: Coordinate Geometry, practice questions & worked solutions

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Questions & worked solutions

Frequently asked questions

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