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Sec 4 E Maths: Coordinate Geometry, practice questions & worked solutions

Coordinate geometry questions selected from 2023 to 2025 Singapore school E Maths prelim papers, each with the school’s full worked solution.

8 questionsStep-by-step solutionsFree to access

About this topic & key methods

Coordinate geometry in Sec 3 and Sec 4 E Maths covers gradient, the length of a line segment, the equation of a straight line, parallel lines and areas on the coordinate plane. Most questions here are Paper 1, with two Paper 2 questions that link several of these steps.

Attempt each question on paper before opening its worked solution. The solutions are the schools’ own marking schemes, so some steps are brief.

Key methods for coordinate geometry

Gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$
Length $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ ; an unknown coordinate leads to a quadratic, so check which root fits
Equation of a line: $y=mx+c$ , then substitute a known point to find $c$
Parallel lines have equal gradients; equal gradients with different intercepts never meet
Find where two lines meet by solving their equations simultaneously
Area of a triangle on axes: use a horizontal or vertical side as the base
Shortest distance from a point to a line $=\dfrac{2\times\text{area of triangle}}{\text{length of the line segment}}$

For lessons that build these methods week by week, see our E Maths tuition for Sec 3 and Sec 4.

Coordinate Geometry questions with worked solutions

Unless the question specifies otherwise, give non-exact answers to 3 significant figures and angles in degrees to 1 decimal place.

Q1Chung Cheng High School (Main)2024Prelim P1 Q10●○○ Foundational2 marks

Gradient with an unknown coordinate, Chung Cheng High School (Main), 2024 Prelim P1 Q10

The gradient of the line joining the points $(-3+2a,7)$ and $(a+1,2)$ is $-\dfrac23$ . Find the value of $a$ . [2]

Show worked solution▾

$\begin{aligned}\text{gradient}&=\frac{2-7}{a+1-(-3+2a)}\\&=\frac{-5}{-a+4}\end{aligned}$

Given gradient $=-\dfrac23$

$\begin{aligned}\frac{-5}{-a+4}&=-\frac23\\-15&=-2(-a+4)\\a&=-\frac72\end{aligned}$

Answer: $a=-\dfrac72$

Q2Fairfield Methodist School (Secondary)2023Prelim P1 Q5●○○ Foundational3 marks

Equation of a line from its intercepts, Fairfield Methodist School (Secondary), 2023 Prelim P1 Q5

$L$ is a line with a negative gradient and it has positive $x$ - and $y$ -intercepts. The value of $y$ -intercept is five times the value of $x$ -intercept. Given that the $x$ -intercept is $\dfrac25$ , find the equation of $L$ . [3]

Show worked solution▾

$y$ -intercept $=5\left(\dfrac25\right)=2$
Gradient of line graph

$\begin{aligned}&=-\frac{\textit{vert}}{\textit{horizontal}}\\&=-\frac2{\frac25}\\&=-5\end{aligned}$

Equation of line is $y=-5x+2$

Answer: $y=-5x+2$

Q3Presbyterian High School2025Prelim P1 Q16●○○ Foundational4 marks

Length of a segment and a parallel line, Presbyterian High School, 2025 Prelim P1 Q16

$A$ , $B$ and $C$ are points $(-4,3)$ , $(1,9)$ and $(0,2)$ respectively.

Find the length of $AB$ . [2]
Find the equation of the line that passes through $B$ and has the same gradient as $AC$ . [2]

Show worked solution▾

(a)

$\begin{aligned}AB&=\sqrt{(1-(-4))^2+(9-3)^2}\\AB&=7.81\end{aligned}$

(b)

$\begin{aligned}m_{AC}&=\frac{2-3}{0-(-4)}=-0.25\\y&=-0.25x+c\\9&=-0.25+c\\c&=9.25\\y&=-0.25x+9.25\end{aligned}$

Answer: (a) 7.81 units (b) $y=-0.25x+9.25$

Q4Anglo-Chinese School (Independent)2024Prelim P1 Q15●●○ Standard3 marks

Distance between two points with an unknown, Anglo-Chinese School (Independent), 2024 Prelim P1 Q15

The distance between the points $M\ (k,7)$ and $N\ (9,k)$ is $\sqrt{20}$ units.
Given that $k>10$ , find the value of $k$ . [3]

Show worked solution▾

$\begin{aligned}\sqrt{(k-9)^2+(7-k)^2}&=\sqrt{20}\\k^2-18k+81+49-14k+k^2&=20\\2k^2-32k+110&=0\\k^2-16k+55&=0\\(k-5)(k-11)&=0\end{aligned}$

$k=5\quad k=11$
(rej.)

Answer: $k=11$

Q5Catholic High School2025Prelim P1 Q13●●○ Standard4 marks

Area of a triangle on axes and an exact cosine, Catholic High School, 2025 Prelim P1 Q13

The vertices of $A$ and $B$ are $(-3,8)$ and $(7,2)$ .

The line $BC$ is parallel to the $x$ -axis.

The area of triangle $ABC$ is 12 square units.

Catholic High School 2025 P1 Q13: diagram for the question

Find the coordinates of $C$ . [2]
Find the exact value of $\cos\angle ACB$ . [2]

Show worked solution▾

Catholic High School 2025 P1 Q13: worked solution diagram

(a)

$\begin{aligned}\frac12(b)(6)&=12\\b&=4\\C&=(3,2)\end{aligned}$

(b)

$\begin{aligned}AC&=\sqrt{6^2+6^2}=6\sqrt2\\\cos\angle ACB&=-\frac6{6\sqrt2}=-\frac{\sqrt2}2\end{aligned}$

Answer: (a) $(3,2)$ (b) $-\dfrac{\sqrt2}2$

Q6CHIJ St. Nicholas Girls’ School2023Prelim P2 Q3●●○ Standard8 marks

Intersection of two lines and a length ratio, CHIJ St. Nicholas Girls’ School, 2023 Prelim P2 Q3

$J$ is a point $(-2,7)$ and $K$ is a point $(10,1)$ .

Find the length of $JK$ . [2]
Find the equation of $JK$ . [2]
The line with equation $6y-7x=16$ intersects the line $JK$ at point $L$ .
Find the coordinates of $L$ . [3]
State the ratio $JL:JK$ . [1]

Show worked solution▾

(a)

$\begin{aligned}\text{Distance}&=\sqrt{(-2-10)^2+(7-1)^2}\\&=13.41\\&\approx13.4\text{ units (3sf)}\end{aligned}$

(b)

$\begin{aligned}m_{JK}&=\dfrac{7-1}{-2-10}&1&=-\dfrac12(10)+c\\&=-\dfrac12&c&=6\\\therefore y&=-\dfrac12x+6\end{aligned}$

(c)

$\begin{aligned}y&=-\dfrac12x+6&&\quad\text{(1)}\\6y-7x&=16&&\quad\text{(2)}\end{aligned}$

sub (1) into (2)

$\begin{aligned}6\left(-\dfrac12x+6\right)-7x&=16\\-3x+36-7x&=16\\-10x&=-20\\x&=2\\y&=5\\\therefore L(2,5)&\end{aligned}$

(d)

$\begin{aligned}JL&=\sqrt{(-2-2)^2+(7-5)^2}=\sqrt{20}\\JK&=\sqrt{180}\\\therefore JL:JK&=\sqrt{20}:\sqrt{180}\end{aligned}$

or $JL_{run}:JK_{run}=4:12$ or $JL_{rise}:JK_{rise}=2:6$

$\therefore JL:JK=1:3$

Answer: (a) 13.4 (b) $y=-\dfrac12x+6$ (c) $L(2,5)$ (d) $1:3$

Q7Cedar Girls’ Secondary School2025Prelim P1 Q18●●● Challenging9 marks

Intersecting lines and the shortest distance to a line, Cedar Girls’ Secondary School, 2025 Prelim P1 Q18

In the figure below, line $L_1$ and line $L_2$ cut the $x$ -axis at $P$ and $R$ respectively.
The equation of $L_1$ is $y=2x-8$ .
Angle $PRQ$ is $45^\circ$ . $PR$ is 9 units.
Line $L_2$ meets line $L_1$ at $Q$ .

Cedar Girls’ Secondary School 2025 P1 Q18: diagram for the question

By finding the coordinates of $R$ , show that the equation of Line $L_2$ is $y=-x-5$ . [4]
Find the coordinates of $Q$ . [2]
Find the shortest distance from $R$ to line $PQ$ . [3]

Show worked solution▾

(a)

$P=(4,0)$
$R=(-5,0)$

Gradient $=-\tan45^\circ=-1$
Therefore, $y=-x+c$ .

At $(-5,0)$ ,

$0=5+c$

$c=-5$

Therefore, equation of $L_2:y=-x-5$ (Shown).

(b)

$\begin{aligned}y&=2x-8&&\quad\text{(1)}\\y&=-x-5&&\quad\text{(2)}\end{aligned}$

(1) $=$ (2)

$2x-8=-x-5$

$x=1\rightarrow y=-6$

$Q=(1,-6)$

(c)

$\text{Area of triangle}=\frac12\times6\times9$

$=27\text{ units}^2$

$PQ=\sqrt{(4-1)^2+(0+6)^2}$

$=\sqrt{45}\text{ units}$

$\text{Shortest distance}=\frac{27\times2}{\sqrt{45}}=8.05\text{ units (3s.f.)}$

Answer: $\text{(b) }Q=(1,-6)\quad\text{(c) }8.05\text{ units}$

Q8Swiss Cottage Secondary School2024Prelim P2 Q5●●● Challenging10 marks

Parallel lines, intersection and area of a triangle, Swiss Cottage Secondary School, 2024 Prelim P2 Q5

Line $l$ passes through the points $(-4,4)$ and $(8,-2)$ .

Find the equation of line $l$ . [2]
The equation of line $m$ is $6y=18-3x$ .
Show how you can tell that the line $m$ does not intersect the line $l$ . [2]
The equation of line $n$ is $2y=3x-4$ .
Line $n$ and line $l$ intersect at point $A$ . Find the coordinates of $A$ . [3]
Line $l$ intersects the $y$ -axis at point $B$ and line $n$ intersects the $y$ -axis at point $C$ .
Calculate the area of triangle $ABC$ . [3]

Show worked solution▾

(a)

$\begin{aligned}\text{grad of line }l&=\frac{-2-4}{8-(-4)}\\&=-\frac12\end{aligned}$

$y=-\frac12x+c$

subst $(-4,4)$ into $y=-\dfrac12x+c$

$\begin{aligned}4&=-\frac12(-4)+c\\c&=2\end{aligned}$

$y=-\frac12x+2$

(b)

Line $m$

$\begin{aligned}6y&=18-3x\\y&=-\frac12x+3\end{aligned}$

The gradient of line $l$ and line $m$ are equal (gradient $=-\dfrac12$ ) and the $y$ -intercept not equal.
Line $l$ and line $m$ are parallel.
Therefore, line $m$ does not intersect the line $l$ .

(c)

$y=-\dfrac12x+2$ (1) $2y=3x-4$ (2)

Subst (1) into (2)

$\begin{aligned}2\left(-\frac12x+2\right)&=3x-4\\-x+4&=3x-4\\4x&=8\\x&=2\end{aligned}$

Subst $x=2$ into (1)

$\begin{aligned}y&=-\frac12x+2\\y&=-\frac12(2)+2\\y&=1\end{aligned}$

$A(2,1)$

(d)

Area of $\Delta ABC$ $=\dfrac12\times4\times2$

$=4$ units $^2$

Swiss Cottage Secondary School 2024 P2 Q5: worked solution diagram

Answer: (a) $y=-\dfrac12x+2$ (b) line $m$ is $y=-\dfrac12x+3$ : the same gradient as $l$ but a different $y$ -intercept, so the lines are parallel and never meet (c) $A(2,1)$ (d) $4$ units $^2$

Need help with coordinate geometry? Our Sec 4 E Maths tuition classes work through prelim questions like these every week.

More practice related to coordinate geometry

Sec 3 A Maths: Coordinate Geometry

Sec 1 Maths: Functions and Linear Graphs

Sec 4 E Maths: Graphs of Functions

Sec 4 E Maths: Vectors

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