How is the distance between parallel lines found?

Write both lines with the same slope in general form, then divide the difference of constants by the square root of the slope squared plus one.

Why solve simultaneous equations to find a foot of perpendicular?

The foot lies on both the given line and the perpendicular through the point, so the intersection of the two equations is the foot.

Why is the radius squared in a circle equation?

The standard form has r^2 on the right, so radius 3 is written as 9; putting the radius itself on the right is the usual slip.

How is the distance formula linked to a circle equation?

A circle is the set of points at a fixed distance from the centre, so squaring that distance equation removes the root and yields the circle form.

S4 Maths coordinate geometry paper 3: areas, parallel distances and circles

2026-11-10

Tutor Prep Guide · S4 Maths Coordinate Geometry

Ten items combine triangle area, the distance between parallel lines, a foot of perpendicular, a circle equation and an intersection.

Direct answers　Items 01 to 03 cover area, a parallel line and the parallel distance, 04 to 06 a foot of perpendicular, a distance equation and a circle, 07 to 09 an intersection, a point-to-line distance and collinearity, and 10 a slip between adding and subtracting in the midpoint formula.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	25
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S4 Maths Coordinate Geometry
Level	S4

02　Questions (10 items)

FoundationFind the area of the triangle with vertices (0, 0), (6, 0) and (2, 4).
FoundationFind the line through (1, -3) parallel to 2x + y = 4.
FoundationFind the distance between the parallel lines y = 2x + 1 and y = 2x + 7.
IntermediateFind the foot of the perpendicular from (4, 5) to the line y = x.
IntermediateThe distance from (k, 3) to (2, -1) is 5. Find all possible values of k.
IntermediateFind the equation of the circle with centre (2, -1) and radius 3.
ChallengeThe line y = 2x + 1 meets the perpendicular bisector of (0, 0) and (4, 2). Find the point of intersection.
ChallengeFind the shortest distance from the origin to the line 3x + 4y = 10.
ChallengeA(1, 2), B(3, k) and C(7, 14) are collinear. Find k.
Error redoFind the midpoint of (-2, 5) and (4, -1). A student answered (-3, 3).

03　Answers and one-line notes

#	Answer	One-line note
1	12 square units	The base along the x-axis is 6 and the height is 4, so the area is 6 x 4 / 2 = 12.
2	2x + y + 1 = 0	Parallel means the same slope, so write 2x + y + c = 0; substituting (1, -3) gives 2 – 3 + c = 0, hence c = 1.
3	About 2.68	The constants differ by 6 and the square root of the slope squared plus 1 is √5, so the distance is 6 / √5, about 2.68.
4	(4.5, 4.5)	The line y = x has slope 1, so the perpendicular has slope -1 and passes (4, 5) as y = -x + 9; solving with y = x gives (4.5, 4.5).
5	k = 5 or k = -1	(k – 2)^2 + 16 = 25 gives (k – 2)^2 = 9, so k – 2 is 3 or -3 and k is 5 or -1.
6	(x – 2)^2 + (y + 1)^2 = 9	The standard form is (x – a)^2 + (y – b)^2 = r^2, so the centre and radius give (x – 2)^2 + (y + 1)^2 = 9.
7	(1, 3)	The midpoint is (2, 1) with segment slope 0.5, so the bisector is y = -2x + 5; solving with y = 2x + 1 gives (1, 3).
8	2	The shortest distance is 10 divided by the square root of 9 + 16, that is 10 / 5 = 2.
9	6	AC has slope (14 – 2) / (7 – 1) = 2, and collinearity forces AB to share it, so k = 2 + 2 x 2 = 6.
10	(1, 2); the answer (-3, 3) subtracts the coordinates instead of adding them.	The x-coordinate is (-2 + 4) / 2 = 1 and the y-coordinate is (5 – 1) / 2 = 2; subtracting then halving only gives half the displacement.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

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