Can a distance between two points be negative?

No. Squares are never negative and the positive root is taken, so a distance is never below zero.

Which difference goes on top in the slope formula?

The change in y goes on top and the change in x at the bottom, taken in the same order for both points.

How do you get a perpendicular slope?

Take the negative reciprocal, so a slope of 2 pairs with -1/2; a horizontal line is perpendicular to a vertical one.

Why is the section formula 2A + 1B here?

With AP : PB = 1 : 2 the point P is closer to A, so A carries weight 2 and the weights add to 3.

S4 Maths coordinate geometry paper 4: error drills

2026-11-10

Tutor Prep Guide · S4 Maths Coordinate Geometry

Ten items target common slips in distance, midpoint, slope, the perpendicular condition, parallel distances, section points and collinearity.

Direct answers　Items 01 to 03 cover distance, midpoint and slope, 04 to 06 a line equation, a perpendicular slope and distance from the origin, 07 to 09 parallel distance, a section point and collinearity, and 10 the reversed slope ratio.

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 distance between A(-2, 1) and B(3, 13).
FoundationFind the midpoint of (-4, 7) and (6, -3).
FoundationFind the slope of the line through (5, -2) and (1, 6).
IntermediateA line of slope 3/4 passes through (2, -3). Find its equation.
IntermediateThe line L has equation 4x – 2y + 1 = 0. Find the slope of a line perpendicular to L.
IntermediateFind the distance from (3, -4) to the origin.
ChallengeFind the distance between the parallel lines 3x + 4y – 6 = 0 and 3x + 4y + 14 = 0.
ChallengeA(1, 2) and B(10, 8); P lies on AB with AP : PB = 1 : 2. Find P.
ChallengeIf (2, 3), (4, 7) and (k, 13) are collinear, find k.
Error redoError drill: a student answered 1/2 for the slope through A(1, 2) and B(5, 10). What is the correct value?

03　Answers and one-line notes

#	Answer	One-line note
1	13	The horizontal change is 3 – (-2) = 5 and the vertical change is 13 – 1 = 12, so the distance is the square root of 25 + 144, which is 13.
2	(1, 2)	The x-coordinate is (-4 + 6) / 2 = 1 and the y-coordinate is (7 – 3) / 2 = 2.
3	-2	The slope is (6 – (-2)) / (1 – 5) = 8 / (-4) = -2, keeping the same order in both differences.
4	y = (3/4)x – 9/2	Substituting into y = mx + c gives -3 = (3/4)(2) + c, so c = -3 – 3/2 = -9/2.
5	-1/2	Rearranging gives y = 2x + 1/2, so the slope is 2 and the perpendicular slope is the negative reciprocal -1/2.
6	5	The distance is the square root of 3 squared plus -4 squared, which is the square root of 25, so it is 5.
7	4	The constants differ by 20 in absolute value, and dividing by the square root of 9 + 16, which is 5, gives 4.
8	(4, 4)	The section point is (2A + 1B) / 3, giving x = (2 + 10) / 3 = 4 and y = (4 + 8) / 3 = 4.
9	7	The slope between the first two points is 2, so 13 – 3 = 2(k – 2), which gives k = 7.
10	2	The slope is the y-change 8 divided by the x-change 4, which is 2; the answer 1/2 comes from reversing the two differences.

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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