SPM 2020 Add Maths Paper 2, Question 11Coordinate Geometry

Solved on video in English and Bahasa Melayu · Form 4, Coordinate Geometry · 5:41 video

The question

(a)

Diagram 5 shows a triangle ACE. It is given that the equation of the straight line CE is y = -3x + 9, the point B(-2, 1) lies on the straight line AC, and A(-4, 0). Find (a) the equation of the straight line AC.

(b)

Find (b) the coordinates of C.

(c)

Find (c) the coordinates of E such that CD:CE = 3:5 (where D is the point at which line CE crosses the x-axis).

(d)

e. PC = AC).

SPM 2020 Add Maths Paper 2, Question 11: the question as drawn in Eduly's video5:41

Part (a)

Step 1

A(-4, 0) and B(-2, 1) both lie on AC.

3 more steps, each with the reason for it, in the full solution

Final answer

y = ½x + 2

Where students lose marks

Watch out: AC and CE are not perpendicular. Use points A and B for the gradient, and substitute the correct point for the intercept.

Part (b)

4 more steps, each with the reason for it, in the full solution

Final answer

C = (2, 3)

Where students lose marks

Watch out: use your correct AC from part (a), then move −3x across as +3x carefully when solving the simultaneous equations.

Part (c)

6 more steps, each with the reason for it, in the full solution

Final answer

E = (11/3, −2)

Where students lose marks

Watch out: CD:CE = 3:5 is part to whole. Convert to CD:DE = 3:2 first, and put weight 3 on E.

Part (d)

6 more steps, each with the reason for it, in the full solution

Final answer

x² + y² − 4x − 6y − 32 = 0

Where students lose marks

Watch out: the locus is a circle centred at C with radius AC (PC = AC), not the perpendicular bisector from PA = PC.

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