O-LevelMathematicsVectors and transformations4024/21 · Oct/Nov 2018 · Q79 Marks
The position vector, OA, of point A is (-4, 7) and AB = (6, -3).
- (a)Find the position vector, OB, of point B.[1]
- (b)Find |AB|.[2]
- (c)Given that AB = 3CB, find the coordinates of point C.[2]
- (d)(i)Line L is parallel to AB and passes through the point (–2, 5). Find the equation of line L.[3]
- (d)(ii)Line M is perpendicular to line L and passes through the origin. Find the equation of line M.[1]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [1] (2, 4)
- [2] 6.71 or 6.708...
- M1 for 6^2 + (-3)^2 oe
- [2] (0, 5)
- FT their (a) ((their 2 – 2), (their 4 + 1))
- B1 for one value in coordinates correct or for [CB=](-2/3, 1) soi
- [3] y = -1/2 x + 4 oe final answer
- B2 for y = -1/2 x + c oe
- OR
- M1 for gradient = -3/6 soi
- M1 for (-2, 5) substituted into y = their mx + c
- [1] y = 2x oe
- FT their gradient from (d)(i)
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