O-LevelAdditional MathematicsStraight line graphs4037/22 · Oct/Nov 2019 · Q911 Marks
The diagram shows the points A (−3, 5) and B (5, −1). The mid-point of AB is M and the line PM is perpendicular to AB. The point P has coordinates (r, s). [Figure is on page 12]
- (i)Find the equation of the line PM in the form y = mx+c, where m and c are exact constants.[5]
- (ii)Hence find an expression for s in terms of r.[1]
- (iii)Given that the length of PM is 10 units, find the value of r and of s.[5]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.
- B1 Midpoint (1, 2) May be seen on diagram
- B1 Gradient of AB= -3/4
- M1 Gradient of PM -1 / their gradient of AB = 4/3 Use m₁ × m₂ = -1
- M1 dep Attempt to find equation of line with their midpoint and their gradient of PM. If y = mx + c used c must be found.
- A1 y = 4/3 x + 2/3
- B1 FT s = 4/3 r + 2/3
- B1 FT (r− 1)² + (s – 2)² = 100 oe Insert (r, s) into their linear equation to obtain s =...
- M1 Eliminate r or s From one linear and one quadratic expression. Unsimplified
- A1 25r²-50r-875 = 0 oe or 25s²-100s-1500 = 0 oe
- M1 (5r+25)(5r-35) = 0 oe or (5s-50)(5s + 30) = 0 oe Solve three term quadratic Can be implied by correct solution.
- A1 r = 7, s = 10 Do not award if negative values of r and s are also given nfww
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