The line y = 4x-3 meets the curve y = 3+5x-2x² at the points A and B.
- (a)Find the coordinates of A and B.[4]
- (b)The perpendicular bisector of the line AB cuts the coordinate axes at the points P and Q. Given that O is the origin, find the area of the triangle POQ.[5]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.
- [M1] (a) Eliminates one unknown: e.g. 4x-3=3+5x-2x²
- [A1] (a) Writes in solvable form: 2x² - x - 6 [ = 0] or – 2x² + x + 6 [ = 0]
- [M1 dep previous M1] (a) Solves or factorises their 3-term quadratic e.g. (2x + 3)(x – 2)
- [A1] (a) (-1.5, -9) (2, 5)
- [M3 FT their midpoint and perpendicular gradient from correct use of their (-1.5, -9) and their (2, 5)] (b) Equation of their perpendicular bisector soi: y+2=-1/4(x - 1/4) oe, isw or y = -1/4x - 31/16 oe, isw
- [M1 FT] (b) for midpoint: ( (their(-1.5)+their2)/2 , (their(-9)+their5)/2 ) or (0.25, -2)
- [M1 FT] (b) for perpendicular gradient: m_perp = -1 / ( (their(-9)-their5) / (their(-1.5)-their2) ) oe or 1/4
- [OR M1 FT their(−1.5, −9) and their (2, 5)] (b) for (x−2)² + (y − 5)² = (x+1.5)² + (y + 9)² soi
- [M1 FT their(−1.5, −9) and their (2, 5)] (b) for expanding x² + y² - 4x-10y + 29 = x² + y² +3x+18y +83.25 soi
- [A2 not from wrong coordinates] (b) 961/128 isw or 7.51 or 7.507 to 7.508 oe cao
- [A1] (b) for sight or use of e.g. (-31/4,0) and (0,-31/16) nfww
- [or x = 31/4 and y = -31/16 nfww or x = -31/4 and y = 31/16 nfww]
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