The coordinates of points A, B, C and D are as follows. A(-4,3) B(6, -9) C(15,10) D(14, -1) The line L has equation y = 11x-75. The perpendicular bisector of the line AB meets L at the point E. Find the area of triangle CDE.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [M1] m_AB = (-9-3)/(6--4) oe isw or -6/5
2. [M1] FT their m_AB providing that difference in y attempted difference in x [m_perp =] 5/6 or -1/(their m_AB)
3. [M1] [Midpoint] ((-4+6)/2, (3-9)/2) oe isw or (1,-3)
4. [M1] FT their perpendicular gradient providing gradient is -1/(their m_AB) and using their midpoint Finds equation of perpendicular bisector: y - their (-3) = -1/(their -6/5) (x - their 1) oe
5. [M1] Eliminates one variable dep on previous M mark 11x – 75 = their perp bisector equation
6. [A1] E(7, 2)
7. [A1] Dependent on all previous marks [Area CDE =] 40