k(5−2x)21=−3
Equating their dxdy of the form k(5−2x)21 to -3 .
[B is ](2,6)
Gradient AB=m1=−2−232−6,gradient of perpendicular =−m11=264
M1*
For A, y must be 32 .
Clear use of difference in x co-ordinates difference in y co-ordinates for points
A and B, condone inconsistent order, and using
m1m2=−1.
If incorrect values or another complete method used, then working must be clear.
Mid point is (22−2,26+32)=(0,19)
M1*
Finding the midpoint of A B using A and B. If
incorrect values used then all working must be clear.
For A, y must be 32 .
y−19=132(x−0)
Finding the equation of the perpendicular bisector using their midpoint and their perpendicular
gradient.
2 x-13 y+247=0 or ± integer multiples of this.