The coordinates of A are (-3,2) and the coordinates of C are (5,).The mid point of AC is M and the perpendicular bisector of AC cuts the x-axis at B.

(i)Find the equation of MB and the coordinates ofB.

(ii)Show that AB is perpendicular to BC.

(iii)Given that ABCD is a square, find the coordinates of D and the length of AD.


(i) M=(1,4)  \(gradient=\frac{1}{2}\)

Gradient of MB=-2

Equation of MB:\(y-4=-2\left ( x-1 \right )\)

When y=0,x=3 or B=(3,0)

(ii)Gradient of \(AB=-\frac{2}{6}\);Gradient  of \(BC=\frac{6}{2}\)

\(m_{1}m_{2}=-1\left ( \Rightarrow AB  perpendicular AC\right )\)

(iii) D=(-1,8) \(AD=\sqrt{40}\) OR 6.32


C is the mid-point of the line joining A(14,-7) to B(-6,3).Tge line through C perpendicular to AB crosses the y-axis Aat D

(i)Find the equation of the line CD,giving your answer in the form y=mx+c

(ii)Find the distance AD.



\(m_{AB}=-\frac{1}{2}\rightarrow m_{CD}=2\)

Equation of CD is \(y+2=2\left ( x-4 \right )\) 


(ii)\(AD^{2}=\left ( 14-0 \right )^{2}+\left ( -7-\left ( -10 \right ) \right )^{2}\)

\(AD=14.3\) or \(\sqrt{205}\)


The points A(1,1) and B(5,9) lie on the curve\(6y=5x^{2}-18x+19\)

(i)Show that the equation of the perpendicular bisector of AB is \(2y=13-x\)

The perpendicular bisector of AB meets the curve C and D.

(ii)Find ,by calculation ,the distance CD,giving your answer in the form \(\sqrt{\frac{p}{q}}\),where p and q are integers.


(i)Mid-point of AB=(3,5)

Gradient of AB=2

Equation of perpendicular bisector is\( y-5=-\frac{1}{2}\left ( x-3 \right )\rightarrow 2y=13-x\)

(ii)\(-3x+39=5x^{2}-18x+19\rightarrow (5)\left ( x^{2} -3x-4\right )(=0)\)

X=4 or-1

\(y=4\tfrac{1}{2}\) or 7

\(CD^{2}=5^{2}+2\tfrac{1}{2}^{2}\rightarrow CD=\sqrt{\frac{125}{4}}\)


The point A has coordinates (−1, 6) and the point B has coordinates (7, 2).

(i) Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.
(ii) A point C on the perpendicular bisector has coordinates (p, q). The distance OC is 2 units, where O is the origin. Write down two equations involving p and q and hence find the coordinates of the possible positions of C.


(i) mid-point=(3,4)

Gradient AB=\(-\frac{1}{2}\rightarrow \) gradient of perpendicular =2



(ii)q=2p-2   \(  p^{2}+q^{2}=4\)

\(p^{2}+(2p-2)^{2}=4\rightarrow 5p^{2}-8p=0\)


\(\frac{1}{4}\left ( q+2 \right )^{2}+q^{2}=4\rightarrow 5q^{2}+4q-12=0\)

(0,-2) and \(\left ( \frac{8}{5},\frac{6}{5} \right )\)


The diagram shows a circle with centre A and radius r. Diameters CAD and BAE are perpendicular to each other. A larger circle has centre B and passes through C and D.
(i) Show that the radius of the larger circle is \(r\sqrt{2}\).
(ii) Find the area of the shaded region in terms of r.


(i)\(BC^{2}=r^{2}+r^{2}=2r^{2}\rightarrow BC=r\sqrt{2}\)

(ii) Area sector BCFD \(=\frac{1}{4}\pi (r\sqrt{2})^{2}\)

Area\(=\Delta BCAD=\frac{1}{2}(2r)r\)

Area segment\( CFDA=\frac{1}{2}\pi r^{2}-r^{2}\)

Area semi-circle CADE\(=\frac{1}{2}\pi r^{2}\)

Shaded area \(=\frac{1}{2}\pi r^{2}-\left ( \frac{1}{2}\pi r^{2}-r^{2} \right )\)

or \( \pi r^{2}-(\frac{1}{2}\pi r^{2}+\left ( \frac{1}{2}\pi r^{2}-r^{2} \right ))\)




Two points have coordinates A (5, 7) and B (9, -1).

(i) Find the equation of the perpendicular bisector of AB. 

The line through C(1, 2) parallel to AB meets the perpendicular bisector of AB at the point X.  

(ii) Find, by calculation, the distance BX.


Ans(i)Use of \(m_{1}m_{2} =-1\)
Mid-point of \(AB = (7,3)\) 

Grad. of \(AB=-2\)

grad of perp. bisector \(= 1/2\)

Eqn of perp. bisector is \(y-3=\frac{1}{2}(x-7)\)

(ii)equation of CX is \( y-2=-2(x-1)\)


\(x=9/5\) , \(y=2/5\)




A sector of a circle of radius r cm has an area of As cm2 Express the perimeter of the sector in terms of r and A.


Uses A \(\frac{1}{2}r^{2\Theta }\)

\(\Theta \)=\(\frac{2A}{r^{2}}\) p=r+r+r\(\Theta\)


The diagram shows a trapezium ABCD in which the coordinates of A, B and C are 4, 0, 0, 2 and h, 3h respectively. The lines BC and AD are parallel, angle ABC = 90Å and CD is parallel to the x-axis.
(i) Find, by calculation, the value of h.

(ii) Hence find the coordinates of D.


(i) Gradient of AB =−1⁄2→ Gradient of BC = 2

Forms equation in h \(\frac{3h-2}{h}\)=2 h = 2 
Alternative method for question  (i)

Vectors AB.BC=0  Solving h= 2

Alternative method for question (i)
Use of Pythagoras to find 3 lengths Solving h = 2

(ii) y coordinate of
D is 6, (3× ‘their’ h)

\(\frac{6-0}{x-4}=\rightarrow 7\rightarrow d(7,6)\) Vectors: AD.AB=0