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 line 4x + ky = 20 passes through the points A(8, −4) and B(b, 2b), where k and b are constants.
(i) Find the values of k and b. 
(ii) Find the coordinates of the mid-point of AB.



\(4b+3\times 2b=20\)




The equation of a curve is y = 2 cos x.
(i) Sketch the graph of y = 2 cos x for −π ≤ x ≤ π, stating the coordinates of the point of intersection
with the y-axis.
Points P and Q lie on the curve and have x-coordinates of \(\frac{\pi}{3}\) and \(\pi\) respectively.

(ii) Find the length of PQ correct to 1 decimal place.

The line through P and Q meets the x-axis at H (h, 0) and the y-axis at K(0, k).

(iii) Show that \(h =\frac{5\pi}{9}\) and find the value of k.



(ii)\(P(\frac{\pi }{3},1)Q\left ( \pi ,-2 \right )\)

\(\rightarrow PQ^{2}=\left ( \frac{2\pi }{3} \right )^{2}+3^{2}\rightarrow PQ=3.7\)

(iii)Equation of PQ \( y-1=-\frac{9}{2\pi }\left ( x-\frac{\pi }{3} \right )\)

If y=0\(\rightarrow h=\frac{5\pi }{9}\)

If x=0\(\rightarrow k=\frac{5}{2}\)


The diagram shows the curve with equation \(y = 4x^{\frac{1}{2}}\)

(i) The straight line with equation \(y = x + 3\) intersects the curve at points A and B. Find the length
of AB.

(ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T.

(iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB.



\(4x^{\frac{1}{2}}=x+3\rightarrow (x^{\frac{1}{2}})^{2}-4x^{\frac{1}{2}}+3=0\) OR  \(16x=x^{2}+6x+9\)

\(x^{\frac{1}{2}}=1\)  or 3

\(x^{2}-10x+9(=0)\)x=1or 9

y=4 or 12


AB=\(\sqrt{128} or8\sqrt{2} oe or 11.3\)






Equation of normal is y \(-8=-1(x-4)\)

Eliminate \(y(or x)\rightarrow -x+12=x+3 or y-3=12-y\)