Question

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)  $$gradient=\frac{1}{2}$$

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

Question

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

(i)C=(4,-2)

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

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

y=2x+10

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

$$AD=14.3$$ or $$\sqrt{205}$$

Question

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)

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

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

Question

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.

(i)32-4k=20⇒k=3

$$4b+3\times 2b=20$$

b=2

(ii)Mid-point=(5,0)

#### Question.

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.

(i)

(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$$\rightarrow k=\frac{5}{2}$$

Question

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.

10(i)

$$4x^{\frac{1}{2}}=x+3\rightarrow (x^{\frac{1}{2}})^{2}-4x^{\frac{1}{2}}+3=0$$ 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=$$(9-1)^{2}+(12-4)^{2}$$

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

10(ii)

$$\frac{dy}{dx}=2x^{\frac{-1}{2}}$$

$$2x^{\frac{-1}{2}}=1$$

(4,8)

10(iii)

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

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

$$(4^{\frac{1}{2}},7^{\frac{1}{2}})$$