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.
Answer/Explanation
(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
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
(ii)Find the distance AD.
Answer/Explanation
(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.
Answer/Explanation
(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
\(CD^{2}=5^{2}+2\tfrac{1}{2}^{2}\rightarrow CD=\sqrt{\frac{125}{4}}\)
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.
Answer/Explanation
(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.
Answer/Explanation
(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.
Answer/Explanation
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 \(y(or x)\rightarrow -x+12=x+3 or y-3=12-y\)
\((4^{\frac{1}{2}},7^{\frac{1}{2}})\)