**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) Equation of \(MB\) and Coordinates of \(B\):

Given that \(A(-3,2)\) and \(C(5,y_C)\):

The midpoint \(M\) is found as \(M\left(\frac{-3+5}{2}, \frac{2+y_C}{2}\right) = M(1, \frac{y_C + 2}{2})\).

The slope of \(MB\) is the negative reciprocal of the slope of \(AC\), which is \(\frac{y_C – 2}{5 – (-3)} = \frac{y_C – 2}{8}\).

Using the point-slope form of the equation, the equation of \(MB\) is \(y – 4 = -\frac{y_C – 2}{8}(x – 1)\).

For \(MB\) to intersect the x-axis, \(y\) must be \(0\):

\(0 – 4 = -\frac{y_C – 2}{8}(x – 1)\)

Solving for \(x\), we get \(x = 3\). Therefore, the coordinates of \(B\) are \((3, 0)\).

(ii) Proving \(AB\) is Perpendicular to \(BC\):

The slope of \(AB\) is found using the coordinates of \(A\) and \(B\):

\(m_{AB} = \frac{0 – 2}{3 – (-3)} = -\frac{1}{2}\)

The slope of \(BC\) is \(\frac{y_C – 2}{5 – (-3)} = \frac{y_C – 2}{8}\).

To check if \(AB\) is perpendicular to \(BC\), we check if \(m_{AB} \cdot m_{BC} = -1\):

\(-\frac{1}{2} \cdot \frac{y_C – 2}{8} = -1\)

\(y_C – 2 = 4\)

Solving for \(y_C\), we get \(y_C = 6\), confirming that \(AB\) is perpendicular to \(BC\).

(iii) Finding \(D\) and Length of \(AD\):

Since \(D\) lies on the perpendicular bisector (\(x = 1\)), the coordinates of \(D\) are \((1, 2)\). The length of \(AD\) is then calculated using the distance formula:

\(AD = \sqrt{(1 – (-3))^2 + (2 – 2)^2} = \sqrt{40} \approx 6.32\)

**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) Equation of Line CD:

Given that \(C\) is the midpoint of \(AB\), the coordinates of \(C\) are \(\left(\frac{14 + (-6)}{2}, \frac{(-7) + 3}{2}\right) = (4, -2)\).

The slope of \(AB\) is \(\frac{3 – (-7)}{(-6) – 14} = \frac{10}{-20} = -\frac{1}{2}\). Therefore, the slope of the line perpendicular to \(AB\) (and hence the slope of \(CD\)) is \(2\).

Using the point-slope form of the equation, the equation of \(CD\) is \(y – (-2) = 2(x – 4)\).

Simplifying, we get \(y + 2 = 2x – 8\), and finally, \(y = 2x – 10\).

(ii) Distance AD:

The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the distance formula:

\(d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)

In this case, \(A(14, -7)\) and \(D(x_D, 0)\), where \(x_D\) is the x-coordinate of \(D\).

\(AD = \sqrt{(14-x_D)^2 + (-7-(-10))^2}\)

Therefore, the distance \(AD\) is \(\sqrt{205}\).

**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)The midpoint of \(AB\) is given by the average of the x-coordinates and y-coordinates of \(A\) and \(B\):

\(\left(\frac{1+5}{2}, \frac{1+9}{2}\right) = (3, 5)\)

The slope of \(AB\) is given by:

\(m_{AB} = \frac{9 – 1}{5 – 1} = \frac{8}{4} = 2\)

The negative reciprocal of the slope of \(AB\) (since the bisector is perpendicular) is \(m_{\text{bisector}} = -\frac{1}{2}\).

Using the point-slope form, the equation of the perpendicular bisector is given by:

\(y – 5 = -\frac{1}{2}(x – 3)\)

Simplifying, we get \(2y – 10 = -x + 3\), and finally, \(2y = 13 – x\).

(ii)\((x^2 – 3x – 4) = 0\). Solving for \(x\), you found \(x = 4\) and \(x = -1\).

Now, substitute these x-values back into the equation of the curve \(6y = 5x^2 – 18x + 19\) to find the corresponding y-values:

For \(x = 4\): \(6y = 5(4)^2 – 18(4) + 19\), solving this gives \(y = 4 \frac{1}{2}\).

For \(x = -1\): \(6y = 5(-1)^2 – 18(-1) + 19\), solving this gives \(y = 7\).

So, \(C = (4, 5)\) and \(D = (-1, 7)\). Now, use the distance formula to find \(CD\):

\(CD = \sqrt{(-1 – 4)^2 + (2\frac{1}{2})^2} = \sqrt{\frac{125}{4}}\)

Therefore, the distance \(CD\) is \(\sqrt{\frac{125}{4}}\).

**Question**

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.

**▶️Answer/Explanation**

(i)Given points \(A(-1, 6)\) and \(B(7, 2)\), let’s find the equation of the perpendicular bisector.

The midpoint \(M\) of \(AB\):

\(M = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right)\)

\(M = \left(\frac{-1 + 7}{2}, \frac{6 + 2}{2}\right) = (3, 4)\)

Slope of \(AB\):

\(m_{AB} = \frac{y_B – y_A}{x_B – x_A} = \frac{2 – 6}{7 – (-1)} = -\frac{4}{8} = -\frac{1}{2}\)

The negative reciprocal of the slope of \(AB\) is the slope of the perpendicular bisector:

\(m_{\text{perpendicular bisector}} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{2}} = 2\)

Use the point-slope form of a line with the midpoint \(M\):

\(y – y_M = m_{\text{perpendicular bisector}}(x – x_M)\)

\(y – 4 = 2(x – 3)\)

\(y = 2x – 2\)

So, the equation of the perpendicular bisector is \(y = 2x – 2\).

(ii) The point \(C\) on the perpendicular bisector has coordinates \((p, q)\), and the distance \(OC\) is 2 units.

Equation for the distance \(OC\):

\(OC^2 = p^2 + q^2\)

Equation of the perpendicular bisector:

\(y = 2x – 2\)

\(q = 2p – 2\)

Now, substitute the values into the distance equation:

\(2^2 = p^2 + (2p – 2)^2\)

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

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

\(p(5p-8)=0\)

\(p=0 \) or \(p=\frac{8}{5}\)

OR

\(\frac{1}{4}(q+2)^2+q^2=4 \)

\(5 q^2+4 q-12=0\)

\(5q^{2}+10q-6q-12=0\)

\( (q+2)(5q-6)=0\)

\(q=-2\) or\( q=\frac{6}{5}\)

Coordinates of the possible positions of C are (0,-2) and \(\left ( \frac{8}{5},\frac{6}{5} \right )\)

**Question**

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.

**▶️Answer/Explanation**

(I)Given that diameters CAD and BAE are perpendicular to each other, we can see that the diameter of the larger circle is the hypotenuse of right-angled triangle CAB. By the Pythagorean theorem:

\[ AC^2 + AB^2 = CB^2 \]

Since AC and AD are both radii of the smaller circle with length \(r\), we can write:

\( r^2 + r^2 = CB^2 \)

\[ 2r^2 = CB^2 \)

\[ CB = r \sqrt{2}\)

Now, the radius of the larger circle, which is also the hypotenuse of CAB, is \(r \sqrt{2}\).

(ii) The shaded region is the area between the larger and smaller circles. This is essentially a ring or an annulus. The area of an annulus is given by:

\( \text{Area} = \pi (R^2 – r^2) \)

where \(R\) is the radius of the larger circle and \(r\) is the radius of the smaller circle.

\( \text{Area} = \pi \left((r \sqrt{2})^2 – r^2\right) \)

\( \text{Area} = \pi \left(2r^2 – r^2\right) \)

\( \text{Area} = \pi r^2 \)

**Question**

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.

**▶️Answer/Explanation**

(i) The midpoint of line segment AB is the point M with coordinates \(\left(\frac{5+9}{2}, \frac{7+(-1)}{2}\right) = (7, 3)\).

The slope of the line AB is \(\frac{-1-7}{9-5} = -2\). The negative reciprocal of the slope of AB is \(\frac{1}{2}\), which is the slope of the perpendicular bisector.

Using the point-slope form of a line \((y-y_1) = m(x-x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope, we can write the equation of the perpendicular bisector passing through M:

\( (y-3) = \frac{1}{2}(x-7) \)

\( 2y – 6 = x – 7 \)

\( x – 2y = -1 \)

So, the equation of the perpendicular bisector of AB is \(x – 2y = -1\).

(ii) Finding the point of intersection X:

Now, we have a line through C(1,2) with a slope of -2 (parallel to AB) and the equation of the perpendicular bisector. Solving these two equations simultaneously gives the coordinates of the point X:

\( x = \frac{9}{5} \)

\( y = \frac{2}{5} \)

So, the point of intersection is \(X \left(\frac{9}{5}, \frac{2}{5}\right)\).

Now, we can use the distance formula to find the distance BX between B(9,-1) and X:

\( BX = \sqrt{(9 – 9/5)^2 + (-1 – 2/5)^2} \)

Simplifying this expression gives \(BX \approx 7.33\).

**Question**

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.

**▶️Answer/Explanation**

The area of a sector of a circle with radius \(r\) and central angle \(\theta\) is given by:

\( \text{Area} = \frac{\theta}{360^\circ} \pi r^2 \)

The area is \(A\), so

\( A = \frac{\theta}{360^\circ} \pi r^2 \)

\( \theta = \frac{360^\circ A}{\pi r^2} \)

The perimeter (\(P\)) of the sector consists of the arc length (\(L\)) and the two radii. The arc length is a fraction of the circumference (\(C\)) of the whole circle, which is \(2\pi r\):

\( L = \frac{\theta}{360^\circ} \cdot 2\pi r \)

\( L = \frac{\frac{360^\circ A}{\pi r^2}}{360^\circ} \cdot 2\pi r \)

\( L = \frac{2A}{r} \)

The perimeter \(P\) is the sum of the arc length and the two radii:

\( P = L + 2r \)

\( P = \frac{2A}{r} + 2r \)

So, the expression for the perimeter of the sector in terms of \(r\) and \(A\) is:

\( P = \frac{2A}{r} + 2r \)

**Question**

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.

**▶️Answer/Explanation**

– Trapezium $\mathrm{ABCD}$ has coordinates for points $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$ as $ (4,0), (0,2), \text{ and } (h,3h)$ respectively.

– The lines $\mathrm{BC}$ and $\mathrm{AD}$ are parallel.

– Angle $\mathrm{ABC} = 90^\circ$.

– Line segment $\mathrm{CD}$ is parallel to the $\mathrm{x}$-axis.

(i) Finding the value of $h$:

Gradient of $\mathrm{AB}$:

The gradient of a line segment is given by the change in $\mathrm{y}$ divided by the change in $\mathrm{x}$. For line $\mathrm{AB}$:

\( \text{Gradient of } \mathrm{AB} = \frac{\text{change in } \mathrm{y}}{\text{change in } \mathrm{x}} = \frac{0 – 2}{4 – 0} = -\frac{1}{2} \)

Gradient of $\mathrm{BC}$:

Since $\mathrm{BC}$ is perpendicular to $\mathrm{AB}$ and angle $\mathrm{ABC} = 90^\circ$, the product of the gradients of $\mathrm{AB}$ and $\mathrm{BC}$ is equal to -1.

\( \text{Gradient of } \mathrm{BC} = -\frac{1}{\text{Gradient of } \mathrm{AB}} = -\frac{1}{-\frac{1}{2}} = 2 \)

Equation involving $h$:

Using the point-slope form of the equation of a line \((y – y_1) = m(x – x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the gradient:

\( \frac{3h – 2}{h – 0} = 2 \)

Solving this equation for \(h\), we get \(h = 2\).

(ii) Coordinates of $\mathrm{D}$:

Using the point-slope form for the line $\mathrm{AD}$ and the fact that it is perpendicular to $\mathrm{AB}$, we can express it as:

\( \frac{y – 0}{x – 4} = -\frac{1}{2} \)

Solving for \(y\) when \(x = 7\) (which is the x-coordinate for point \(\mathrm{D}\)), we get \(y = 2\).

So, the coordinates of \(\mathrm{D}\) are (7, 2).

Therefore, \(h = 2\) and the coordinates of \(\mathrm{D}\) are (7, 2).