**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) Equation of the Line \(CD\):

The midpoint \(C\) of the line segment \(\overline{AB}\) is given by the average of the corresponding coordinates of \(A\) and \(B\):

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

For points \(A(14, -7)\) and \(B(-6, 3)\), the coordinates of \(C\) are:

\(C\left(\frac{14 + (-6)}{2}, \frac{(-7) + 3}{2}\right) = (4, -2)\)

Now, we know that the line \(CD\) is perpendicular to \(\overline{AB}\), so the slope of \(CD\) is the negative reciprocal of the slope of \(\overline{AB}\).

The slope of \(\overline{AB}\) is given by:

\(m_{AB} = \frac{y_B – y_A}{x_B – x_A}\)

\(m_{AB} = \frac{3 – (-7)}{(-6) – 14} = \frac{10}{-20} = -\frac{1}{2}\)

The slope of \(CD\) is the negative reciprocal:

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

Now, by point-slope form of the equation to find the equation of \(CD\):

\(y – y_C = m_{CD}(x – x_C)\)

\(y – (-2) = 2(x – 4)\)

\(y + 2 = 2x – 8\)

\(y = 2x – 10\)

So, the equation of the line \(CD\) is \(y = 2x – 10\).

(ii) The distance \(AD\) is the length of the line segment \(\overline{AD}\) and can be found using the distance formula:

\(AD = \sqrt{(x_D – x_A)^2 + (y_D – y_A)^2}\)

For \(A(14, -7)\) and \(D(x_D, 0)\) (since it lies on the y-axis), the distance \(AD\) is:

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

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

Since \(D\) lies on the y-axis, \(x_D = 0\):

\(AD = \sqrt{(14)^2 + (3)^{2}} = \sqrt{205}\)

So, 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) Equation of the Perpendicular Bisector:

The equation of the perpendicular bisector of \(AB\) can be found by first determining the midpoint of \(AB\), which is the point \(M\), and then finding the negative reciprocal of the slope of \(AB\) to get the slope of the perpendicular bisector.

The coordinates of the midpoint \(M\) are given by the average of the \(x\)-coordinates and the \(y\)-coordinates of \(A\) and \(B\):

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

Substituting in the coordinates of \(A(1, 1)\) and \(B(5, 9)\):

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

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

\(m_{AB} = \frac{y_B – y_A}{x_B – x_A}\)

Substituting in the coordinates of \(A\) and \(B\):

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

The slope of the perpendicular bisector (\(m_{\text{perpendicular}}\)) is the negative reciprocal of \(m_{AB}\):

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

Equation 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, substitute \(M(3, 5)\) and

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

\(y – 5 = -\frac{1}{2}(x – 3)\)

\(2y – 10 = 3 – x\)

\(2y = 13 – x\)

Therefore, the equation of the perpendicular bisector is \(2y = 13 – x\).

(ii) Finding \(C\) and \(D):

The equation of the curve:

\(6y = 5x^2 – 18x + 19\)

\(6\left(\frac{13 – x}{2}\right) = 5x^2 – 18x + 19\)

\(-3x + 39 = 5x^2 – 18x + 19\)

\(5x^2 – 15x – 20 = 0\)

\(5(x^2 – 3x – 4) = 0\)

\(5(x – 4)(x + 1) = 0\)

This gives two solutions for \(x\): \(x = 4\) or \(x = -1\).

Now, substitute these values back into the equation of the perpendicular bisector to find the corresponding \(y\)-coordinates.

For \(x = 4\):

\(2y = 13 – 4\)

\(2y = 9\)

\(y = 4.5\)

For \(x = -1\):

\(2y = 13 – (-1)\)

\(2y = 14\)

\(y = 7\)

So, the points \(C\) and \(D\) are \((4, 4.5)\) and \((-1, 7)\), respectively.

The distance formula to find the distance \(CD\):

\(CD = \sqrt{(x_D – x_C)^2 + (y_D – y_C)^2}\)

Substitute the coordinates of \(C\) and \(D\):

\(CD = \sqrt{(-1 – 4)^2 + (7 – 4.5)^2}\)

\(CD = \sqrt{(-5)^2 + (2.5)^2}\)

\(CD = \sqrt{25 + 6.25}\)

\(CD = \sqrt{\frac{125}{4}}\)

So, the distance \(CD\) is \(\frac{\sqrt{125}}{2}\).

**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) Given that the line \(4x + ky = 20\) passes through point \(A(8, -4)\), we can substitute the coordinates of \(A\) into the equation to find \(k\):

\(4(8) + k(-4) = 20\)

\(32 – 4k = 20\)

\(-4k = 20 – 32\)

\(-4k = -12\)

\(k = 3\)

Now that we have found \(k\), we can use it to find the value of \(b\) using point \(B(b, 2b)\). Substitute the coordinates of \(B\) into the equation:

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

\(4b + 6b = 20\)

\(10b =20\)

\(b = 2\

So, the values of \(k\) and \(b\) are \(k = 3\) and \(b = 2\).

(ii) Finding the Mid-point of \(AB):

The coordinates of the mid-point \((M)\) of a line segment \(AB\) with endpoints \(A(x_1, y_1)\) and \(B(x_2, y_2)\) are given by:

\(M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\)

For \(A(8, -4)\) and \(B(2, 4)\), the coordinates of the mid-point \(M\) are:

\(M\left(\frac{8 + 2}{2}, \frac{(-4) + (2 \times 4)}{2}\right)\)

\(M(5, 0)\)

So, the coordinates of the mid-point \(M\) are \((5, 0)\).

**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) The graph of \(y = 2\cos x\) for \(-\pi \leq x \leq \pi\) is a cosine function that oscillates between \(-2\) and \(2\). The amplitude is \(2\) and the period is \(2\pi\). The graph starts at its maximum value at \(x = 0\) and repeats every \(\pi\) radians.

The point of intersection with the \(y\)-axis occurs when \(x = 0\). Substitute \(x = 0\) into the equation:

\(y = 2\cos 0 = 2\)

So, the point of intersection with the \(y\)-axis is \((0, 2)\).

(ii) Given \(P\left(\frac{\pi}{3}, 1\right)\) and \(Q(\pi,-2)\), the length of PQ is given by the distance formula:

\(PQ = \sqrt{\left(\pi – \frac{\pi}{3}\right)^2 + (-2 – 1)^2} = \sqrt{\left(\frac{2\pi}{3}\right)^2 + 3^2} \approx 3.7\)

So, the length of PQ is approximately \(3.7\) (correct to 1 decimal place).

(iii) Finding \(h\) and \(k\):

The equation of the line PQ can be found using the two given points P and Q:

\(y – 1 = -\frac{9}{2\pi}\left(x – \frac{\pi}{3}\right)\)

Setting \(y = 0\) to find \(h\):

\(0 – 1 = -\frac{9}{2\pi}\left(x – \frac{\pi}{3}\right)\)

\(-1 = -\frac{9}{2\pi}\left(x – \frac{\pi}{3}\right)\)

\(\frac{2\pi}{9} = x – \frac{\pi}{3}\)

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

So, \(h = \frac{5\pi}{9}\).

\(x = 0\) to find \(k\):

\(y – 1 = -\frac{9}{2\pi}\left(0 – \frac{\pi}{3}\right)\)

\(y – 1 = \frac{3}{2}\)

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

So, \(k = \frac{5}{2}\).

Therefore, \(h = \frac{5\pi}{9}\) and \(k = \frac{5}{2}\).

**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**

(i) Length of \(AB\):

To find the points of intersection between the curve \(y = 4x^{1/2}\) and the line \(y = x + 3\), set the two equations equal to each other:

\(4x^{1/2} = x + 3\)

\(4x^{1/2} – x = 3\)

\((4x^{1/2} – x)^2 = 3^2\)

Solving this equation gives \(x = 1\) and \(x = 9\). Substituting these values back into either equation gives \(y = 4\) and \(y = 12\).

Now, using distance formula to find the length of \(AB\):

\(AB = \sqrt{(x_B – x_A)^2 + (y_B – y_A)^2}\)

Substituting in the coordinates of \(A(1, 4)\) and \(B(9, 12)\):

\(AB = \sqrt{(9 – 1)^2 + (12 – 4)^2} = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}\)

So, the length of \(AB\) is \(8\sqrt{2}\).

(ii) We are looking for the point on the curve where the slope of the tangent is equal to 1. The slope of the tangent is given by the derivative of the curve \(y = 4x^{1/2}\), which is \(\frac{dy}{dx} = 2x^{-1/2}\).

Setting \(\frac{dy}{dx} = 1\), we have:

\(2x^{-1/2} = 1\)

\(x^{-\frac{1}{2}}=\frac{1}{2}\)

\(x^{-1}=\frac{1}{4}\)

x=4

Substituting this back into the original equation gives the corresponding \(y\)-coordinate:

\(y = 4x^{1/2} = 4(4^{1/2}) = 4 \cdot 2 = 8\)

So, the correct coordinates of the point \(T\) where the tangent is parallel to \(AB\) are \((4, 8)\).

(iii) Coordinates of the point of intersection of the normal at \(T\) with \(AB\):

The slope of the normal at \(T\) is the negative reciprocal of the slope of the tangent at \(T\). The slope of the tangent at \(T\) is 1, so the slope of the normal is -1.

Equation of the Normal:

The equation of the normal to the curve at \(T(4, 8)\) is given by:

\(y – 8 = -1(x – 4)\)

Eliminating \(y\) (or \(x\)):

Now, we need to find the point where this normal intersects the line \(AB\), which has the equation \(y = x + 3\).

Substitute the equation of the normal into the equation of \(AB\) to solve for the intersection point:

\(y – 8 = -1(x – 4)\)

\(y – 8 = -x + 4\)

\(y = -x + 12\)

\(-x + 12 = x + 3\)

\(2x = 9\)

\(x = \frac{9}{2}\)

Substitute this value back into either equation to find \(y\). Using the equation of the normal:

\(y – 8 = -1\left(\frac{9}{2} – 4\right)\)

\(y – 8 = -1\left(-\frac{1}{2}\right)\)

\(y = 7^{1/2}\)

So, the coordinates of the point of intersection of the normal at \(T\) with \(AB\) are \((\sqrt{4}, \sqrt{7})\), which simplifies to \((2, \sqrt{7})\).