Past Exam Questions

Answer: \(A=\left(-\frac32,-9\right)\), \(B=(2,5)\), and the area of \(\triangle POQ\) is \(\dfrac{961}{128}\).

At the points of intersection, the two expressions for \(y\) are equal:

\(4x-3=3+5x-2x^2\).

Rearranging gives

\(2x^2-x-6=0\).

Factorising,

\((2x+3)(x-2)=0\).

Hence

\(x=-\frac32\quad\text{or}\quad x=2\).

Using the straight line \(y=4x-3\), the corresponding coordinates are

\(x=-\frac32:\quad y=4\left(-\frac32\right)-3=-9\),

and

\(x=2:\quad y=4(2)-3=5\).

So the two points are

\(A=\left(-\frac32,-9\right)\), \(B=(2,5)\).

The gradient of \(AB\) is

\(\dfrac{5-(-9)}{2-\left(-\frac32\right)}=\dfrac{14}{\frac72}=4\).

Therefore the gradient of the perpendicular bisector is \(-\dfrac14\).

The midpoint of \(AB\) is

\(\left(\dfrac{-\frac32+2}{2},\dfrac{-9+5}{2}\right)=\left(\frac14,-2\right)\).

So the perpendicular bisector is

\(y+2=-\dfrac14\left(x-\frac14\right)\).

Simplifying,

\(y=-\dfrac14x-\dfrac{31}{16}\).

When \(y=0\), \(x=-\dfrac{31}{4}\), so \(P=\left(-\dfrac{31}{4},0\right)\).

When \(x=0\), \(y=-\dfrac{31}{16}\), so \(Q=\left(0,-\dfrac{31}{16}\right)\).

The triangle \(POQ\) is right-angled at \(O\), hence

\(\text{area}=\dfrac12\left|OP\right|\left|OQ\right|=\dfrac12\cdot\dfrac{31}{4}\cdot\dfrac{31}{16}=\dfrac{961}{128}\).

Answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Substitute \(y=2x+1\) into \(y+xy+3x^2=15\):

\(2x+1+x(2x+1)+3x^2=15.\)

This simplifies to

\(5x^2+3x-14=0.\)

Solving gives \(x=\frac75\) or \(x=-2\). The corresponding \(y\)-values are \(\frac{19}{5}\) and \(-3\).

The midpoint of \(AB\) is

\(\left(\frac{\frac75-2}{2},\frac{\frac{19}{5}-3}{2}\right)=\left(-\frac{3}{10},\frac25\right).\)

The line \(AB\) has gradient \(2\), so the perpendicular bisector has gradient \(-\frac12\). Its equation is

\(y-\frac25=-\frac12\left(x+\frac{3}{10}\right).\)

Since \(C=\left(\frac{21}{10},k\right)\) lies on this line,

\(k-\frac25=-\frac12\left(\frac{21}{10}+\frac{3}{10}\right)=-\frac65.\)

Thus \(k=-\frac45\).

The vector from the midpoint to \(C\) is

\(\left(\frac{21}{10},-\frac45\right)-\left(-\frac{3}{10},\frac25\right)=\left(\frac{12}{5},-\frac65\right).\)

A point twice as far from \(AB\) on the same perpendicular line can lie on either side of the midpoint. Hence

\(D=\left(-\frac{3}{10},\frac25\right)+2\left(\frac{12}{5},-\frac65\right)=\left(\frac92,-2\right)\),

or

\(D=\left(-\frac{3}{10},\frac25\right)-2\left(\frac{12}{5},-\frac65\right)=\left(-\frac{51}{10},\frac{14}{5}\right).\)

This gives the required answer: \(k=-\frac45\). The possible coordinates of \(D\) are \(\left(\frac92,-2\right)\) and \(\left(-\frac{51}{10},\frac{14}{5}\right)\).

Answer: \(P=(\sqrt6,\frac{\sqrt6}{3})\), \(Q=(-\sqrt6,-\frac{\sqrt6}{3})\); \(PQ=\frac{4\sqrt{15}}{3}\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

(a) Substitute \(y=\frac2x\) into

\(4x^2-3y^2+xy=24.\)

This gives

\(4x^2-3\left(\frac2x\right)^2+x\left(\frac2x\right)=24.\)

So

\(4x^2-\frac{12}{x^2}+2=24.\)

Multiply by \(x^2\):

\(4x^4-12+2x^2=24x^2.\)

Hence

\(4x^4-22x^2-12=0,\)

or

\(2x^4-11x^2-6=0.\)

Let \(u=x^2\). Then

\(2u^2-11u-6=0.\)

Factorise:

\((2u+1)(u-6)=0.\)

Since \(u=x^2\geq0\), we take \(u=6\). Thus

\(x=\pm\sqrt6.\)

Using \(y=\frac2x\), the points are

\(\left(\sqrt6,\frac{\sqrt6}{3}\right)\quad\text{and}\quad \left(-\sqrt6,-\frac{\sqrt6}{3}\right).\)

(b) The distance between these two points is

\(PQ=\sqrt{(2\sqrt6)^2+\left(\frac{2\sqrt6}{3}\right)^2}.\)

So

\(PQ=\sqrt{24+\frac83}=\sqrt{\frac{80}{3}} =\frac{4\sqrt{15}}{3}.\)

This completes the solution and gives the required result.

Answer: \(k=-\dfrac{19}{8}\), and \(D=\left(\dfrac{29}{8},-\dfrac98\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

Substitute \(y=3x-2\) into

\(2x^2-xy+y^2=2.\)

This gives

\(2x^2-x(3x-2)+(3x-2)^2=2.\)

Expanding and simplifying,

\(4x^2-5x+1=0.\)

Factorise:

\((4x-1)(x-1)=0.\)

So

\(x=\frac14\quad\text{or}\quad x=1.\)

Using \(y=3x-2\), the intersection points are

\(\left(\frac14,-\frac54\right)\quad\text{and}\quad(1,1).\)

The midpoint of \(AB\) is

\(\left(\frac{\frac14+1}{2},\frac{-\frac54+1}{2}\right) =\left(\frac58,-\frac18\right).\)

The gradient of \(AB\) is \(3\), because \(AB\) lies on \(y=3x-2\).

Therefore the perpendicular bisector has gradient \(-\frac13\).

Its equation is

\(y+\frac18=-\frac13\left(x-\frac58\right).\)

The point \(C=\left(k,\frac78\right)\) lies on this line, so substitute \(y=\frac78\):

\(\frac78+\frac18=-\frac13\left(k-\frac58\right).\)

So

\(1=-\frac13\left(k-\frac58\right).\)

Hence

\(k-\frac58=-3\), and

\(k=-\frac{19}{8}.\)

For part (b), \(D\) is the reflection of \(C\) in the line \(AB\).

The perpendicular bisector meets \(AB\) at the midpoint of \(CD\), which is also the midpoint of \(AB\):

\(M=\left(\frac58,-\frac18\right).\)

Since \(M\) is the midpoint of \(C\) and \(D\),

\(D=2M-C.\)

Thus

\(D=\left(2\cdot\frac58-\left(-\frac{19}{8}\right),\ 2\cdot\left(-\frac18\right)-\frac78\right).\)

Therefore

\(D=\left(\frac{29}{8},-\frac98\right).\)

This completes the solution and gives the required result.

Answer: (a) \(3\sqrt3\); (b) \(\left(-10+\sqrt{17},-\dfrac{10+\sqrt{17}}{83}\right)\) and \(\left(-10-\sqrt{17},-\dfrac{10-\sqrt{17}}{83}\right)\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

For part (a), the curve cuts the \(y\)-axis where \(x=0\).

Substitute \(x=0\) into

\(9(x-1)^2+4(y-3)^2=36.\)

This gives

\(9+4(y-3)^2=36.\)

So

\(4(y-3)^2=27\), and hence

\(y=3\pm\sqrt{\frac{27}{4}}=3\pm\frac{3\sqrt3}{2}.\)

The distance between the two points on the \(y\)-axis is therefore

\(\left(3+\frac{3\sqrt3}{2}\right)-\left(3-\frac{3\sqrt3}{2}\right)=3\sqrt3.\)

For part (b), substitute \(y=\frac1x\) into the other curve:

\(2x^2+83x\left(\frac1x\right)=x^3\left(\frac1x\right)-20x.\)

So

\(2x^2+83=x^2-20x.\)

Hence

\(x^2+20x+83=0.\)

Using the quadratic formula,

\(x=\frac{-20\pm\sqrt{20^2-4(1)(83)}}{2}.\)

Thus

\(x=\frac{-20\pm\sqrt{68}}{2}=-10\pm\sqrt{17}.\)

Now \(y=\frac1x\). For \(x=-10+\sqrt{17}\),

\(y=\frac{1}{-10+\sqrt{17}}\cdot\frac{-10-\sqrt{17}}{-10-\sqrt{17}} =\frac{-10-\sqrt{17}}{100-17} =-\frac{10+\sqrt{17}}{83}.\)

For \(x=-10-\sqrt{17}\),

\(y=\frac{1}{-10-\sqrt{17}}\cdot\frac{-10+\sqrt{17}}{-10+\sqrt{17}} =\frac{-10+\sqrt{17}}{100-17} =-\frac{10-\sqrt{17}}{83}.\)

Therefore the two points are

\(\left(-10+\sqrt{17},-\frac{10+\sqrt{17}}{83}\right)\)

and

\(\left(-10-\sqrt{17},-\frac{10-\sqrt{17}}{83}\right).\)

This completes the solution and gives the required result.

Answer: \(a=3\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At the point of contact, the line and curve have equal \(y\)-values:

\(ax^2-5x+2=(2a+1)x-10.\)

Bring all terms to one side:

\(ax^2-(2a+6)x+12=0.\)

For the line to be a tangent, this quadratic in \(x\) must have one repeated root.

So its discriminant is zero:

\(\big(-(2a+6)\big)^2-4(a)(12)=0.\)

Hence

\((2a+6)^2-48a=0.\)

Expanding gives

\(4a^2+24a+36-48a=0.\)

So

\(4a^2-24a+36=0.\)

Divide by \(4\):

\(a^2-6a+9=0.\)

Therefore

\((a-3)^2=0,\)

so

\(a=3.\)

This completes the solution and gives the required result.

Answer: \(k=4\).

We start with the main method. Substitute the line into the curve; a tangent corresponds to a repeated root, so the discriminant is zero.

At points of intersection,

\(x(x+2k)=-2x-6k-1.\)

So

\(x^2+2kx+2x+6k+1=0,\)

or

\(x^2+(2k+2)x+6k+1=0.\)

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root.

Therefore its discriminant is zero:

\((2k+2)^2-4(6k+1)=0.\)

Expanding gives

\(4k^2+8k+4-24k-4=0.\)

So

\(4k^2-16k=0.\)

Factorising,

\(4k(k-4)=0.\)

The question asks for the non-zero value, so

\(k=4.\)

This completes the solution and gives the required result.

Answer: \(\displaystyle k=\frac{1+\sqrt{13}}{6}\) or \(\displaystyle k=\frac{1-\sqrt{13}}{6}\).

At a tangent point, the line and the curve meet at one repeated point. After equating the two expressions for \(y\), use the discriminant condition \(b^2-4ac=0\).

At a point of intersection, the line and the curve have the same \(y\)-value:

\(kx^2+x+2k=1-k-x.\)

Rearranging gives

\(kx^2+2x+3k-1=0.\)

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root. Therefore its discriminant is zero:

\(2^2-4(k)(3k-1)=0.\)

So

\(4=4k(3k-1).\)

Dividing by 4,

\(1=3k^2-k.\)

Hence

\(3k^2-k-1=0.\)

Using the quadratic formula,

\(k=\frac{1\pm\sqrt{(-1)^2-4(3)(-1)}}{2(3)} =\frac{1\pm\sqrt{13}}{6}.\)

Answer: \(x=-5,\ \frac32,\ 2\).

Answer: \(x=-5,\ \frac32,\ 2\).

At the points of intersection, the two expressions for \(y\) are equal:

\(2x^3+3x^2-26x+22=3x-8.\)

Rearranging gives

\(2x^3+3x^2-29x+30=0.\)

Testing simple integer roots gives \(x=2\):

\(2(2)^3+3(2)^2-29(2)+30=0.\)

So \((x-2)\) is a factor. Dividing gives

\(2x^3+3x^2-29x+30=(x-2)(2x^2+7x-15).\)

Now factorise the quadratic:

\(2x^2+7x-15=(2x-3)(x+5).\)

Therefore

\((x-2)(2x-3)(x+5)=0.\)

The required \(x\)-coordinates are

\(x=2,\quad x=\frac32,\quad x=-5.\)

Answer: \(x=1,\ -\frac32,\ \frac43\).

Answer: \(x=1,\ -\frac32,\ \frac43\).

At the points of intersection, the two expressions for \(y\) are equal:

\(7x^3-7x^2-17x-4=x^3-2x^2-4x-16.\)

Rearrange:

\(6x^3-5x^2-13x+12=0.\)

Testing simple roots, \(x=1\) is a root, so \((x-1)\) is a factor. Dividing gives

\(6x^3-5x^2-13x+12=(x-1)(6x^2+x-12).\)

Now factorise the quadratic:

\(6x^2+x-12=(2x+3)(3x-4).\)

Therefore

\((x-1)(2x+3)(3x-4)=0.\)

Hence

\(x=1, \quad x=-\frac32, \quad x=\frac43.\)

Answer: \(k=3\), and the remaining quadratic factor has no real roots.

Answer: \(k=3\), and the line intersects the curve at only one point.

At an intersection,

\(kx+6=x^3-4x^2+3kx+2.\)

Substitute \(x=2\):

\(2k+6=8-16+6k+2.\)

So

\(2k+6=6k-6,\)

which gives

\(k=3.\)

Now substitute \(k=3\) into the intersection equation:

\(3x+6=x^3-4x^2+9x+2.\)

Hence

\(x^3-4x^2+6x-4=0.\)

Since \(x=2\) is a root, factorise:

\(x^3-4x^2+6x-4=(x-2)(x^2-2x+2).\)

The quadratic factor has discriminant

\((-2)^2-4(1)(2)=4-8=-4\lt 0.\)

So it has no real roots. Therefore the only real intersection occurs at \(x=2\), and the line intersects the curve at only one point.

(i) To find the value of h, we use the fact that ∠ABC = 90° and lines BC and AD are parallel. The gradient of AB is \(-\frac{1}{2}\) and the gradient of BC is \(\frac{3h - 2}{h}\). Since BC is perpendicular to AB, the product of their gradients is \(-1\):

\(\left(-\frac{1}{2}\right) \times \frac{3h - 2}{h} = -1\)

Solving for h, we get:

\(\frac{3h - 2}{h} = 2\)

\(3h - 2 = 2h\)

\(h = 2\)

(ii) With h = 2, the coordinates of C are (2, 6). Since CD is parallel to the x-axis, the y-coordinate of D is the same as C, which is 6. The x-coordinate of D can be found using the fact that AD is parallel to BC, so the gradient of AD is also 2:

\(\frac{6 - 0}{x - 4} = 2\)

\(6 = 2(x - 4)\)

\(6 = 2x - 8\)

\(2x = 14\)

\(x = 7\)

Thus, the coordinates of D are (7, 6).

(i) To find the equation of the perpendicular bisector of AC:

The midpoint of AC is \(\left( \frac{-1 + 5}{2}, \frac{2 + 4}{2} \right) = (2, 3)\).

The gradient of AC is \(\frac{4 - 2}{5 + 1} = \frac{1}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal, \(-3\).

Using the point-slope form, the equation is \(y - 3 = -3(x - 2)\).

(ii) To find the coordinates of B and D:

Since B lies on the y-axis, its x-coordinate is 0. Using the perpendicular bisector equation, \(y - 3 = -3(0 - 2)\), we find \(y = 9\). Thus, \(B(0, 9)\).

To find D, use the vector move from A to C, which is \((5 - (-1), 4 - 2) = (6, 2)\).

Since D is opposite B, apply the vector move to B: \((0 + 4, 9 - 12) = (4, -3)\). Thus, \(D(4, -3)\).

(iii) To find the area of the rhombus:

The length of AC is \(\sqrt{(5 - (-1))^2 + (4 - 2)^2} = \sqrt{40}\).

The length of BD is \(\sqrt{(4 - 0)^2 + (-3 - 9)^2} = \sqrt{160}\).

The area of the rhombus is \(\frac{1}{2} \times \sqrt{40} \times \sqrt{160} = 40\).

The gradient of OB is calculated as follows:

Gradient of OB = \(\frac{1 - 0}{3 - 0} = \frac{1}{3}\).

Since L₁ is parallel to OB, it has the same gradient: \(\frac{1}{3}\).

The equation of L₁ passing through A (-1, 3) is:

\(y - 3 = \frac{1}{3}(x + 1)\).

The gradient of AB is:

Gradient of AB = \(\frac{1 - 3}{3 + 1} = -\frac{1}{2}\).

The gradient of L₂, being perpendicular to AB, is the negative reciprocal:

\(Gradient of L₂ = 2.\)

The equation of L₂ passing through B (3, 1) is:

\(y - 1 = 2(x - 3)\).

Solving the simultaneous equations:

1. \(y = \frac{1}{3}x + \frac{10}{3}\)

2. \(y = 2x - 5\)

Equating the two expressions for \(y\):

\(\frac{1}{3}x + \frac{10}{3} = 2x - 5\)

Multiply through by 3 to clear the fraction:

\(x + 10 = 6x - 15\)

Rearrange to solve for \(x\):

\(5x = 25\)

\(x = 5\)

Substitute \(x = 5\) back into one of the equations to find \(y\):

\(y = 2(5) - 5 = 5\)

Thus, the coordinates of C are (5, 5).

(i) The gradient of AB is calculated as:

\(\frac{22 - (-2)}{15 - 3} = \frac{24}{12} = 2\)

Since the gradient of AB is given as 2m, equating gives:

\(2m = 2\)

Thus, \(m = 1\).

(ii) The equation of line AC is:

\(y + 2 = -2(x - 3)\)

\(y = -2x + 6 - 2\)

\(y = -2x + 4\)

The equation of line BC is:

\(y - 22 = 1(x - 15)\)

\(y = x + 7\)

Solving the simultaneous equations:

\(y = -2x + 4\)

\(y = x + 7\)

Equating gives:

\(-2x + 4 = x + 7\)

\(-3x = 3\)

\(x = -1\)

Substitute \(x = -1\) into \(y = x + 7\):

\(y = -1 + 7 = 6\)

Thus, \(C = (-1, 6)\).

(iii) The midpoint M of AB is:

\(M = \left( \frac{3 + 15}{2}, \frac{-2 + 22}{2} \right) = (9, 10)\)

The gradient of the perpendicular bisector is:

\(-\frac{1}{2}\)

The equation of the perpendicular bisector is:

\(y - 10 = -\frac{1}{2}(x - 9)\)

\(2y - 20 = -x + 9\)

\(2y + x = 29\)

Solving the equations \(y = x + 7\) and \(2y + x = 29\):

Substitute \(y = x + 7\) into \(2y + x = 29\):

\(2(x + 7) + x = 29\)

\(2x + 14 + x = 29\)

\(3x = 15\)

\(x = 5\)

Substitute \(x = 5\) into \(y = x + 7\):

\(y = 5 + 7 = 12\)

Thus, \(D = (5, 12)\).

(i) The y-coordinate of D is the same as the y-coordinate of the midpoint of AC. The midpoint of AC is \(\left( \frac{0+12}{2}, \frac{-2+14}{2} \right) = (6, 6)\). Therefore, the y-coordinate of D is 6.

(ii) The gradient of AD is calculated as \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-2)}{h - 0} = \frac{8}{h}\) or \(\frac{h-12}{8}\). The gradient of CD is \(\frac{6 - 14}{h - 12} = \frac{-8}{h-12}\) or \(\frac{-h}{8}\).

(iii) Since BD is parallel to the x-axis, the product of the gradients of AD and CD is -1. Solving \(\frac{8}{h} \times \frac{-8}{h-12} = -1\) gives \(h^2 - 12h - 64 = 0\). Solving this quadratic equation, we find \(h = 16\) or \(h = -4\). Thus, \(x_D = 16\) and \(x_B = -4\).

(iv) The area of rectangle ABCD can be calculated using the lengths of the sides. The length of AC is \(\sqrt{(12-0)^2 + (14-(-2))^2} = 20\). The length of BD is \(\sqrt{(16-(-4))^2 + (6-6)^2} = 20\). The area is \(20 \times 8 = 160\).

The line AC has the equation \(2y = x + 4\), which simplifies to \(y = \frac{1}{2}x + 2\). The gradient of AC is \(\frac{1}{2}\).

The line BD is perpendicular to AC, so its gradient is the negative reciprocal of \(\frac{1}{2}\), which is \(-2\).

The equation of line BD passing through \(D(10, -3)\) with gradient \(-2\) is:

\(y + 3 = -2(x - 10)\)

Simplifying gives:

\(y + 3 = -2x + 20\)

\(y = -2x + 17\)

Equating the equations of lines AC and BD:

\(\frac{1}{2}x + 2 = -2x + 17\)

Solving for \(x\):

\(\frac{1}{2}x + 2x = 17 - 2\)

\(\frac{5}{2}x = 15\)

\(x = 6\)

Substitute \(x = 6\) back into \(y = \frac{1}{2}x + 2\):

\(y = \frac{1}{2}(6) + 2 = 3 + 2 = 5\)

Thus, B is \((6, 5)\).

Since \(AB = BC\), the vector from A to B is the same as from B to C. If A is \((0, 2)\), then the vector AB is \((6 - 0, 5 - 2) = (6, 3)\).

Adding this vector to B gives C:

C = \((6 + 6, 5 + 3) = (12, 8)\).

(i) The gradient of AC is \(-\frac{1}{2}\). The perpendicular gradient is 2. The equation of line BX is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\). Solving the equations \(2y + x = 16\) and \(y = 2x - 2\) simultaneously gives the coordinates of X as \((4, 6)\).

(ii) Since X is the midpoint of BD, and using the coordinates of X \((4, 6)\), we find D as \((6, 10)\).

(iii) Using the distance formula, \(AB = \sqrt{(14)^2 + (2)^2} = \sqrt{200}\) and \(BC = \sqrt{(2)^2 + (6)^2} = \sqrt{40}\). The perimeter is \(2\sqrt{200} + 2\sqrt{40} = 40.9\).

First, calculate the gradient of line AB:

Gradient of AB = \(\frac{2 - 8}{6 - 3} = -2\).

Since DC is parallel to AB, the gradient of DC is also -2. Using point C (10, 2), the equation of line DC is:

\(y - 2 = -2(x - 10)\)

Simplifying gives:

\(y = -2x + 20 + 2\)

\(y = -2x + 22\)

For line DA, since it is perpendicular to AB, the gradient of DA is the negative reciprocal of -2, which is \(\frac{1}{2}\). Using point A (3, 8), the equation of line DA is:

\(y - 8 = \frac{1}{2}(x - 3)\)

Simplifying gives:

\(y = \frac{1}{2}x + \frac{1}{2} \times (-3) + 8\)

\(y = \frac{1}{2}x - \frac{3}{2} + 8\)

\(y = \frac{1}{2}x + \frac{13}{2}\)

Now, solve the simultaneous equations:

1. \(y = -2x + 22\)

2. \(y = \frac{1}{2}x + \frac{13}{2}\)

Equating the two expressions for \(y\):

\(-2x + 22 = \frac{1}{2}x + \frac{13}{2}\)

Multiply through by 2 to eliminate the fraction:

\(-4x + 44 = x + 13\)

\(44 - 13 = 4x + x\)

\(31 = 5x\)

\(x = \frac{31}{5} = 6.2\)

Substitute \(x = 6.2\) back into equation 1:

\(y = -2(6.2) + 22\)

\(y = -12.4 + 22\)

\(y = 9.6\)

Thus, the coordinates of D are (6.2, 9.6).

(i) To find the equation of BC, first calculate the slope of AB. The slope \(m\) of AB is given by:

\(m = \frac{14 - 8}{2 - (-2)} = \frac{6}{4} = \frac{3}{2}\)

Since BC is perpendicular to AB, the slope of BC is the negative reciprocal of \(\frac{3}{2}\), which is \(-\frac{2}{3}\).

Using point B (-2, 8) and the slope \(-\frac{2}{3}\), the equation of line BC is:

\(y - 8 = -\frac{2}{3}(x + 2)\)

Simplifying, we get:

\(3y + 2x = 20\)

(ii) Since C lies on the x-axis, its y-coordinate is 0. Substitute \(y = 0\) into the equation of BC:

\(3(0) + 2x = 20\)

\(2x = 20\)

\(x = 10\)

Thus, the coordinates of C are (10, 0).

To find D, use the vector move from B to C and apply it to A. The vector from B to C is:

\((10 - (-2), 0 - 8) = (12, -8)\)

Apply this vector to A (2, 14):

\((2 + 12, 14 - 8) = (14, 6)\)

Thus, the coordinates of D are (14, 6).