Past Exam Questions

Step-by-step Solution:

1. Substitute point A (8, -4) into the line equation 4x + ky = 20:
   \(4(8) + k(-4) = 20\)
   \(32 - 4k = 20\)
   Solve for \(k\):
   \(32 - 20 = 4k\)
   \(12 = 4k\)
   \(k = 3\)
2. Substitute point B (b, 2b) into the line equation:
   \(4b + 3(2b) = 20\)
   \(4b + 6b = 20\)
   \(10b = 20\)
   Solve for \(b\):
   \(b = 2\)
3. Find the mid-point of AB:
   Mid-point formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
   \(A (8, -4), B (2, 4)\)
   Mid-point: \(\left( \frac{8 + 2}{2}, \frac{-4 + 4}{2} \right) = (5, 0)\)

1. Find the midpoint of the line segment joining (2, 7) and (10, 3):

\(\left( \frac{2 + 10}{2}, \frac{7 + 3}{2} \right) = (6, 5)\)

2. Calculate the gradient (slope) of the line segment:

\(m = \frac{3 - 7}{10 - 2} = -\frac{1}{2}\)

3. Determine the gradient of the perpendicular bisector:

\(m_{\text{perp}} = -\frac{1}{m} = 2\)

4. Write the equation of the perpendicular bisector using point-slope form:

\(y - 5 = 2(x - 6)\)

5. Set \(y = 0\) to find the x-intercept:

\(0 - 5 = 2(x - 6)\)

\(-5 = 2x - 12\)

\(2x = 7\)

\(x = \frac{7}{2}\)

6. The coordinates where the perpendicular bisector meets the x-axis are:

\(\left( \frac{7}{2}, 0 \right)\)

1. Use the distance formula between points \((a, 2)\) and \((3, b)\):

\(\sqrt{(a - 3)^2 + (2 - b)^2} = \sqrt{125}\)

Squaring both sides:

\((a - 3)^2 + (2 - b)^2 = 125\)

2. Use the gradient formula for the line \(AB\):

\(\frac{2 - b}{a - 3} = 2\)

Rearrange to find \(b\):

\(2 - b = 2(a - 3)\)

\(b = 2 - 2a + 6\)

\(b = 8 - 2a\)

3. Substitute \(b = 8 - 2a\) into the distance equation:

\((a - 3)^2 + (2 - (8 - 2a))^2 = 125\)

\((a - 3)^2 + (2a - 6)^2 = 125\)

4. Expand and simplify:

\((a - 3)^2 + (2a - 6)^2 = 125\)

\((a^2 - 6a + 9) + (4a^2 - 24a + 36) = 125\)

\(5a^2 - 30a + 45 = 125\)

\(5a^2 - 30a - 80 = 0\)

5. Factorize:

\(5(a^2 - 6a - 16) = 0\)

\(5(a + 2)(a - 8) = 0\)

6. Solve for \(a\):

\(a = -2\) or \(a = 8\)

7. Substitute back to find \(b\):

If \(a = -2\), \(b = 8 - 2(-2) = 12\)

If \(a = 8\), \(b = 8 - 2(8) = -8\)

Thus, the possible values are \(a = -2\) or \(8\), \(b = 12\) or \(-8\).

1. Find the midpoint M of the points (3, 7) and (-1, 1):

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

2. Find the slope of the line \(\frac{x}{3} + \frac{y}{2} = 1\):

Rewriting in slope-intercept form:

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

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

The slope is \(-\frac{2}{3}\).

3. Use the point-slope form to find the equation of the line through M (1, 4) with the same slope:

\(y - 4 = -\frac{2}{3}(x - 1)\)

(i) The slope of the line y = 2x - 7 is 2. The slope of a line perpendicular to this is the negative reciprocal, which is -\frac{1}{2}. Using the point-slope form of a line equation, y - y_1 = m(x - x_1), where m is the slope and (x_1, y_1) is a point on the line, we have:

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

(ii) The midpoint C of AB is calculated as:

C = \left(\frac{3 + (-21)}{2}, \frac{1 + 11}{2}\right) = (-9, 6)

The distance AC is calculated using the distance formula:

AC = \sqrt{(3 - (-9))^2 + (1 - 6)^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13

(i) To find the perpendicular bisector of AB:

1. Calculate the midpoint of AB:

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

2. Find the slope of AB:

Slope of AB = \(\frac{2 - 6}{7 + 1} = -\frac{1}{2}\)

3. The slope of the perpendicular bisector is the negative reciprocal:

Slope = 2

4. Use the point-slope form to find the equation:

\(y - 4 = 2(x - 3)\)

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

\(y = 2x - 2\)

(ii) For point C on the perpendicular bisector:

1. The equation of the bisector is \(q = 2p - 2\)

2. The distance OC is 2 units:

\(p^2 + q^2 = 4\)

3. Substitute \(q = 2p - 2\) into \(p^2 + q^2 = 4\):

\(p^2 + (2p - 2)^2 = 4\)

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

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

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

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

4. Find q for each p:

If \(p = 0\), \(q = 2(0) - 2 = -2\)

If \(p = \frac{8}{5}\), \(q = 2\left(\frac{8}{5}\right) - 2 = \frac{16}{5} - \frac{10}{5} = \frac{6}{5}\)

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

1. The equation of the line is given as \(3y + 2x = 33\). Rearranging gives \(y = -\frac{2}{3}x + 11\).

2. The gradient of the line is \(-\frac{2}{3}\). The gradient of the perpendicular line is the negative reciprocal, \(\frac{3}{2}\).

3. The equation of the line perpendicular to 3y + 2x = 33 and passing through (-1, 3) is \(y - 3 = \frac{3}{2}(x + 1)\).

4. Simplifying gives the equation \(y = \frac{3}{2}x + \frac{9}{2}\).

5. Solving the system of equations \(3y + 2x = 33\) and \(y = \frac{3}{2}x + \frac{9}{2}\) gives the intersection point (3, 9).

6. The reflection of (-1, 3) across the line is found by using the midpoint formula. The midpoint of (-1, 3) and the reflection point R is (3, 9). Solving gives R = (7, 15).

To find the points of intersection, set the equations equal: \(x^2 - 4x + 4 = x\).

Rearrange to form a quadratic equation: \(x^2 - 5x + 4 = 0\).

Factor the quadratic: \((x - 1)(x - 4) = 0\).

Thus, \(x = 1\) and \(x = 4\).

Substitute back to find \(y\) coordinates: \(y = 1\) when \(x = 1\) and \(y = 4\) when \(x = 4\).

Points \(A\) and \(B\) are \((1, 1)\) and \((4, 4)\).

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

1. Substitute \(k = 8\) into the line equation: \(2y + x = 8\).

2. Solve for \(x\) in terms of \(y\): \(x = 8 - 2y\).

3. Substitute \(x = 8 - 2y\) into the curve equation \(xy = 6\):

\(y(8 - 2y) = 6\)

4. Simplify and solve the quadratic equation:

\(2y^2 - 8y + 6 = 0\)

5. Factor or use the quadratic formula to find \(y\):

\((y - 1)(y - 3) = 0\)

6. The solutions are \(y = 1\) and \(y = 3\).

7. Find corresponding \(x\) values:

For \(y = 1\), \(x = 8 - 2(1) = 6\).

For \(y = 3\), \(x = 8 - 2(3) = 2\).

8. Points \(A\) and \(B\) are \((6, 1)\) and \((2, 3)\).

9. Find the midpoint \(M\) of \(AB\):

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

10. Find the slope of \(AB\):

\(m = \frac{3 - 1}{2 - 6} = -\frac{1}{2}\)

11. The perpendicular slope is \(2\) (since \(m_1 \cdot m_2 = -1\)).

12. Use point-slope form for the perpendicular bisector:

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

1. Find the midpoint M of AB:

\(M = \left( \frac{-1 + 7}{2}, \frac{-5 + 1}{2} \right) = (3, -2)\)

2. Calculate the gradient of AB:

\(\text{Gradient of } AB = \frac{1 - (-5)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\)

3. The gradient of the perpendicular bisector is the negative reciprocal:

\(\text{Perpendicular gradient} = -\frac{4}{3}\)

4. Equation of the perpendicular bisector through M:

\(y + 2 = -\frac{4}{3}(x - 3)\)

5. Find where it meets the x-axis (y = 0):

\(0 + 2 = -\frac{4}{3}(x - 3)\)

\(2 = -\frac{4}{3}x + 4\)

\(\frac{4}{3}x = 2\)

\(x = \frac{3}{2}\)

\(C = \left( \frac{3}{2}, 0 \right)\)

6. Find where it meets the y-axis (x = 0):

\(y + 2 = -\frac{4}{3}(0 - 3)\)

\(y + 2 = 4\)

\(y = 2\)

\(D = (0, 2)\)

7. Calculate the length of CD using the distance formula:

\(CD = \sqrt{\left( \frac{3}{2} - 0 \right)^2 + (0 - 2)^2}\)

\(CD = \sqrt{\left( \frac{3}{2} \right)^2 + (-2)^2}\)

\(CD = \sqrt{\frac{9}{4} + 4}\)

\(CD = \sqrt{\frac{9}{4} + \frac{16}{4}}\)

\(CD = \sqrt{\frac{25}{4}}\)

\(CD = \frac{5}{2} = 2.5\)

(i) The midpoint M of AC is calculated as:

\(M = \left( \frac{-3 + 5}{2}, \frac{2 + 6}{2} \right) = (1, 4)\)

The gradient of AC is:

\(\text{Gradient of } AC = \frac{6 - 2}{5 + 3} = \frac{1}{2}\)

The perpendicular bisector has a gradient of:

\(-2\)

The equation of MB is:

\(y - 4 = -2(x - 1)\)

Simplifying gives:

\(y = -2x + 6\)

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

\(0 = -2x + 6\) \(x = 3\)

So, B is (3, 0).

(ii) The gradient of AB is:

\(\frac{2 - 0}{-3 - 3} = -\frac{2}{6} = -\frac{1}{3}\)

The gradient of BC is:

\(\frac{6 - 0}{5 - 3} = \frac{6}{2} = 3\)

Since:

\(-\frac{1}{3} \times 3 = -1\)

AB is perpendicular to BC.

(iii) Since ABCD is a square, D is calculated as:

\(D = (-1, 8)\)

The length of AD is:

\(AD = \sqrt{(-1 + 3)^2 + (8 - 2)^2} = \sqrt{4 + 36} = \sqrt{40}\)

or approximately 6.32.

First, substitute \(k = 2\) into the equations:

Line: \(y = 2x + 6\)

Curve: \(y = x^2 + 3x + 4\)

Set the equations equal to find intersection points:

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

Simplify to: \(x^2 + x - 2 = 0\)

Factorize: \((x - 1)(x + 2) = 0\)

Solutions: \(x = 1\) and \(x = -2\)

Find corresponding \(y\)-values:

For \(x = 1\), \(y = 2(1) + 6 = 8\)

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

Points \(A(1, 8)\) and \(B(-2, 2)\)

Distance \(AB\):

\(AB = \sqrt{(1 - (-2))^2 + (8 - 2)^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\)

Midpoint of \(AB\):

\(\left(\frac{1 + (-2)}{2}, \frac{8 + 2}{2}\right) = \left(-\frac{1}{2}, 5\right)\)

The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis at \(P(a, 0)\) and the y-axis at \(Q(0, b)\).

The distance \(PQ\) is given by:

\(PQ = \sqrt{a^2 + b^2} = \sqrt{45}\)

Squaring both sides, we get:

\(a^2 + b^2 = 45 \quad (1)\)

The gradient of the line \(PQ\) is:

\(\text{Gradient} = \frac{0 - b}{a - 0} = -\frac{b}{a} = -\frac{1}{2}\)

Solving for \(b\), we have:

\(\frac{b}{a} = \frac{1}{2}\)

\(b = \frac{a}{2} \quad (2)\)

Substitute equation (2) into equation (1):

\(a^2 + \left(\frac{a}{2}\right)^2 = 45\)

\(a^2 + \frac{a^2}{4} = 45\)

\(\frac{4a^2 + a^2}{4} = 45\)

\(5a^2 = 180\)

\(a^2 = 36\)

\(a = 6\)

Substitute \(a = 6\) back into equation (2):

\(b = \frac{6}{2} = 3\)

Thus, \(a = 6\) and \(b = 3\).

To solve the problem, follow these steps:

1. Find the gradient of \(L_1\) using points \(A(2, 5)\) and \(B(10, 9)\):

\(\text{Gradient of } L_1 = \frac{9 - 5}{10 - 2} = \frac{1}{2}\)

2. Since \(L_2\) is parallel to \(L_1\), its equation is:

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

3. The line \(AC\) is perpendicular to \(L_2\), so its gradient is the negative reciprocal:

\(\text{Gradient of } AC = -2\)

4. Using point \(A(2, 5)\), the equation of line \(AC\) is:

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

5. Solve the equations \(y = \frac{1}{2}x\) and \(y - 5 = -2(x - 2)\) simultaneously to find \(C\):

Substitute \(y = \frac{1}{2}x\) into \(y - 5 = -2(x - 2)\):

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

\(\frac{1}{2}x - 5 = -2x + 4\)

\(\frac{1}{2}x + 2x = 9\)

\(\frac{5}{2}x = 9\)

\(x = \frac{18}{5} = 3.6\)

\(y = \frac{1}{2}(3.6) = 1.8\)

Thus, \(C(3.6, 1.8)\).

6. Calculate the distance \(AC\):

\(AC = \sqrt{(3.6 - 2)^2 + (1.8 - 5)^2}\)

\(AC = \sqrt{1.6^2 + 3.2^2}\)

\(AC = \sqrt{2.56 + 10.24}\)

\(AC = \sqrt{12.8}\)

\(AC \approx 3.58\)

Step 1: Find the coordinates of D.

The midpoint D of AB is given by:

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

Step 2: Write an equation for CD² = 20.

The distance squared from C to D is:

\((x - 5)^2 + (y - 1)^2 = 20\)

Step 3: Find an equation for AC = BC and simplify.

The distances squared are:

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

Expanding both sides:

\(x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 - 18x + 81 + y^2 + 2y + 1\)

Simplifying gives:

\(-2x - 6y + 10 = -18x + 2y + 82\)

\(16x - 8y = 72\)

\(y = 2x - 9\)

Step 4: Find the possible coordinates of C.

Substitute \(y = 2x - 9\) into \((x - 5)^2 + (y - 1)^2 = 20\):

\((x - 5)^2 + (2x - 10)^2 = 20\)

\((x - 5)^2 + (2x - 10)^2 = 20\)

\(5x^2 - 50x + 105 = 0\)

Solving gives \(x = 3\) or \(x = 7\).

For \(x = 3\), \(y = 2(3) - 9 = -3\).

For \(x = 7\), \(y = 2(7) - 9 = 5\).

Thus, the possible coordinates of C are (3, -3) and (7, 5).

To find the coordinates of \(Q\), substitute \(y = x + 4\) into the curve equation:

\(x + 4 = 2x^2 - 4x + 1\)

Simplify to:

\(2x^2 - 5x - 3 = 0\)

Factorize:

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

Solutions are \(x = -\frac{1}{2}\) and \(x = 3\). Since \(P = (3, 7)\), \(Q = \left( -\frac{1}{2}, \frac{3}{2} \right)\).

For part (iii), find the mid-point of \(AP\):

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

Gradient of line \(Q\) to mid-point:

\(\frac{3/2 - 3}{-1/2 - 2} = \frac{-3/2}{-5/2} = \frac{1}{5}\)

Equation of line:

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

(i) To find the midpoint of AB, use the midpoint formula:

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

Substitute \(A(2, 5)\) and \(B(5, -1)\):

\(\left( \frac{2 + 5}{2}, \frac{5 + (-1)}{2} \right) = \left( \frac{7}{2}, 2 \right)\)

(ii) First, find the slope of AB:

\(m = \frac{-1 - 5}{5 - 2} = \frac{-6}{3} = -2\)

The slope of the line perpendicular to AB is the negative reciprocal:

\(m_{\perp} = \frac{1}{2}\)

Using point C(8, 6) and the point-slope form:

\(y - 6 = \frac{1}{2}(x - 8)\)

Simplify to get the equation in the form \(ax + by + c = 0\):

\(y - 6 = \frac{1}{2}x - 4\)

\(2y - 12 = x - 8\)

\(x - 2y + 4 = 0\)

First, solve the system of equations:

1. \(y^2 = 12x\)

2. \(3y = 4x + 6\)

From equation (2), express \(x\) in terms of \(y\):

\(x = \frac{3y - 6}{4}\)

Substitute \(x\) in equation (1):

\(y^2 = 12 \left( \frac{3y - 6}{4} \right)\)

Simplify:

\(y^2 = 9y - 18\)

Rearrange to form a quadratic equation:

\(y^2 - 9y + 18 = 0\)

Factor the quadratic:

\((y - 3)(y - 6) = 0\)

Thus, \(y = 3\) or \(y = 6\).

Find corresponding \(x\) values:

For \(y = 3\):

\(x = \frac{3(3) - 6}{4} = \frac{9 - 6}{4} = \frac{3}{4}\)

For \(y = 6\):

\(x = \frac{3(6) - 6}{4} = \frac{18 - 6}{4} = 3\)

The points of intersection are \(\left( \frac{3}{4}, 3 \right)\) and \((3, 6)\).

Calculate the distance between the points:

\(\text{Distance} = \sqrt{\left( 3 - \frac{3}{4} \right)^2 + (6 - 3)^2}\)

\(= \sqrt{\left( \frac{9}{4} \right)^2 + 3^2}\)

\(= \sqrt{\frac{81}{16} + 9}\)

\(= \sqrt{\frac{81}{16} + \frac{144}{16}}\)

\(= \sqrt{\frac{225}{16}}\)

\(= \frac{15}{4}\)

\(= 3.75\)

To solve the problem, we follow these steps:

1. Find the midpoint \(M\) of \(AB\):

\(M = \left( \frac{2 + 8}{2}, \frac{6 + 10}{2} \right) = (5, 8)\)

2. Calculate the gradient of \(AB\):

\(\text{Gradient of } AB = \frac{10 - 6}{8 - 2} = \frac{2}{3}\)

3. Find the gradient of the perpendicular bisector:

\(\text{Perpendicular gradient} = -\frac{3}{2}\)

4. Use the point-slope form to find the equation of the perpendicular bisector:

\(y - 8 = -\frac{3}{2}(x - 5)\)

Expanding and rearranging gives:

\(2y + 3x = 31\)

5. Find the equation of line \(BC\):

\(y = 5(x - 6) \Rightarrow y = 5x - 30\)

6. Solve the system of equations:

\(2y + 3x = 31\)

\(y = 5x - 30\)

Substitute \(y = 5x - 30\) into \(2y + 3x = 31\):

\(2(5x - 30) + 3x = 31\)

\(10x - 60 + 3x = 31\)

\(13x = 91\)

\(x = 7\)

Substitute \(x = 7\) back into \(y = 5x - 30\):

\(y = 5(7) - 30 = 5\)

Thus, the coordinates of \(D\) are \((7, 5)\).

(i) To find the intersection points \(A\) and \(B\), solve the equations simultaneously:

\(x^2 - 4x + 7 = 9 - 3x\)

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

Solving this quadratic equation gives \(x = 2\) or \(x = -1\).

For \(x = 2\), \(y = 3\). For \(x = -1\), \(y = 12\).

Thus, \(A = (2, 3)\) and \(B = (-1, 12)\).

The midpoint \(M\) is:

\(M = \left( \frac{2 + (-1)}{2}, \frac{3 + 12}{2} \right) = \left( \frac{1}{2}, \frac{15}{2} \right)\)

(ii) The derivative of the curve is:

\(\frac{dy}{dx} = 2x - 4\)

Set \(\frac{dy}{dx} = -3\) (slope of the line):

\(2x - 4 = -3\)

\(2x = 1\)

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

Substitute \(x = \frac{1}{2}\) into the curve equation:

\(y = \left( \frac{1}{2} \right)^2 - 4 \left( \frac{1}{2} \right) + 7 = \frac{1}{4} - 2 + 7 = \frac{5}{4}\)

So, \(Q = \left( \frac{1}{2}, \frac{5}{4} \right)\).

(iii) The distance \(MQ\) is:

\(MQ = \sqrt{\left( \frac{1}{2} - \frac{1}{2} \right)^2 + \left( \frac{7}{2} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{0 + \left( \frac{14}{4} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{\left( \frac{9}{4} \right)^2} = \frac{9}{4}\)