2025 Cambridge IGCSE Mathematics (0580) 模擬試題連答案詳解

The prime numbers between 20 and 30 are integers greater than 1 that have no positive divisors other than 1 and themselves. Checking the numbers: 21 (divisible by 3), 22 (divisible by 2), 23 (prime), 24 (divisible by 2), 25 (divisible by 5), 26 (divisible by 2), 27 (divisible by 3), 28 (divisible by 2), 29 (prime). Thus, the prime numbers are 23 and 29.

B2 for both correct and no extras. B1 for one correct and no extras, or two correct and one incorrect.

Cost of the pencils: 3 * $1.40 = $4.20. Total cost of pencils and notebook: $4.20 + $2.50 = $6.70. Change from $10: $10.00 - $6.70 = $3.30.

M1 for 10 - (3 * 1.40 + 2.50) or for total cost of 6.70. A1 for 3.30 (accept 3.3).

The first significant figure is 8 and the second is 4. The digit after 4 is 7, which is 5 or more, so we round up. This gives 0.085.

B1 for 0.085.

\(4^{-2} = \frac{1}{4^2} = \frac{1}{16}\).

B1 for 1/16.

\(\frac{7}{20} = \frac{7 \times 5}{20 \times 5} = \frac{35}{100} = 35\%\).

B1 for 35 or 35%.

Add 7 to both sides: \(5x = 25\). Divide by 5: \(x = 5\).

M1 for 5x = 25 or 18 + 7. A1 for 5.

Expanding: \(3(2x - 5) = 6x - 15\) and \(-2(x - 4) = -2x + 8\). Combining like terms: \(6x - 2x - 15 + 8 = 4x - 7\).

M1 for 6x - 15 or -2x + 8. A1 for 4x - 7.

The angles in a triangle add up to \(180^\circ\). Sum of known angles: \(48^\circ + 72^\circ = 120^\circ\). Third angle: \(180^\circ - 120^\circ = 60^\circ\).

B1 for 60.

First, order the numbers from smallest to largest: 8, 11, 12, 14, 15, 20. Since there are 6 numbers, the median is the average of the two middle numbers, which are 12 and 14. Median = \( \frac{12 + 14}{2} = 13 \).

M1 for ordering the numbers or for identifying 12 and 14 as the middle numbers.
A1 for 13.

Expand the brackets: \( 8x - 12 - 3x + 15 \). Combine like terms: \( 8x - 3x = 5x \) and \( -12 + 15 = 3 \). The simplified expression is \( 5x + 3 \).

M1 for correct expansion of at least one bracket, e.g. \( 8x - 12 \) or \( -3x + 15 \).
A1 for \( 5x + 3 \).

To find \( \frac{3}{8} \) of 120, first divide 120 by 8: \( 120 \div 8 = 15 \). Then multiply the result by 3: \( 15 \times 3 = 45 \).

M1 for \( 120 \div 8 \times 3 \) or showing a correct partial calculation.
A1 for 45.

The sum of angles in a triangle is \( 180^\circ \). The total number of parts in the ratio is \( 2 + 3 + 5 = 10 \). Each part represents \( 180^\circ \div 10 = 18^\circ \). The largest angle has 5 parts: \( 5 \times 18^\circ = 90^\circ \).

M1 for \( 180 \div (2 + 3 + 5) \) or showing a correct method to find the size of one part.
A1 for 90.

To write \( 0.0000305 \) in standard form, move the decimal point 5 places to the right to obtain \( 3.05 \). Since the decimal point was moved to the right, the index is negative: \( 3.05 \times 10^{-5} \).

M1 for \( 3.05 \times 10^k \) where \( k \neq -5 \).
A1 for \( 3.05 \times 10^{-5} \).

Find \( 15\% \) of \( \$85 \): \( 10\% \text{ of } 85 = 8.50 \), and \( 5\% \text{ of } 85 = 4.25 \). Total reduction = \( 8.50 + 4.25 = 12.75 \). Sale price = \( 85 - 12.75 = 72.25 \).

M1 for a correct method to find \( 15\% \) of 85 or \( 85\% \) of 85.
A1 for 72.25.

Area of a trapezium = \( \frac{1}{2}(a + b)h \). Substituting the given values: \( \text{Area} = \frac{1}{2}(6 + 10) \times 7 = \frac{1}{2}(16) \times 7 = 8 \times 7 = 56\text{ cm}^2 \).

M1 for substituting correctly into the trapezium area formula: \( \frac{1}{2}(6 + 10) \times 7 \).
A1 for 56.

Multiply both sides by 4: \( 3x - 5 = 28 \). Add 5 to both sides: \( 3x = 33 \). Divide by 3: \( x = 11 \).

M1 for isolating the numerator: \( 3x - 5 = 28 \).
A1 for 11.

First, expand the brackets:
\(12x - 8 = 28\)

Add 8 to both sides:
\(12x = 36\)

Divide by 12:
\(x = 3\)

Alternatively, divide both sides by 4 first:
\(3x - 2 = 7\)

Add 2 to both sides:
\(3x = 9\)

Divide by 3:
\(x = 3\)

M1 for \(12x - 8 = 28\) or \(3x - 2 = 7\)
A1 for \(3\)

To divide by a fraction, multiply by its reciprocal:
\(\frac{3}{5} \div \frac{9}{10} = \frac{3}{5} \times \frac{10}{9}\)

Multiply the numerators and denominators:
\(\frac{3 \times 10}{5 \times 9} = \frac{30}{45}\)

Simplify the fraction by dividing the numerator and denominator by 15:
\(\frac{30 \div 15}{45 \div 15} = \frac{2}{3}\)

M1 for \(\frac{3}{5} \times \frac{10}{9}\) or \(\frac{30}{45}\) or equivalent unsimplified fraction
A1 for \(\frac{2}{3}\)

Let \(l\) be the length of the rectangle.
The perimeter of a rectangle is given by:
\(2(l + w) = 32\)

Substitute the given width \(w = 6\):
\(2(l + 6) = 32\)

Divide by 2:
\(l + 6 = 16\)

Subtract 6:
\(l = 10\text{ cm}\)

Now, calculate the area:
\(\text{Area} = \text{length} \times \text{width} = 10 \times 6 = 60\text{ cm}^2\)

M1 for finding the length, \(10\text{ cm}\), or for a correct expression for the area such as \(\left(\frac{32 - 2 \times 6}{2}\right) \times 6\)
A1 for \(60\)

First, arrange the temperatures in ascending order:
\(-3\), \(-3\), \(-1\), \(2\), \(5\)

The median is the middle value in the ordered list.
Since there are 5 values, the middle (3rd) value is \(-1\).

M1 for ordering the numbers: \(-3, -3, -1, 2, 5\) (with at least 4 correct numbers in correct relative positions)
A1 for \(-1\)

Multiply the second equation by 2:
\(8x - 2y = 22\)

Add this equation to the first equation:
\((3x + 2y) + (8x - 2y) = 11 + 22\)
\(11x = 33\)
\(x = 3\)

Substitute \(x = 3\) back into the second equation:
\(4(3) - y = 11\)
\(12 - y = 11\)
\(y = 1\)

M1 for multiplying the second equation by 2 to get \(8x - 2y = 22\) (or alternative valid method to equate coefficients)
M1 for adding equations to eliminate \(y\) to get \(11x = 33\) (or alternative correct elimination of one variable)
A1 for \(x = 3\)
A1 for \(y = 1\)

Area of the rectangular lawn:
\(\text{Area}_{\text{rectangle}} = 12 \times 8 = 96\text{ m}^2\)

Area of the circular flowerbed:
\(\text{Area}_{\text{circle}} = \pi \times 3^2 = 9\pi\text{ m}^2\)

Remaining area of the lawn:
\(\text{Remaining Area} = 96 - 9\pi\text{ m}^2\)

M1 for area of the rectangle = \(12 \times 8\) or \(96\)
M1 for area of the circle = \(\pi \times 3^2\) or \(9\pi\)
M1 for subtracting their area of the circle from their area of the rectangle
A1 for \(96 - 9\pi\)

(a) The sequence increases by \(4\) each time:
\(15 + 4 = 19\)
\(19 + 4 = 23\)
The next two terms are \(19\) and \(23\).

(b) Since the common difference is \(4\), the \(n\)th term is of the form \(4n + c\).
Using the first term where \(n = 1\):
\(4(1) + c = 3\)
\(c = -1\)
So, the \(n\)th term is \(4n - 1\).

B1 for 19
B1 for 23
M1 for \(4n + c\) (where \(c\) is any constant)
A1 for \(4n - 1\)

Find the sum of the products of Goals scored \(\times\) Frequency:
\((0 \times 3) + (1 \times 5) + (2 \times 6) + (3 \times 4) + (4 \times 2)\)
\(= 0 + 5 + 12 + 12 + 8\)
\(= 37\)

Total number of matches (sum of frequencies):
\(3 + 5 + 6 + 4 + 2 = 20\)

Mean number of goals scored per match:
\(\text{Mean} = \frac{37}{20} = 1.85\)

M1 for attempting to calculate the sum of products of goals and frequency (at least 3 correct products shown)
A1 for 37
M1 for dividing their total goals by 20
A1 for 1.85

First write out the elements of each set:
\(\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\)
\(A = \{2, 3, 5, 7\}\)
\(B = \{1, 3, 5, 7, 9\}\)

(a) \(A \cap B\) is the set of elements in both \(A\) and \(B\):
\(A \cap B = \{3, 5, 7\}\)

(b) \(A \cup B\) is the set of elements in \(A\) or \(B\) (or both):
\(A \cup B = \{1, 2, 3, 5, 7, 9\}\)
\((A \cup B)'\) consists of elements in \(\mathcal{E}\) that are not in \(A \cup B\):
\((A \cup B)' = \{4, 6, 8, 10\}\)

B2 for (a) 3, 5, 7 (B1 for 2 correct elements, or 3 correct elements and 1 extra)
B2 for (b) 4, 6, 8, 10 (B1 for 2 or 3 correct elements, or 4 correct and 1 extra)

(a) Find the highest common factor of \(6x^2y\) and \(-9xy^2\), which is \(3xy\):
\(6x^2y - 9xy^2 = 3xy(2x - 3y)\)

(b) Expand the brackets using the distributive property:
\((x + 4)(x - 7) = x^2 - 7x + 4x - 28\)
Simplify by combining the like terms:
\(x^2 - 3x - 28\)

M1 for \(3(2x^2y - 3xy^2)\) or \(xy(6x - 9y)\) or \(3xy(\text{two term algebraic expression})\)
A1 for \(3xy(2x - 3y)\)
M1 for 3 out of 4 terms correct in expansion: \(x^2 - 7x + 4x - 28\)
A1 for \(x^2 - 3x - 28\)

Round each number in the calculation to 1 significant figure:
\(19.8 \approx 20\)
\(5.03 \approx 5\)
\(0.197 \approx 0.2\)

Substitute the rounded values into the calculation:
$$\frac{20 \times 5}{0.2} = \frac{100}{0.2} = 500$$

B1 for \(19.8\) rounded to \(20\)
B1 for \(5.03\) rounded to \(5\)
B1 for \(0.197\) rounded to \(0.2\)
B1 for 500 (dependent on all roundings being to 1 s.f.)

We can write \( \frac{1}{125} \) as a power of 5:
\( \frac{1}{125} = 5^{-3} \).

Therefore, we have:
\( 5^{2x - 1} = 5^{-3} \).

Equating the indices:
\( 2x - 1 = -3 \)
\( 2x = -2 \)
\( x = -1 \).

M1 for \( 125 = 5^3 \) or \( 5^{-3} \) or \( 2x - 1 = -3 \) (or equivalent method)
A1 for \( -1 \)

Rearranging the given line equation into slope-intercept form:
\( y = -3x + 7 \).

The gradient is \( -3 \). Since the required line is parallel, its gradient is also \( -3 \).

Using the point \( (2, -3) \) in the line equation \( y = mx + c \):
\( -3 = -3(2) + c \)
\( -3 = -6 + c \)
\( c = 3 \).

Thus, the equation is \( y = -3x + 3 \).

M1 for gradient \( = -3 \) or for substituting \( (2, -3) \) into \( y = -3x + c \)
A1 for \( y = -3x + 3 \)

Divide the numerical coefficients and subtract the indices:
\( \frac{1.2}{3} = 0.4 \)
\( 10^4 \div 10^{-3} = 10^{4 - (-3)} = 10^7 \)

This gives:
\( 0.4 \times 10^7 \).

Converting to standard form:
\( 4 \times 10^6 \).

M1 for \( 0.4 \times 10^7 \) or showing division of the coefficients and subtraction of exponents
A1 for \( 4 \times 10^6 \)

First, factor out the highest common factor of 2:
\( 50x^2 - 8 = 2(25x^2 - 4) \).

Next, recognise \( 25x^2 - 4 \) as a difference of two squares:
\( 25x^2 - 4 = (5x - 2)(5x + 2) \).

Therefore, the completely factorised expression is:
\( 2(5x - 2)(5x + 2) \).

M1 for \( 2(25x^2 - 4) \) or for identifying difference of two squares structure, e.g., \( (5x\sqrt{2} - 2\sqrt{2})(5x\sqrt{2} + 2\sqrt{2}) \)
A1 for \( 2(5x - 2)(5x + 2) \) or equivalent, e.g., \( (10x - 4)(5x + 2) \)

Convert the mixed numbers to improper fractions:
\( 3\frac{1}{4} = \frac{13}{4} \)
\( 1\frac{2}{3} = \frac{5}{3} \)

Find a common denominator of 12:
\( \frac{13 \times 3}{4 \times 3} - \frac{5 \times 4}{3 \times 4} = \frac{39}{12} - \frac{20}{12} \)
\( = \frac{19}{12} \)

Convert back to a mixed number:
\( \frac{19}{12} = 1\frac{7}{12} \).

M1 for correct conversions to improper fractions with common denominators, e.g. \( \frac{39}{12} - \frac{20}{12} \)
A1 for \( 1\frac{7}{12} \)

The total number of marbles is \( 5 + 3 = 8 \).

The probability of picking a blue marble first is \( \frac{3}{8} \).

Since there is no replacement, 2 blue marbles are left out of 7 total marbles.
The probability of picking a blue marble second is \( \frac{2}{7} \).

The probability that both are blue is:
\( \frac{3}{8} \times \frac{2}{7} = \frac{6}{56} = \frac{3}{28} \).

M1 for \( \frac{3}{8} \times \frac{2}{7} \)
A1 for \( \frac{3}{28} \) or equivalent fraction

Let the original price be \( x \).
A 15% reduction means the sale price is 85% of the original price.
\( 0.85x = 68 \)
\( x = \frac{68}{0.85} = \frac{6800}{85} \)

Dividing both numerator and denominator by 17:
\( \frac{68}{17} = 4 \)
\( \frac{85}{17} = 5 \)

Therefore:
\( x = \frac{400}{5} = 80 \).

M1 for \( 68 \div 0.85 \) or for equating \( 85\% = 68 \)
A1 for \( 80 \)

The sequence is arithmetic with first term \( a = 7 \) and common difference \( d = 4 - 7 = -3 \).

The formula for the \( n \)th term is:
\( a + (n - 1)d \)
\( = 7 + (n - 1)(-3) \)
\( = 7 - 3n + 3 \)
\( = 10 - 3n \).

M1 for any expression of the form \( -3n + k \) where \( k \) is a constant, or for \( 7 + (n - 1)(-3) \)
A1 for \( 10 - 3n \) or equivalent

First, rewrite the negative exponent as a reciprocal:
\(27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}}\)

Next, evaluate the fractional power:
\(27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9\)

Therefore, \(27^{-\frac{2}{3}} = \frac{1}{9}\).

M1 for \((\sqrt[3]{27})^{-2}\) or \(\left(\frac{1}{27}\right)^{\frac{2}{3}}\) or \(\frac{1}{9}\) seen in working
A1 for \(\frac{1}{9}\) or equivalent fraction

We need two numbers that multiply to \(3 \times 8 = 24\) and add to \(-10\). These numbers are \(-6\) and \(-4\).

Rewrite the middle term:
\(3x^2 - 6x - 4x + 8\)

Factor by grouping:
\(3x(x - 2) - 4(x - 2)\)

Factor out the common bracket:
\((3x - 4)(x - 2)\)

M1 for a correct partial factorisation or split of the middle term, e.g., \(3x(x - 2) - 4(x - 2)\) or \((3x + a)(x + b)\) where \(ab=8\) or \(3b+a=-10\)
A1 for \((3x - 4)(x - 2)\) or equivalent

Use the midpoint formula \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\):

\(x\)-coordinate: \(\frac{-3 + 5}{2} = \frac{2}{2} = 1\)
\(y\)-coordinate: \(\frac{8 + (-2)}{2} = \frac{6}{2} = 3\)

So, the midpoint is \((1, 3)\).

M1 for a correct substitution into the midpoint formula for at least one coordinate, e.g., \(\frac{-3+5}{2}\) or \(\frac{8+(-2)}{2}\)
A1 for \((1, 3)\)

Convert both numbers to the same power of 10:
\(4.2 \times 10^3 = 4200\)
\(1.5 \times 10^2 = 150\)

Add the values:
\(4200 + 150 = 4350\)

Convert back to standard form:
\(4350 = 4.35 \times 10^3\)

M1 for converting to ordinary numbers: \(4200 + 150\) or equal powers of 10: \(42 \times 10^2 + 1.5 \times 10^2\) or \(4.2 \times 10^3 + 0.15 \times 10^3\)
A1 for \(4.35 \times 10^3\)

Find the total number of parts:
\(2 + 3 + 7 = 12\)

Find the value of one part:
\(\frac{180}{12} = 15\)

Find the largest share (which has 7 parts):
\(7 \times 15 = 105\)

M1 for \(\frac{180}{2+3+7}\) or \(180 \div 12\)
A1 for 105

Find the total number of marbles:
\(6 + 4 + 5 = 15\)

The number of marbles that are not blue (red or green) is:
\(6 + 5 = 11\)

Therefore, the probability of selecting a marble that is not blue is \(\frac{11}{15}\).

M1 for finding total number of marbles (15) and number of non-blue marbles (11), or for \(1 - \frac{4}{15}\)
A1 for \(\frac{11}{15}\) or equivalent decimal/percentage

Find the radius of the semicircle:
\(r = \frac{14}{2} = 7\text{ cm}\)

Find the area of the full circle:
\(A_{\text{circle}} = \pi r^2 = \pi \times 7^2 = 49\pi\)

Find the area of the semicircle:
\(A_{\text{semicircle}} = \frac{49\pi}{2} = 24.5\pi\text{ cm}^2\)

M1 for \(\frac{1}{2} \times \pi \times 7^2\)
A1 for \(24.5\pi\) or \(\frac{49}{2}\pi\) or \(49\pi/2\)

Total students = 40.
Students studying Biology, Chemistry, or both = 40 - 5 = 35.
Using the set formula: n(B \cup C) = n(B) + n(C) - n(B \cap C)
35 = 23 + 18 - x
35 = 41 - x
x = 6.

For part (b):
n(B \cap C') = n(B) - n(B \cap C) = 23 - 6 = 17.

(a) M1 for 23 + 18 - x + 5 = 40 oe
A1 for x = 6
(b) M1 for 23 - [their x]
A1 for 17

(a) 12x^2 - 27y^2 = 3(4x^2 - 9y^2)
Using the difference of two squares:
4x^2 - 9y^2 = (2x - 3y)(2x + 3y)
So, the factorised expression is 3(2x - 3y)(2x + 3y).

(b) Group terms in 6ab - 8b + 9ac - 12c:
2b(3a - 4) + 3c(3a - 4)
Factor out the common bracket (3a - 4):
(2b + 3c)(3a - 4).

(a) B1 for 3(4x^2 - 9y^2) oe
B1 for 3(2x - 3y)(2x + 3y)
(b) M1 for grouping, e.g., 2b(3a - 4) or 3c(3a - 4)
A1 for one correct partial factorisation
A1 for (2b + 3c)(3a - 4) oe

(a) Scale factor of lengths, k = 18/12 = 3/2 = 1.5.
Ratio of areas = k^2 = (3/2)^2 = 9/4.
Area of larger container = 80 * 9/4 = 180 cm^2.

(b) Ratio of volumes = k^3 = (3/2)^3 = 27/8.
Capacity of larger container = 1.35 litres = 1350 ml.
Capacity of smaller container = 1350 / (27/8) = 1350 * 8 / 27 = 50 * 8 = 400 ml.

(a) M1 for 80 * (18/12)^2 oe
A1 for 180
(b) M1 for 1.35 litres = 1350 ml
M1 for 1350 / (18/12)^3 oe
A1 for 400

(a) Let sector area A = (\theta/360) * \pi * r^2 = 15\pi, and arc length L = (\theta/360) * 2 * \pi * r = 3\pi.
Using the formula A = 0.5 * L * r:
15\pi = 0.5 * (3\pi) * r
15 = 1.5 * r
r = 10 cm.

(b) Using L = (\theta/360) * 2 * \pi * r:
3\pi = (\theta/360) * 20\pi
3 = 20\theta / 360
3 = \theta / 18
\theta = 54.

(c) Perimeter = Arc length + 2r = 3\pi + 2(10) = 3\pi + 20 cm.

(a) M1 for 15\pi = 0.5 * (3\pi) * r oe
A1 for completing the show that to get r = 10
(b) B1 for 54
(c) M1 for 3\pi + 2 * 10 oe
A1 for 3\pi + 20

Substitute y = 2x - 3 into x^2 + y^2 = 13:
x^2 + (2x - 3)^2 = 13
x^2 + 4x^2 - 12x + 9 = 13
5x^2 - 12x - 4 = 0
Factorise the quadratic:
(5x + 2)(x - 2) = 0
So, x = 2 or x = -0.4.

When x = 2:
y = 2(2) - 3 = 1.

When x = -0.4:
y = 2(-0.4) - 3 = -3.8.

The solutions are: x = 2, y = 1 and x = -0.4, y = -3.8.

M1 for substitute y = 2x - 3 into second equation
M1 for expand and simplify to 5x^2 - 12x - 4 = 0
M1 for factorise or use formula to solve their quadratic
A1 for x = 2, x = -0.4 oe
A1 for y = 1, y = -3.8 oe corresponding to correct x

Terms: 3, 10, 21, 36, 55
First differences: 7, 11, 15, 19
Second differences: 4, 4, 4
(a) The next first difference is 19 + 4 = 23.
The next term is 55 + 23 = 78.

(b) The second difference is constant at 4, so the quadratic term is an^2, where a = 4/2 = 2.
Subtract 2n^2 from terms:
3 - 2(1) = 1
10 - 2(4) = 2
21 - 2(9) = 3
36 - 2(16) = 4
55 - 2(25) = 5
The remaining sequence is 1, 2, 3, 4, 5, which is n.
So, the nth term is 2n^2 + n.

(a) B1 for 78
(b) M1 for finding first differences (7, 11, 15, 19) and second differences (4)
M1 for identifying the an^2 term where a = 2
M1 for subtracting 2n^2 from terms to find the linear part n (or using simultaneous equations)
A1 for 2n^2 + n

4\sin^2(x) = 3
\sin^2(x) = 3/4
\sin(x) = \pm \sqrt{3}/2

If \sin(x) = \sqrt{3}/2:
x = 60^\circ or x = 180^\circ - 60^\circ = 120^\circ.

If \sin(x) = -\sqrt{3}/2:
x = 180^\circ + 60^\circ = 240^\circ or x = 360^\circ - 60^\circ = 300^\circ.

Solutions: 60^\circ, 120^\circ, 240^\circ, 300^\circ.

M1 for \sin^2(x) = 3/4
M1 for \sin(x) = \pm\sqrt{3}/2
A1 for any two correct angles
A1 for the remaining two correct angles
B1 for all four correct solutions and no extras in range

(a) Midpoint of AB = ((2+6)/2, (7-1)/2) = (4, 3).

(b) Gradient of AB = (-1 - 7)/(6 - 2) = -8/4 = -2.
Gradient of the perpendicular line = -1 / (-2) = 0.5.
Equation of perpendicular bisector through (4, 3) with gradient 0.5:
y - 3 = 0.5(x - 4)
y - 3 = 0.5x - 2
y = 0.5x + 1.

(a) M1 for (2+6)/2 or (7-1)/2 oe
A1 for (4, 3)
(b) M1 for finding gradient of AB = -2
M1 for perpendicular gradient = 0.5
A1 for y = 0.5x + 1 oe

First, find the midpoint, $M$, of the line segment $PQ$: $M = \left(\frac{-1 + 5}{2}, \frac{8 + (-2)}{2}\right) = (2, 3)$. Next, find the gradient, $m$, of the line segment $PQ$: $m = \frac{-2 - 8}{5 - (-1)} = \frac{-10}{6} = -\frac{5}{3}$. The gradient of the perpendicular bisector, $m_{\perp}$, is the negative reciprocal of $m$: $m_{\perp} = -\frac{1}{-\frac{5}{3}} = \frac{3}{5}$. Now, use the point-slope form with the midpoint $(2, 3)$ and gradient $m_{\perp} = \frac{3}{5}$ to find the equation of the perpendicular bisector: $y - 3 = \frac{3}{5}(x - 2) \Rightarrow y = \frac{3}{5}x - \frac{6}{5} + 3 \Rightarrow y = \frac{3}{5}x + \frac{9}{5}$.

M1 for finding the midpoint $(2, 3)$ of $PQ$ or showing a correct method.
M1 for finding the gradient of $PQ$ as $-\frac{5}{3}$ or $-\frac{10}{6}$.
M1 for gradient of perpendicular line $= \frac{3}{5}$ (FT their gradient of $PQ$).
M1 for substituting their midpoint and their perpendicular gradient into a linear equation form, e.g. $y - y_1 = m(x - x_1)$.
A1 for $y = \frac{3}{5}x + \frac{9}{5}$ oe (such as $y = 0.6x + 1.8$).