Questions on straight lines

Question 1

Find the equation of the line through (1, 3) and (4, −3), giving your answer in the form ax + by + c = 0 with integer coefficients.

Accepted Answer

The gradient is m = (−3 − 3)/(4 − 1) = −2. Point-gradient form with (1, 3) gives y − 3 = −2(x − 1), which tidies to 2x + y − 5 = 0. The answer is asked for in integer form, so rearrange y = −2x + 5 rather than leaving it. M1 for the gradient, A1 for the equation, A1 for the integer form. Substituting the second point, 8 − 3 − 5 = 0, is a one-line check that catches most slips.

Question 2

Write down the gradient of any line perpendicular to a line of gradient 2/3, and of any line perpendicular to a line of gradient −4.

Accepted Answer

Perpendicular gradients multiply to −1, so each is the negative reciprocal. That gives −3/2 for the first and 1/4 for the second. B1 B1 for the two gradients.

Question 3

Write down the gradient of a line parallel to y = 3x − 7, and find the equation of the line through (2, 1) parallel to it, in the form y = mx + c.

Accepted Answer

A parallel line shares the same gradient, 3. B1. Through (2, 1): y − 1 = 3(x − 2), which gives y = 3x − 5. M1 for the equation through the point, A1 for the simplified form.

Question 4

Show that the points (1, 1), (3, 5) and (6, 11) are collinear.

Accepted Answer

The gradient from (1, 1) to (3, 5) is (5 − 1)/(3 − 1) = 2. The gradient from (3, 5) to (6, 11) is (11 − 5)/(6 − 3) = 2. M1 for both gradients, A1 for both equal to 2. The two gradients agree, and the middle point (3, 5) is shared by both, so all three points lie on a single line: they are collinear. B1 for the conclusion.

Question 5

Find the equation of the perpendicular bisector of the segment joining (1, 2) and (7, 10).

Accepted Answer

The midpoint is (4, 6) and the segment's gradient is 8/6 = 4/3, so the bisector's gradient is −3/4. Through (4, 6): y − 6 = −(3/4)(x − 4), which clears to 3x + 4y − 36 = 0. B1 for the midpoint, M1 for the perpendicular gradient, M1 for the equation through (4, 6), A1 for 3x + 4y − 36 = 0.

Question 6

Find the equation of the line through (4, −1) perpendicular to the line 2x + 3y − 6 = 0.

Accepted Answer

Rearranging the given line, y = 2 − (2/3)x, so its gradient is −2/3 and the perpendicular gradient is 3/2. Through (4, −1): y + 1 = (3/2)(x − 4), which doubles to 2y + 2 = 3x − 12, giving 3x − 2y − 14 = 0. M1 for the gradient of the given line, A1 for the perpendicular gradient 3/2, A1 for 3x − 2y − 14 = 0. Reading the gradient straight off the coefficients as −a/b saves a line, provided the sign is quoted correctly.

Question 7

The height in centimetres of a burning candle after t hours is modelled by H = 30 − 2.5t. Interpret the numbers 30 and 2.5, find when the height is 10 cm, and state when the model must stop applying.

Accepted Answer

The intercept 30 is the starting height in centimetres. The gradient −2.5 says the candle burns down 2.5 cm every hour. Setting H = 10 gives 2.5t = 20, so t = 8 hours. At t = 12 the model gives H = 0, and past that a negative height, so it cannot apply beyond 12 hours. B1 for the starting height, B1 for the rate with its direction, M1 for setting H = 10, A1 for t = 8, B1 for the model failing beyond 12 hours. An interpretation needs units and a direction, so “gradient is −2.5” on its own is not enough.

Question 8

Find the coordinates of the point of intersection of the lines 2x + y = 7 and x − y = −1.

Accepted Answer

Adding the two equations eliminates y: (2x + y) + (x − y) = 7 + (−1), so 3x = 6 and x = 2. M1 for eliminating y, A1 for x = 2. Substituting into x − y = −1 gives 2 − y = −1, so y = 3. A1 for y = 3. The lines meet at (2, 3).

Question 9

Find the equation of the line through (5, 2) parallel to 4x − 3y + 1 = 0, giving your answer in the form ax + by + c = 0.

Accepted Answer

Rearranging, y = (4x + 1)/3, so the gradient is 4/3. A parallel line shares this gradient. M1 for the gradient, A1 for 4/3. Through (5, 2): y − 2 = (4/3)(x − 5), which triples to 3y − 6 = 4x − 20, giving 4x − 3y − 14 = 0. M1 for the equation through the point, A1 for the demanded form.

Question 10

A company models the total cost £C of printing n leaflets with a straight line. Printing 200 leaflets costs £90, and printing 500 leaflets costs £180. Find the equation of the model in the form C = mn + c, and interpret the values of m and c.

Accepted Answer

The gradient is m = (180 − 90)/(500 − 200) = 90/300 = 0.3. M1 for the gradient, A1 for 0.3. Using the first point, 90 = 0.3(200) + c, so c = 90 − 60 = 30. The model is C = 0.3n + 30. A1 for c = 30, completing the equation. The gradient 0.3 is the cost, in pounds, of printing one more leaflet, and the intercept 30 is the fixed cost, in pounds, charged however many leaflets are printed. B1 for both interpretations, with units.

Question 11

The lines x + y = 2 and x + y = 10 are parallel. Find the shortest distance between them.

Accepted Answer

Take a point on the first line, say (2, 0), since 2 + 0 = 2. The perpendicular to both lines has gradient 1, the negative reciprocal of −1. Through (2, 0): y = x − 2. M1 for the point and the perpendicular gradient. Substituting into the second line: x + (x − 2) = 10, so 2x = 12 and x = 6, y = 4, meeting it at (6, 4). M1 for the intersection. The distance is √((6 − 2)² + (4 − 0)²) = √32 = 4√2. M1 for the distance formula, A1 for 4√2. Any other starting point on the first line gives the same distance, since parallel lines stay a constant distance apart.

Question 12

The line l₁ has equation 3x + 4y − 12 = 0. The line l₂ passes through the origin and is perpendicular to l₁. Find the coordinates of the point where the two lines meet, and hence find the shortest distance from the origin to l₁.

Accepted Answer

l₁ has gradient −3/4, so l₂ has gradient 4/3 and, through the origin, equation y = (4/3)x. Substituting into 3x + 4y = 12 gives 3x + (16/3)x = 12, so 25x = 36 and x = 36/25, y = 48/25. The meeting point is (36/25, 48/25). Its distance from the origin is √(36² + 48²)/25 = 60/25 = 12/5 = 2.4. M1 for the perpendicular gradient, A1 for y = (4/3)x, M1 for substituting, A1 for (36/25, 48/25), M1 for the distance, A1 for 12/5. The word shortest is what forces the perpendicular; a distance measured along any other line to l₁ is longer.

Question 13

A vertical line and a horizontal line are perpendicular, yet the rule m₁m₂ = −1 cannot be applied to them. Explain why not.

Accepted Answer

A vertical line has no gradient. The run is zero, so the fraction is undefined, and the rule compares two gradients that both have to exist. The lines are perpendicular by geometry, not by passing the product test. B1 for a vertical line having no gradient, B1 for the rule needing two gradients that exist.

Question 14

Line l₁ has gradient k and line l₂ has gradient 2 − 3k. Find the value of k for which l₁ and l₂ are parallel, and the values of k for which they are perpendicular.

Accepted Answer

Parallel lines share a gradient: k = 2 − 3k, so 4k = 2 and k = 1/2. M1 for the equation, A1 for k = 1/2. Perpendicular gradients multiply to −1: k(2 − 3k) = −1, that is 2k − 3k² = −1, so 3k² − 2k − 1 = 0. M1 for this equation. Factorising, (3k + 1)(k − 1) = 0, so k = 1 or k = −1/3. M1 for factorising or solving, A1 for both values. A single linear condition gives one value of k for parallel lines, but the perpendicular condition is quadratic in k and can give two.

Question 15

The line has equation 2x + (k − 1)y = 3k + 7 for a constant k. Show that the line passes through a fixed point for every value of k, and find the coordinates of that point.

Accepted Answer

Expanding, 2x + (k − 1)y = 3k + 7 becomes 2x + ky − y − 3k − 7 = 0. M1 for expanding. Grouping the terms that carry k, k(y − 3) + (2x − y − 7) = 0. A1 for the grouped form. This holds for every value of k only if both brackets vanish together, since otherwise changing k would change a left side that has to stay 0: y − 3 = 0 and 2x − y − 7 = 0. M1 for that argument. So y = 3, and substituting gives 2x − 3 − 7 = 0, so x = 5. A1 for y = 3, A1 for x = 5. The fixed point is (5, 3), and checking directly, the left side there is 10 + 3(k − 1) = 7 + 3k, matching the right side for every k.