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Straight lines questions
One point and one gradient determine a straight line, and everything else follows from them. Gradients also settle whether two lines are parallel or perpendicular. Straight-line models then describe quantities changing at a constant rate, where the gradient gives the rate and the intercept gives the initial value.
7 original questions · 25 marks · the straight lines notes · Coordinate geometry
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
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.
Worked 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. Marks go to the gradient, the equation and the demanded integer form, so a correct y = −2x + 5 left unrearranged still loses one. 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.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.
Worked 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. Flip the fraction, change the sign, and check the product really is −1.Find the equation of the perpendicular bisector of the segment joining (1, 2) and (7, 10).
Worked 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. Both ingredients carry a mark. A bisector that uses one endpoint instead of the midpoint scores the gradient and nothing else.Find the equation of the line through (4, −1) perpendicular to the line 2x + 3y − 6 = 0.
Worked 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.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.
Worked 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. Interpretation marks want units and a direction, so “gradient is −2.5” on its own is not enough.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.
Worked 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. The foot of the perpendicular has to be found first, and that extra stage is what makes this a six-mark question rather than a one-liner.A vertical line and a horizontal line are perpendicular, yet the rule m1m2 = −1 cannot be applied to them. Explain why not.
Worked 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. Writing m = ∞ and claiming ∞ × 0 = −1 earns nothing.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise straight lines one question at a time
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