Quadratic graphs and coordinate geometry · applications

Applications: Quadratic Graphs and Coordinate Geometry

10 question types · Secondary 3 · each worked step by step with a figure that follows the steps

O-Level · SAT · GCSE Higher

01

A Water Jet from a Nozzle on a Pool Wall: Where It Lands and How High It Rises

methodFind Where the Curve Meets the x-Axis by Factorizing, Then Use the Axis of Symmetry Halfway Between the Two Roots to Find the Turning Point

A nozzle on the wall of a pool sends out a jet of water. At a horizontal distance of x m from the wall the jet is y m above the water, where y = −14x2 + x + 3. (a) How far from the wall does the jet land on the water? (b) Find the greatest height of the jet above the water, and the distance from the wall at which the jet reaches it.

24−2246meters from the wall, xheight (m), ynozzlethe jet lands where y = 0−1/4 x2+ x + 3 = 0, so x2− 4x − 12 = 0
The jet lands where its height is zero. Put y = 0 and multiply both sides by −4: x2 − 4x − 12 = 0.
The jet lands where its height is zero, so put y = 0: −14x2 + x + 3 = 0. Multiply both sides by −4 to clear the fraction and make the x2 term positive: x2 − 4x − 12 = 0.
step 1 of 6

The graph of a quadratic is a parabola, and a parabola is symmetrical about a vertical line through its turning point. The jet lands where the curve meets the x-axis, so that part is a quadratic equation. The axis of symmetry is halfway between the two roots, even when one of the roots has no meaning for the jet, and the greatest height is the value of y on that line.

  1. The jet lands where its height is zero, so put y = 0: −14x2 + x + 3 = 0. Multiply both sides by −4 to clear the fraction and make the x2 term positive: x2 − 4x − 12 = 0.
  2. Factorize. Two numbers that multiply to −12 and add to −4 are −6 and 2, so (x − 6)(x + 2) = 0 and x = 6 or x = −2.
  3. (a) The root x = −2 is behind the wall, where there is no jet, so it is rejected. The jet lands 6 m from the wall. Check: −14 × 36 + 6 + 3 = −9 + 9 = 0.
  4. The axis of symmetry is halfway between the two roots of the curve: x = −2 + 62 = 2. The coefficient of x2 is negative, so the parabola opens downward and its turning point is the highest point.
  5. Put x = 2 into the equation: y = −14 × 4 + 2 + 3 = −1 + 5 = 4.
  6. (b) The greatest height is 4 m, reached 2 m from the wall. Check the symmetry: at x = 0 and at x = 4, which are 2 m on either side of the axis, the height is 3 m both times.

answer(a) 6 m from the wall; (b) 4 m above the water, reached 2 m from the wall

techniqueQuadratic Graphs

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Taking the greatest height to be halfway along the jet, at x = 3. The jet starts 3 m above the water, not at the water, so the axis of symmetry is halfway between the two roots −2 and 6, which is x = 2.
  • Reading the constant term 3 as the greatest height. The constant term is the height at x = 0, which is the height of the nozzle. The jet rises after it leaves the nozzle.
02

The Main Cable of a Footbridge: Its Lowest Point and the Length of a Hanger

methodFind Two Points of the Curve at the Same Height, Take the Axis of Symmetry Halfway Between Them for the Turning Point, Then Use Equal Heights at Equal Distances from the Axis

The main cable of a footbridge hangs between the tops of two towers. At a horizontal distance of x m from the left tower the cable is y m above the deck, where y = 18x2 − 2x + 10. The top of each tower is 10 m above the deck. (a) Find the distance between the towers, and the height of the lowest point of the cable above the deck. (b) A vertical hanger joins the cable to the deck 2 m from the left tower. Find the length of this hanger, and the position of the other hanger that has the same length.

5100481216meters from the left tower, xheight above the deck (m), yy = 10the cable is at the tower tops where y = 101/8 x2− 2x + 10 = 10, so x2− 16x = 0
The cable meets the tower tops where y = 10, so 18x2 − 2x = 0. Multiply both sides by 8: x2 − 16x = 0.
The cable meets the tower tops where y = 10: 18x2 − 2x + 10 = 10, so 18x2 − 2x = 0. Multiply both sides by 8: x2 − 16x = 0.
step 1 of 6

A parabola is symmetrical about the vertical line through its turning point. Any two points of the curve at the same height are the same distance from that line, so the axis of symmetry is halfway between them. Here the two tower tops are at the same height, which gives the axis without finding the roots. The curve opens upward, so the turning point is the lowest point of the cable.

  1. The cable meets the tower tops where y = 10: 18x2 − 2x + 10 = 10, so 18x2 − 2x = 0. Multiply both sides by 8: x2 − 16x = 0.
  2. Factorize: x(x − 16) = 0, so x = 0 or x = 16. The left tower is at x = 0 and the right tower is at x = 16, so the towers are 16 m apart.
  3. The axis of symmetry is halfway between the towers, at x = 8. The coefficient of x2 is positive, so the parabola opens upward and its turning point is the lowest point. Put x = 8: y = 18 × 64 − 16 + 10 = 8 − 16 + 10 = 2.
  4. (a) The towers are 16 m apart, and the lowest point of the cable is 2 m above the deck, midway between them.
  5. For the hanger, put x = 2: y = 18 × 4 − 4 + 10 = 0.5 + 6 = 6.5. The hanger is 6.5 m long.
  6. (b) The hanger at x = 2 is 6 m to the left of the axis x = 8. The point 6 m to the right of the axis is x = 14, and the cable has the same height there. Check: 18 × 196 − 28 + 10 = 24.5 − 18 = 6.5.

answer(a) 16 m apart, and the lowest point is 2 m above the deck; (b) 6.5 m, and the hanger 14 m from the left tower has the same length

techniqueQuadratic Graphs

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Solving y = 0 to find the towers. The cable never reaches the deck: 18x2 − 2x + 10 = 0 has no real roots. The towers are where the cable is at the height of the tower tops, y = 10.
  • Taking the other hanger of the same length to be at x = 4, twice as far from the tower. Equal heights are at equal distances from the axis of symmetry x = 8, so the matching hanger is at 8 + 6 = 14.
03

A Wheelchair Ramp Checked Against the Steepest Gradient the Building Code Allows

methodFind the Gradient from Two Surveyed Points as Rise over Run, Compare It with the Limit, Then Keep the Rise and Find the Run That the Limit Needs

A surveyor draws the side view of a wheelchair ramp on axes, with x m the horizontal distance from a gate post and y m the height above the pavement. The ramp is a straight line from its foot at (3, 0.1) to the door at (9, 0.7). The building code says that the gradient of a ramp must not be more than 112. (a) Find the gradient of the ramp and decide whether the code allows it. (b) The door cannot be moved. Where must the foot of the ramp be, still at a height of 0.1 m, for the gradient to be exactly 112?

0.510246810meters from the gate post, xheight above the pavement (m), yrun 6rise 0.6(3, 0.1)(9, 0.7)rise = 0.7 − 0.1 = 0.6 m, run = 9 − 3 = 6 mgradient = 0.6/6 = 0.1 = 1/10
The rise is 0.7 − 0.1 = 0.6 m and the run is 9 − 3 = 6 m, so the gradient is 0.66 = 110. The height axis is stretched so that the rise can be seen.
Find the rise and the run from the two points. The rise is 0.7 − 0.1 = 0.6 m and the run is 9 − 3 = 6 m, so the gradient is 0.66 = 0.1 = 110.
step 1 of 5

The gradient of a line through two points is the change in y divided by the change in x, the rise over the run. For a ramp it says how many meters the ramp climbs for each meter along the ground. To make a ramp less steep without changing the rise, the run must be longer, and a gradient of 112 means that the run is 12 times the rise.

  1. Find the rise and the run from the two points. The rise is 0.7 − 0.1 = 0.6 m and the run is 9 − 3 = 6 m, so the gradient is 0.66 = 0.1 = 110.
  2. (a) Compare 110 with 112. The numerators are equal and 10 is less than 12, so 110 is the larger fraction. The ramp is steeper than the code allows.
  3. The door stays at (9, 0.7) and the foot stays at a height of 0.1 m, so the rise is still 0.6 m. For a gradient of 112 the run must be 12 times the rise: 12 × 0.6 = 7.2 m.
  4. The foot is 7.2 m before the door, so its x-coordinate is 9 − 7.2 = 1.8.
  5. (b) The foot of the ramp must be at (1.8, 0.1), which is 3 − 1.8 = 1.2 m further from the door than it is now. Check: 0.67.2 = 672 = 112.

answer(a) the gradient is 110, which is more than 112, so the code does not allow it; (b) at (1.8, 0.1), which makes the run 7.2 m

techniqueGradient from Two Points · Coordinate Geometry

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing the run by the rise, which gives 10 and makes the ramp seem far too steep. The gradient is the change in y divided by the change in x: the height gained for each meter along the ground.
  • Deciding that 110 is less than 112 because 10 is less than 12. When the numerators are equal, the fraction with the smaller denominator is the larger one: a tenth of a meter is more than a twelfth of a meter.
04

A Relay Station Halfway Between Two Towns, and a Track Continued Through It

methodAverage the Two x-Coordinates and the Two y-Coordinates for the Midpoint, Then Use the Same Formula with an Unknown End When the Midpoint Is Known

On a map with a grid in kilometers, town A is at (1, 2) and town B is at (17, 14). A relay station M is built exactly halfway along the straight line from A to B. (a) Find the coordinates of M and its distance from each town. (b) A straight track runs from a farm at C(6, 12) through M and carries on for the same distance to a depot D, so that M is also the midpoint of CD. Find the coordinates of D.

481216061218km east, xkm north, yA(1, 2)B(17, 14)M(9, 8)M = ((1 + 17)/2, (2 + 14)/2)M = (9, 8)
The midpoint has the mean of the two x-coordinates and the mean of the two y-coordinates: M = (1 + 172, 2 + 142) = (9, 8).
Take the mean of the x-coordinates and the mean of the y-coordinates: M = (1 + 172, 2 + 142) = (9, 8).
step 1 of 6

The midpoint of a segment has the mean of the two x-coordinates and the mean of the two y-coordinates. The distance between two points is the hypotenuse of a right-angled triangle whose shorter sides are the difference in x and the difference in y. When the midpoint and one end are known, the same midpoint formula gives an equation for each coordinate of the other end.

  1. Take the mean of the x-coordinates and the mean of the y-coordinates: M = (1 + 172, 2 + 142) = (9, 8).
  2. From A(1, 2) to M(9, 8) the difference in x is 9 − 1 = 8 and the difference in y is 8 − 2 = 6. By Pythagoras' theorem, AM = √82 + 62 = √100 = 10.
  3. (a) The relay station is at M(9, 8), and it is 10 km from each town. Check with B: the differences are 17 − 9 = 8 and 14 − 8 = 6 again.
  4. Let the depot be D(p, q). The midpoint of C(6, 12) and D(p, q) is M(9, 8), so 6 + p2 = 9 and 12 + q2 = 8.
  5. Multiply both sides of each equation by 2: 6 + p = 18, so p = 12, and 12 + q = 16, so q = 4.
  6. (b) The depot is at D(12, 4). Check: from C to M the track goes 3 across and 4 down, and from M to D it goes 3 across and 4 down again.

answer(a) M(9, 8), which is 10 km from each town; (b) D(12, 4)

techniqueThe Midpoint of a Segment · The Distance Between Two Points

examsO-Level · GCSE Higher

Common pitfalls

  • Subtracting the coordinates and halving, which gives (8, 6). That is half of the journey from A to B, not a place on the map. The midpoint is the mean of the coordinates, so they are added before halving.
  • Taking the midpoint of C and M for the depot, which gives (7.5, 10). The station M is the midpoint, so D is on the far side of M, as far beyond it as C is before it.
05

A Delivery Drone's Straight Flight and the Nearer of Two Charging Depots

methodMake a Right-Angled Triangle from the Difference in x and the Difference in y, Use Pythagoras' Theorem for Each Distance, and Compare the Squares of Two Distances When a Root Is Not Exact

A map has a grid in kilometers. A drone flies in a straight line from its base at L(2, 2) to a customer at S(14, 7). It then flies straight to a depot to recharge. Depot P is at (8, 15) and depot Q is at (21, 15). (a) How far does the drone fly from L to S? (b) Which depot is nearer to S? The battery lasts for 25 km of flight. Is that enough for the flight from the base to the customer and on to the nearer depot?

5101505101520km east, xkm north, y125L(2, 2)S(14, 7)P(8, 15)Q(21, 15)from L to S: 14 − 2 = 12 across and 7 − 2 = 5 upthe flight is the hypotenuse of this triangle
From L to S the map goes 14 − 2 = 12 across and 7 − 2 = 5 up. The flight is the hypotenuse of this right-angled triangle.
From L(2, 2) to S(14, 7) the difference in x is 14 − 2 = 12 and the difference in y is 7 − 2 = 5. These are the two shorter sides of a right-angled triangle whose hypotenuse is the flight.
step 1 of 5

The straight distance between two points on a grid is the hypotenuse of a right-angled triangle. Its shorter sides are the difference in the x-coordinates and the difference in the y-coordinates, so the distance is √(x2 − x1)2 + (y2 − y1)2. To decide which of two distances is smaller it is enough to compare their squares, which avoids a root that is not exact.

  1. From L(2, 2) to S(14, 7) the difference in x is 14 − 2 = 12 and the difference in y is 7 − 2 = 5. These are the two shorter sides of a right-angled triangle whose hypotenuse is the flight.
  2. (a) By Pythagoras' theorem, LS = √122 + 52 = √144 + 25 = √169 = 13. The drone flies 13 km to the customer.
  3. From S(14, 7) to P(8, 15) the differences are 8 − 14 = −6 and 15 − 7 = 8, so SP2 = (−6)2 + 82 = 36 + 64 = 100 and SP = 10 km.
  4. From S(14, 7) to Q(21, 15) the differences are 21 − 14 = 7 and 15 − 7 = 8, so SQ2 = 72 + 82 = 49 + 64 = 113. Since 113 is more than 100, SQ is longer than SP.
  5. (b) Depot P is nearer, at 10 km from S. The whole flight is 13 + 10 = 23 km, which is less than 25 km, so the battery lasts with 2 km to spare.

answer(a) 13 km; (b) depot P, which is 10 km from S, because SQ2 = 113 is more than SP2 = 100; the whole flight is 23 km, so 25 km is enough

techniqueThe Distance Between Two Points · Coordinate Geometry

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Adding the two differences, 12 + 5 = 17 km, for the flight. That is the distance along the grid lines. The drone flies straight, along the hypotenuse, which is √122 + 52 = 13 km.
  • Writing −62 = −36 for the flight to P, which gives SP2 = 28. The difference −6 is squared as a whole, (−6)2 = 36. A square is never negative, so the order of the subtraction does not change the distance.
06

The Area of a Four-Sided Plot of Land from the Coordinates of Its Corners

methodList the Corners in Order Round the Plot, Cross-Multiply Along the List and Halve the Difference of the Two Sums, Then Check with the Enclosing Rectangle Minus Four Triangles

A surveyor records the corners of a plot of land on a grid in meters: A(40, 10), B(90, 30), C(70, 70) and D(10, 50), in order round the plot. (a) Find the area of the plot. (b) One bag of grass seed covers 60 m2. How many bags are needed to seed the whole plot?

20406080020406080100meters east, xmeters north, yABCDthe corners in order, with the first one again at the end(40, 10) (90, 30) (70, 70) (10, 50) (40, 10)
Write the corners in order round the plot, counterclockwise, and repeat the first corner at the end of the list.
Write the corners in order, counterclockwise, and repeat the first corner at the end: (40, 10), (90, 30), (70, 70), (10, 50), (40, 10).
step 1 of 6

The area of a polygon can be found from the coordinates of its corners alone. Write the corners in order round the polygon, with the first corner repeated at the end. Multiply each x by the y of the next corner and add the products; then multiply each y by the x of the next corner and add those. The area is half of the difference between the two sums. A rectangle drawn round the plot gives an independent check.

  1. Write the corners in order, counterclockwise, and repeat the first corner at the end: (40, 10), (90, 30), (70, 70), (10, 50), (40, 10).
  2. Multiply each x by the y of the next corner and add: 40 × 30 + 90 × 70 + 70 × 50 + 10 × 10 = 1200 + 6300 + 3500 + 100 = 11100.
  3. Multiply each y by the x of the next corner and add: 10 × 90 + 30 × 70 + 70 × 10 + 50 × 40 = 900 + 2100 + 700 + 2000 = 5700.
  4. (a) The area is half of the difference: 12(11100 − 5700) = 12 × 5400 = 2700 m2.
  5. Check with the rectangle from x = 10 to x = 90 and from y = 10 to y = 70, whose area is 80 × 60 = 4800. The plot leaves a right-angled triangle in each corner, with areas 12 × 50 × 20 = 500, 12 × 20 × 40 = 400, 12 × 60 × 20 = 600 and 12 × 30 × 40 = 600. Then 4800 − 2100 = 2700, which agrees.
  6. (b) The number of bags is 2700 ÷ 60 = 45.

answer(a) 2700 m2; (b) 45 bags

techniqueThe Area of a Polygon from Coordinates

examsO-Level

Common pitfalls

  • Listing the corners out of order, such as A, C, B, D. The list must go round the edge of the plot. Out of order, the path crosses itself and the formula gives the area of a different shape.
  • Forgetting to halve the difference, which gives 5400 m2. Each cross-multiplication measures a parallelogram, which is twice a triangle, so the total is twice the area of the plot.
07

A Straight Pipeline Through Two Wells, and Whether Two Other Wells Lie on It

methodFind the Gradient from the Two Points, Use One of Them to Reach y = mx + c, Then Test Any Other Point by Substituting Its x-Coordinate

On a map with a grid in kilometers, a straight water pipeline passes through well A at (2, 3) and well B at (8, 6). (a) Find the equation of the pipeline. (b) Two more wells are at C(12, 9) and D(14, 9). Which of them lies on the pipeline, and how far due north or due south of the pipeline is the other one?

48120481216km east, xkm north, yrun 6rise 3A(2, 3)B(8, 6)gradient = rise over runm = (6 − 3)/(8 − 2) = 3/6 = 1/2
Find the gradient from the two wells: m = 6 − 38 − 2 = 36 = 12.
Find the gradient from A(2, 3) and B(8, 6): m = 6 − 38 − 2 = 36 = 12.
step 1 of 6

Two points fix a straight line. Find the gradient from the two points first, then use the gradient and one of the points in y − y1 = m(x − x1) and simplify to y = mx + c. A point lies on the line only if its coordinates satisfy the equation, so substitute its x-coordinate and compare the result with its y-coordinate.

  1. Find the gradient from A(2, 3) and B(8, 6): m = 6 − 38 − 2 = 36 = 12.
  2. Use the point A(2, 3): y − 3 = 12(x − 2). Expand the bracket: y − 3 = 12x − 1, so y = 12x + 2.
  3. (a) The equation of the pipeline is y = 12x + 2. Check with B: 12 × 8 + 2 = 6.
  4. Test C(12, 9). At x = 12 the pipeline is at y = 12 × 12 + 2 = 8, but the well is at y = 9. The well C is not on the pipeline. It is 9 − 8 = 1 km due north of it.
  5. Test D(14, 9). At x = 14 the pipeline is at y = 12 × 14 + 2 = 9, which is the y-coordinate of D.
  6. (b) The well D lies on the pipeline. The well C is 1 km due north of the pipeline.

answer(a) y = 12x + 2; (b) D lies on the pipeline, and C is 1 km due north of it

techniqueThe Equation of a Line Through Two Points · Gradient from Two Points

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Deciding that C is on the pipeline because A, B and C look as if they are in a line on the map. From B to C the gradient is 9 − 612 − 8 = 34, which is not 12, so the three wells are not in a straight line.
  • Writing the equation as y = 12x + 3, with the y-coordinate of A as the intercept. The intercept is the value of y at x = 0, and A is at x = 2, so the point has to be substituted to find c.
08

A New Road Through a School, Parallel to an Existing Road

methodParallel Lines Have Equal Gradients: Find the Gradient of the Existing Road, Then Write the Line with That Gradient Through the Given Point

On a town plan with a grid in kilometers, an existing straight road passes through (0, 2) and (6, 6). A new straight road is to be built parallel to it, passing a school at S(3, 8). (a) Find the equation of the new road. (b) A river runs along the line x = 12. Find the point where the new road reaches the river, and the distance along the river bank between the two roads.

4812160481216km east, xkm north, yriverrun 6rise 4(0, 2)(6, 6)S(3, 8)existing road: m = (6 − 2)/(6 − 0) = 4/6 = 2/3it meets the y-axis at 2, so y = 2/3 x + 2
The existing road rises 4 while it runs 6, so its gradient is 46 = 23. It meets the y-axis at 2, so its equation is y = 23x + 2.
Find the gradient of the existing road from (0, 2) and (6, 6): m = 6 − 26 − 0 = 46 = 23. It meets the y-axis at (0, 2), so its equation is y = 23x + 2.
step 1 of 5

Parallel lines are equally steep, so they have the same gradient. The gradient of the new road is therefore the gradient of the existing road, and the school gives a point on it. Two parallel lines stay the same vertical distance apart everywhere, and that distance is the difference between their y-intercepts.

  1. Find the gradient of the existing road from (0, 2) and (6, 6): m = 6 − 26 − 0 = 46 = 23. It meets the y-axis at (0, 2), so its equation is y = 23x + 2.
  2. The new road is parallel, so its gradient is also 23. It passes through S(3, 8): y − 8 = 23(x − 3), so y − 8 = 23x − 2 and y = 23x + 6.
  3. (a) The equation of the new road is y = 23x + 6. Check with the school: 23 × 3 + 6 = 8.
  4. At the river x = 12. The new road is at y = 23 × 12 + 6 = 14, and the existing road is at y = 23 × 12 + 2 = 10.
  5. (b) The new road reaches the river at (12, 14). The two roads are 14 − 10 = 4 km apart along the bank. This equals the difference between the intercepts, 6 − 2 = 4, because parallel lines stay the same vertical distance apart.

answer(a) y = 23x + 6; (b) at (12, 14), and the roads are 4 km apart along the river bank

techniqueGradients of Parallel and Perpendicular Lines · The Equation of a Line Through Two Points

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Keeping the intercept of the existing road and writing y = 23x + 2 for the new road. That is the existing road itself, and it does not pass the school: at x = 3 it is at y = 4, not 8. Parallel lines share the gradient, not the intercept.
  • Using the negative reciprocal −32 for the gradient of the new road. That gradient belongs to a perpendicular line, which would cross the existing road at a right angle.
09

The Shortest Track from a Farmhouse to a Straight Road

methodThe Shortest Path Meets the Road at a Right Angle: Use the Negative Reciprocal of the Road's Gradient, Solve the Two Equations Together for the Foot of the Perpendicular, Then Find the Length

On a map with a grid in kilometers, a straight road passes through (3, 6) and (9, 14). A farmhouse is at H(10, 7). The farmer wants the shortest possible straight track from the farmhouse to the road. (a) Find the equation of the line along which the track must run. (b) Find the point F where the track meets the road, and the length of the track.

4812160481216km east, xkm north, y(3, 6)(9, 14)H(10, 7)road: m = (14 − 6)/(9 − 3) = 8/6 = 4/3y − 6 = 4/3 (x − 3), so y = 4/3 x + 2
The road has gradient 14 − 69 − 3 = 43. Through (3, 6) its equation is y − 6 = 43(x − 3), which simplifies to y = 43x + 2.
Find the equation of the road. Its gradient is m = 14 − 69 − 3 = 86 = 43. Use the point (3, 6): y − 6 = 43(x − 3), so y − 6 = 43x − 4 and y = 43x + 2.
step 1 of 6

The shortest path from a point to a line is the perpendicular from the point to the line. The gradients of two perpendicular lines multiply to −1, so the gradient of the track is the negative reciprocal of the gradient of the road. The track meets the road at the point whose coordinates satisfy both equations, and the distance formula gives the length of the track.

  1. Find the equation of the road. Its gradient is m = 14 − 69 − 3 = 86 = 43. Use the point (3, 6): y − 6 = 43(x − 3), so y − 6 = 43x − 4 and y = 43x + 2.
  2. The shortest track is perpendicular to the road. Perpendicular gradients multiply to −1, so the gradient of the track is the negative reciprocal, −34. Check: 43 × (−34) = −1.
  3. (a) The track passes through H(10, 7): y − 7 = −34(x − 10), so y − 7 = −34x + 7.5 and y = −34x + 14.5.
  4. At F both equations hold: 43x + 2 = −34x + 14.5. Multiply both sides by 12: 16x + 24 = −9x + 174, so 25x = 150 and x = 6. Then y = 43 × 6 + 2 = 10.
  5. From F(6, 10) to H(10, 7) the difference in x is 4 and the difference in y is −3, so FH = √42 + (−3)2 = √25 = 5.
  6. (b) The track meets the road at F(6, 10) and is 5 km long. Check that F is on the track: −34 × 6 + 14.5 = −4.5 + 14.5 = 10.

answer(a) y = −34x + 14.5; (b) F(6, 10), and the track is 5 km long

techniqueGradients of Parallel and Perpendicular Lines · The Equation of a Line Through Two Points · The Distance Between Two Points

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Building the track due west from the farmhouse, along y = 7. It meets the road where 43x + 2 = 7, at x = 3.75, so it is 10 − 3.75 = 6.25 km long. The perpendicular track is 5 km long, which is shorter.
  • Using 34 for the gradient of the track, without the negative sign. Then 43 × 34 = 1, not −1. One of two perpendicular lines slopes upward and the other slopes downward, so their gradients have opposite signs.
10

Four Fence Posts Shown to Be the Corners of a Rectangle

methodEqual Gradients Show That Opposite Sides Are Parallel, Gradients That Multiply to −1 Show a Right Angle, and the Distance Formula Gives the Lengths of the Sides

Four fence posts are marked on a plan with a grid in meters: P(5, 2), Q(13, 8), R(10, 12) and S(2, 6). (a) Show that PQRS is a rectangle. (b) Find the length of fencing needed to go all the way round PQRS, and the area that the fence encloses.

48120481216meters east, xmeters north, y3/43/4P(5, 2)Q(13, 8)R(10, 12)S(2, 6)gradient of PQ = (8 − 2)/(13 − 5) = 6/8 = 3/4gradient of SR = (12 − 6)/(10 − 2) = 6/8 = 3/4
The gradient of PQ is 8 − 213 − 5 = 34 and the gradient of SR is 12 − 610 − 2 = 34. Equal gradients mean that PQ is parallel to SR.
Find the gradients of one pair of opposite sides. For PQ: 8 − 213 − 5 = 68 = 34. For SR: 12 − 610 − 2 = 68 = 34. The gradients are equal, so PQ is parallel to SR.
step 1 of 5

A quadrilateral whose opposite sides are parallel is a parallelogram, and a parallelogram with one right angle is a rectangle. On a grid, parallel sides have equal gradients, and two sides meet at a right angle when their gradients multiply to −1. The lengths of the sides come from the distance formula.

  1. Find the gradients of one pair of opposite sides. For PQ: 8 − 213 − 5 = 68 = 34. For SR: 12 − 610 − 2 = 68 = 34. The gradients are equal, so PQ is parallel to SR.
  2. Find the gradients of the other pair. For QR: 12 − 810 − 13 = 4−3 = −43. For PS: 6 − 22 − 5 = 4−3 = −43. So QR is parallel to PS, and PQRS is a parallelogram.
  3. (a) The gradients of PQ and QR multiply to 34 × (−43) = −1, so the angle at Q is a right angle. A parallelogram with a right angle is a rectangle.
  4. Find the lengths of two neighboring sides: PQ = √82 + 62 = √100 = 10 m and QR = √(−3)2 + 42 = √25 = 5 m.
  5. (b) The fencing is the perimeter, 2 × (10 + 5) = 30 m. The area enclosed is 10 × 5 = 50 m2.

answer(a) the gradients of PQ and SR are both 34, the gradients of QR and PS are both −43, and 34 × (−43) = −1; (b) 30 m of fencing, and an area of 50 m2

techniqueGradients of Parallel and Perpendicular Lines · Gradient from Two Points · The Distance Between Two Points

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Stopping once the opposite sides are shown to be parallel. That proves only that PQRS is a parallelogram, which may lean over. A rectangle also needs a right angle, which is shown by two gradients that multiply to −1.
  • Finding the area as the width of the plan times its height, 11 × 10 = 110 m2. The rectangle is tilted on the grid, so its sides are not along the grid lines. Their lengths, 10 m and 5 m, come from the distance formula.
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