Flashcards · 0580 · Topic 3

Coordinate geometry flashcards

32 cards on Topic 3, subtopics 3.1 to 3.7. Pick a subtopic or study them all.

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What does the point (x, y) mean?: x: how far across (right positive, left negative). y: how far up or down (up positive). The origin O is (0, 0). Across first, then up.
Signs of x and y in each region of the grid: Top right (+, +); top left (−, +); bottom left (−, −); bottom right (+, −).
Points with the same x-coordinate, and points with the same y-coordinate, lie on what kind of line?: Same x-coordinate: a vertical line. Same y-coordinate: a horizontal line.
Does the point (2, 3) lie on the line y = 2x − 1?: Yes: 2(2) − 1 = 3. A point lies on a line if its coordinates make the equation true.

How do you draw a straight-line graph from its equation?: Make a table of values with at least three x-values, plot the points, and rule one straight line across the whole range asked for.
Complete the table for y = 2x − 1 at x = −2, 0 and 2: x = −2: y = −5 x = 0: y = −1 x = 2: y = 3
Why plot three points rather than two?: The third point is a check. If the points do not lie in a straight line, a value in the table is wrong: fix the table rather than bending the line.
Describe the lines x = 3 and y = −2: x = 3: a vertical line through (3, 0). y = −2: a horizontal line through (0, −2).
Draw 4x + 3y = 12: which points are quickest to find?Extended: The intercepts: x = 0 gives y = 4, and y = 0 gives x = 3. Plot (0, 4) and (3, 0) and a third point as a check, e.g. x = 6 gives y = −4.

Gradient of a straight line: gradient = vertical change ÷ horizontal change. On a grid, count squares up (or down) and across between two points on the line.
Gradient of a line sloping up, a line sloping down, a horizontal line and a vertical line: Up from left to right: positive. Down from left to right: negative. Horizontal: 0. Vertical: undefined.
Gradient from the coordinates of two pointsExtended: m = (y₂ − y₁)/(x₂ − x₁) Subtract the coordinates in the same order on the top and the bottom.
Find the gradient of the line through (0, 3) and (4, 11)Extended: (11 − 3)/(4 − 0) = 8/4 = 2
Exam trap when finding a gradient: Dividing the change in x by the change in y. It is change in y ÷ change in x.

Length of the line segment from (x₁, y₁) to (x₂, y₂)Extended: length = √((x₂ − x₁)² + (y₂ − y₁)²) This is Pythagoras’ theorem on the horizontal and vertical distances.
Midpoint of the line segment from (x₁, y₁) to (x₂, y₂)Extended: ((x₁ + x₂)/2, (y₁ + y₂)/2): the mean of the x-coordinates and the mean of the y-coordinates.
A is (1, 2) and B is (7, 10). Find the length of AB and its midpoint.Extended: Length = √(6² + 8²) = √100 = 10 Midpoint = (4, 6)
Two exam traps with length and midpointExtended: Leaving the length as the sum of the squares (forgetting the square root). Adding the coordinates for a midpoint but not dividing by 2.

In y = mx + c, what do m and c tell you?: m is the gradient. c is the y-intercept: the line crosses the y-axis at (0, c).
State the gradient and y-intercept of y = 3x − 4: Gradient 3, y-intercept −4.
Write the equation of the line with gradient −2 through the origin: y = −2x (c = 0 for a line through the origin).
Find the equation of the line through (2, 3) and (4, 9)Extended: m = (9 − 3)/(4 − 2) = 3 Substitute (2, 3): 3 = 3(2) + c, so c = −3. y = 3x − 3
Find the gradient and y-intercept of 2x + 3y = 12Extended: Rearrange: 3y = 12 − 2x, so y = −(2/3)x + 4. Gradient −2/3, y-intercept 4.

What is true about the gradients of parallel lines?: Parallel lines have the same gradient.
Find the equation of the line parallel to y = 3x − 7 through (1, 4): The gradient stays 3: 4 = 3(1) + c, so c = 1. y = 3x + 1
Which two of these lines are parallel? y = 2x + 1, y = −2x + 1, y = 2x − 5: y = 2x + 1 and y = 2x − 5: both have gradient 2. (y = −2x + 1 has gradient −2.)
Find the equation of a line parallel to a given line through a new point: the two steps: 1. Keep the gradient m of the given line. 2. Substitute the new point into y = mx + c and solve for the new c.
Is 2y = 6x + 1 parallel to y = 3x − 2?Extended: Yes. Rearranged, 2y = 6x + 1 is y = 3x + ½, which has gradient 3.

Gradients of perpendicular linesExtended: They are negative reciprocals: if one gradient is m, the other is −1/m, so the two gradients multiply to −1.
Gradient perpendicular to a line of gradient 2, and to a line of gradient −2/3Extended: Gradient 2: perpendicular gradient −½. Gradient −2/3: perpendicular gradient 3/2.
Find the equation of the line perpendicular to y = 4x − 1 through (2, 5)Extended: Perpendicular gradient −¼: 5 = −¼(2) + c, so c = 5.5. y = −¼x + 5.5
Find the perpendicular bisector of A(0, 0) and B(4, 2)Extended: Midpoint (2, 1). Gradient of AB = 2/4 = ½, so the perpendicular gradient is −2. 1 = −2(2) + c, c = 5: y = −2x + 5