Straight-line graphs flashcards

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Gradient of a straight line
The increase in y for each increase of 1 in x; the same between any two points on the line.
y-intercept
The y-coordinate of the point where a line crosses the y-axis; c in y = mx + c.
Perpendicular bisector of AB
The line through the midpoint of AB at right angles to AB. Every point on it is equidistant from A and B.
Collinear points
Points that all lie on one straight line.
Foot of the perpendicular from P to a line
The point where the line through P perpendicular to the given line meets it. P to the foot is the shortest distance from P to the line.
Negative reciprocal of m
−1/m: flip the fraction and change the sign. It is the gradient of any line perpendicular to a line of gradient m.
Gradient through (x₁, y₁) and (x₂, y₂)
m = (y₂ − y₁) / (x₂ − x₁). (No — learn it)
Gradient–intercept form of a line
y = mx + c (gradient m, crosses the y-axis at (0, c)). (No — learn it)
Line with gradient m through (x₁, y₁)
y − y₁ = m(x − x₁). (No — learn it)
Gradient of ax + by + c = 0
−a/b (make y the subject: y = −(a/b)x − c/b). (No — derive it)
Condition for perpendicular lines
m₁m₂ = −1 (neither line vertical). (No — learn it)
Midpoint of (x₁, y₁) and (x₂, y₂)
((x₁ + x₂)/2, (y₁ + y₂)/2). (No — derive it)
Length of the segment from (x₁, y₁) to (x₂, y₂)
√((x₂ − x₁)² + (y₂ − y₁)²) — Pythagoras on the run and rise. (No — derive it)
Sketch: positive, negative, zero and undefined gradient
Rises left to right; falls left to right; horizontal (y = k); vertical (x = k).
Family y = ½x + c for different c
Parallel lines, all with gradient ½, shifted up or down; each crosses the y-axis at (0, c).
Triangle formed by ax + by = k and the axes
Right-angled at O, with vertices O, (k/a, 0) and (0, k/b); area = ½ × |k/a| × |k/b|.
Quadrilateral between two lines and the axes
Find all four vertices, then split along a diagonal into two triangles or subtract one triangle from a larger one.
Two perpendicular lines meeting at D, with A on one and C on the other
Triangle ACD is right-angled at D, so area = ½ × AD × CD.
Why does gradient AB = gradient BC prove A, B, C are collinear?
Equal gradients make AB and BC parallel, and they share the point B, so they lie on the same line.
Why can a vertical line not be written as y = mx + c?
Its run is zero, so it has no gradient. Write it as x = k (ax + by + c = 0 with b = 0).
How do you test whether a point lies on a line?
Substitute its coordinates into the equation; the point lies on the line only if the equation balances exactly.
What do the solutions of two simultaneous linear equations mean graphically?
The coordinates of the point where the two lines meet. No solution: the lines are parallel.
Why use the foot of the perpendicular for the distance from a point to a line?
The perpendicular is the shortest route from the point to the line; any other route is a hypotenuse, so it is longer.
How do you find a line parallel to ax + by + c = 0 through a point?
Keep a and b, write ax + by + d = 0, and substitute the point to find d.
Line through (−1, −4) and (5, 8), in the form y = mx + c?
m = 12/6 = 2; y + 4 = 2(x + 1), so y = 2x − 2.
Line through (−2, 5) with gradient −¾, with integer coefficients?
y − 5 = −¾(x + 2) → 4y − 20 = −3x − 6 → 3x + 4y − 14 = 0.
Perpendicular bisector of A(−3, 2) and B(5, 6)?
Midpoint (1, 4); gradient AB = ½, so use −2: y = −2x + 6.
Length of PQ for P(2, −3), Q(8, 5)?
√(6² + 8²) = √100 = 10.
Foot of the perpendicular from P(5, −2) to x − 2y + 6 = 0?
Perpendicular: y = −2x + 8. Substitute: 5x − 10 = 0, foot (2, 4); distance √45 = 3√5.
Area of the triangle formed by 3x + 4y = 24 and the axes?
Intercepts (8, 0) and (0, 6); area = ½ × 8 × 6 = 24.
A(−2, 7), B(1, 1), C(6, −9): collinear? If D(k, 13) is on the same line, k?
Gradients AB and BC are both −2 with B common, so yes. Line y = −2x + 3, so 13 = −2k + 3, k = −5.
For which k is kx + 4y − 1 = 0 perpendicular to 2x − 3y + 4 = 0?
Gradients −k/4 and 2/3: (−k/4)(2/3) = −1, so k = 6.
“In the form ax + by + c = 0, where a, b and c are integers”
No fractions and everything on one side: multiply through by the denominators, then check with a known point.
Subtracting a negative coordinate in the gradient formula
Bracket it: (8 − (−4)) ÷ (5 − (−1)), not (8 − 4) ÷ (5 − 1).
Perpendicular gradient written as the reciprocal only
Wrong sign: check that m₁ × m₂ = −1. For ⅖ the perpendicular gradient is −⁵⁄₂, not ⁵⁄₂.
Leaving a length as √80
Simplify surds: √80 = √(16 × 5) = 4√5. Keep exact values through later parts.
Finishing a “show that” about perpendicular lines or collinear points
End with a sentence: “product = −1, so the lines are perpendicular” or “equal gradients and common point B, so collinear”.

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