Flashcards · 11 cards

Coordinate geometry

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Rule
The distance between $(x_1, y_1)$ and $(x_2, y_2)$ ?
Answer
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ : Pythagoras on the rise and the run.
DistanceAB² = (x₂ − x₁)² + (y₂ − y₁)²
From the lesson: Points, distance and midpoint
Rule
The midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ ?
Answer
$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$ : add and halve. Don't subtract.
MidpointM = ((x₁ + x₂) ÷ 2, (y₁ + y₂) ÷ 2)
From the lesson: Points, distance and midpoint
Rule
The gradient of the line through $(x_1, y_1)$ and $(x_2, y_2)$ ?
Answer
$m = \dfrac{y_2 - y_1}{x_2 - x_1}$ : rise over run.
Gradientm = rise ÷ run = (y₂ − y₁) ÷ (x₂ − x₁)
From the lesson: Gradient and the equation of a line
Rule
In $y = mx + c$ , what are $m$ and $c$ ?
Answer
$m$ is the gradient; $c$ is where the line crosses the $y$ -axis.
y = mx + cm is the rise for 1 across; c is where it crosses the y-axis
From the lesson: Gradient and the equation of a line
Know it
What is the gradient of $3x + 4y = 12$ ?
Answer
Make $y$ the subject first: $y = -\frac34 x + 3$ , so $m = -\frac34$ , not 3.
From the lesson: Gradient and the equation of a line
Know it
The equation of the line with gradient $m$ through $(x_1, y_1)$ ?
Answer
$y - y_1 = m(x - x_1)$ .
From the lesson: Gradient and the equation of a line
Rule
What do parallel lines have in common?
Answer
The same gradient.
ParallelSame gradient m
From the lesson: Gradient and the equation of a line
Rule
How are the gradients of perpendicular lines related?
Answer
$m_1 m_2 = -1$ : turn the gradient upside down and change its sign. Perpendicular to $m = \frac23$ is $-\frac32$ .
Perpendicularm × (−1/m) = −1
From the lesson: Gradient and the equation of a line
Rule
How is a line's gradient related to the angle it makes with the $x$ -axis?
Answer
$m = \tan\theta$ .
Gradient and anglem = rise ÷ run = tan θ
From the lesson: Gradient and the equation of a line
Rule
The angle between two lines with gradients $m_1$ and $m_2$ ?
Answer
$\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|$ , with $1 + m_1 m_2$ on the bottom.
The angle between two linestan θ = |(m₁ − m₂) ÷ (1 + m₁m₂)|
From the lesson: Gradient and the equation of a line
Which method?
WAEC 2023 · Paper 2 · Q7 (b)
Given the points $P(3, 5)$ and $Q(-5, 7)$ on the Cartesian plane such that $R$ is the midpoint of $\overline{PQ}$ , find the equation of the line that passes through $R$ and is perpendicular to $\overline{PQ}$ .
What two things do you need to write the line?
Answer
A point and a gradient. The point is the midpoint $R$ ; the gradient is perpendicular to $PQ$ , so $m \times m_{PQ} = -1$ .
From the lesson: Gradient and the equation of a line Try the question