9709 · 1.3
Coordinate geometry flashcards
Revision flashcards for Cambridge 9709 Coordinate geometry (syllabus 1.3). Flip, recall, then mark a real past-paper question.
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What is the formula for the gradient, $m$, of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$?
$m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}$
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What is the condition for two lines with gradients $m_1$ and $m_2$ to be parallel?
They have the same gradient: $m_1 = m_2$.
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What is the condition for two lines with gradients $m_1$ and $m_2$ to be perpendicular?
The product of their gradients is -1: $m_1 m_2 = -1$. This means one gradient is the negative reciprocal of the other, e.g., $m_2 = -\frac{1}{m_1}$.
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What is the formula for the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$?
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. This is an application of Pythagoras' theorem.
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What is the formula for the midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$?
$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$. It's the average of the x-coordinates and the average of the y-coordinates.
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What are the two main forms for the equation of a straight line?
1. Gradient-intercept form: $y = mx + c$. 2. Point-gradient form: $y - y_1 = m(x - x_1)$.
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What is a 'perpendicular bisector'?
A line that passes through the midpoint of a line segment and is perpendicular to it.
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Trap: A line has a gradient of $3$. What is the gradient of a perpendicular line?
The negative reciprocal is $-\frac{1}{3}$. A common mistake is to forget the negative sign and just write $\frac{1}{3}$.
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How do you find the equation of a line when you are only given two points, A and B?
1. Calculate the gradient, $m$, using the two points. 2. Substitute one of the points (A or B) and the gradient $m$ into $y - y_1 = m(x - x_1)$ to find the equation.
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Trap: When using the distance or midpoint formula, what is a common error?
Mixing up the x and y coordinates, or making sign errors with negative coordinates. For distance, a common mistake is $(x_2 - x_1)^2 + (y_2 - y_1)^2$ without the square root.
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How can you show that a triangle ABC is isosceles?
Calculate the lengths of the three sides (AB, BC, AC) using the distance formula. If two of the lengths are equal, the triangle is isosceles.
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How can you show that a triangle ABC is a right-angled triangle?
Method 1: Calculate the gradients of the three sides. If two gradients multiply to give -1, those two sides are perpendicular. Method 2: Calculate the lengths of the three sides and check if they satisfy Pythagoras' theorem ($a^2 + b^2 = c^2$).