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Coordinate geometry is a very important study as it allows us to develop graphical representations for different things such as parallel and perpendicular lines and curves we couldn't normally graph.
We split Coordinate Geometry into three key sections:
- STRAIGHT LINE GRAPHS - Understanding how gradients work and how we can use this in modelling. As well as understanding gradients of parallel and perpendicular lines.
- CIRCLES - Understand how algebraic methods such as completing the square can help us find the radius and centre of a circle. Also understanding how to find a tangent to a circle using methods adopted from straight line graphs.
- PARAMETRIC EQUATIONS - Understanding how we can use one variable to describe what two variables do and understanding how we can find equations for graphs we normally wouldn't be able to find by just looking at the graph.
Let's look at these in a bit more detail.
Straight line graphs
In order to understand coordinate geometry, we will look at straight line graphs in a lot of detail, starting with calculating gradients and intercepts. Then we will move on to parallel and perpendicular lines. Finally, we will start modeling using straight line graphs.
Here's is an example of a question involving straight line graphs. This question will require calculating the gradient.
The amount of money an ice-cream van makes in a day can be modeled as \(y = 5s-12\). Where s is the amount of ice creams sold and y is the amount of money made in pounds.
Find the price of each ice cream.
Calculate the amount of ice cream that needs to be sold so that the ice cream van doesn't make a loss.
SOLUTION: 1. The gradient of this line is the money made from sales. Remember, in a graph, m is the gradient.
\(y = mx+c\)Therefore the gradient of this graph is 5. So each sale is £5. To not make a loss \(5s-12 \geq 0\) We can solve this by saying \(5s \geq 12\)
Circles
Circles are an important part of coordinate geometry. We can use information about circles along with other theories of coordinate geometry to solve more complicated problems.
Remember, a circle with radius r and center (a, b) has an equation: \((x-a)^2 + (y-b)^2 = r^2\)
A circle has an equation \((x-2)^2 + (y-4)^2 = 25 \qquad (5,8)\)
SOLUTION: First, we need to find the equation of the radius. The centre of the circle is (2, 4), the point we are concerned about is. Find the gradient:\(\frac{8-4}{5-2} = \frac{4}{3}\)So general equation of radius is \(y = \frac{4}{3}x +c\) . So where c is some constant is the equation of the radius. Now we use the fact that the tangent is perpendicular to the radius. When two lines are perpendicular to their gradients product -1 . So the gradient of the radius and the gradient of the tangent must multiply to make -1 . Therefore if we call the gradient of the tangent p\(\frac{4}{3}p = -1p = -\frac{3}{4}\)So our equation is:\(y = -\frac{3}{4}x+c\)To find our constant c simply substitute values \(x =5 \text{ and }y = 8\)\(8 = -\frac{3}{4}(5)+c8 = -\frac{15}{4} + cc = 11.75\)So our final answer is:\(y = -\frac{3}{4}x + 11.75\)This is a graphical representation of the circle and perpendicular line:
Parametric equations
Parametric equations represent everything in terms of one variable. The variable normally used is t.
This is because there are a lot of more complicated equations where it is better to represent each x and y in terms of the same variable.
Here's an example of a set of parametric equations.
\(x = 2\cos(t); \space y = 2\sin(t)\)This is the parameterization of a circle as:
\(x^2+y^2 = (2\cos(t))^2 + (2\sin(t))^2 = 4 \cos^2(t) + 4 \sin^2(t) = 4(\sin^2(t) + \cos^2(t)) = 4(1)= 4\)
Below is an example of a parametric equations question.
A curve C contains the following parametric equations.
\(x = 4\cos(t+\frac{\pi}{6}); \space y = 2\sin(t)\)Prove that \(x + y = 2\sqrt3 \cos(t)\)
Show that the Cartesian equation of C is \((x+y)^2 +ay^2 = b\) where a and b are constants to be found.
SOLUTION:
Well \(x+y = 4\cos(t+\frac{\pi}{6}) + 2\sin t\).
By addition formula
\(4 \cos(t+\frac{\pi}{6}) = 4\cos(t)\cos(\frac{\pi}{6})-4\sin(t)\sin(\frac{\pi}{6})4\cos(t + \frac{\pi}{6}) = 4\frac{\sqrt3}{2} \cos(t) 2\sin(t) 4\cos(t+\frac{\pi}{6}) = 2\sqrt{3} \cos(t) -2\sin(t)4\cos(t+\frac{\pi}{6}) +2\sin(t) = 2\sqrt3 \cos(t) - 2\sin(t)+2\sin(t) = 2 \sqrt3 \cos(t)\)
2.
\((x+y)^2 = (2\sqrt3 \cos(t))^2 = 12 \cos^2{t}y^2 = 4\sin^2{t} 12\cos^2{t}+4a(\sin^2{t}) = b\)
By \(\sin^2{t} + \cos^2{t} = 1\):
\(12\cos^2{t}+12\sin^2{t} = 124a = 12 \rightarrow a = 3b =12\)
Coordinate Geometry - Key takeaways
Straight line graphs are decided by a gradient and the y-intercept.
Parallel and perpendicular lines are decided by gradients.
Parallel lines contain the same gradient.
Perpendicular lines have gradients which product -1.
Circle theorems can be used to help find equations of lines on a Cartesian plane.
Coordinate geometry ties together geometrical concepts and rules of lines in Cartesian coordinates.
Parametric Equations involve writing everything in terms of one variable.
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Frequently Asked Questions about Coordinate Geometry
What is coordinate geometry?
Coordinate geometry is the study of the Cartesian plane.
What is a straight line graph?
A graph with the equation y=mx+c.
What is a set of parallel lines?
Two lines that contain the same gradient and never meet.
What is a tangent perpendicular to in a circle?
A radius.
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