C1

Coordinate Geometry in the (x, y) Plane

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A

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Equation of a Straight Line

There are many ways of writing an equation to show the relationship between the x- and y-coordinates. The most well-known up to A-level is y = mx + c, but this can sometimes be a bit untidy if there are negatives and fractions. Another common way to write the equation of a straight line is ax + by + c = 0.

We will now look at two different situations when we will be required to find the equation of a straight line. The first is an equation of the line through two points, the second is the equation of the line through a point with a given gradient. I will use the same technique for both of these.

Given 2 Points

Start by finding the gradient of the line joining the points. This is straightforward.

Gradient = (difference in y) / (difference in x).

Remember that you must subtract the y-coordinates in the same order that you subtract the x-coordinates. For example, find the gradient of the line joining (5, 8) to (3, 10)

The gradient is (diff in y)/(diff in x) = (5-3) / (8-10) = 2 / -2 = 1 NOT 1...

Once you know the gradient then use the method for finding an equation given a point and the gradient.

Given a Point and the Gradient

The equation of the line has the form y = mx + c. In this form "m" is the gradient, so substitute that in straight away. Now all we need is "c". We find "c" by substituting the x and y coordinates that the line goes through.

For example: find an equation of the straight line through (3, 7) that has gradient 2.

We shall find it in the form y=mx+c. We are told the gradient is 2, so our equation is y=2x+c. All we need to do is find c. If we substitute the values (3, 7) in we get... 7=2*3+c 7= 6 + c c=1 so... y=2x+1 is our equation.

Last example: find an equation in the form ax + by + c = 0, where a, b and c are integers, of the line through (5, 2) and (8, -2)

Step 1 - Find gradient m=(diff y)/(diff x) = (5-8) / (2--2) = -3 / 4 so 3/4 x + c Now substitute values (either coordinate will do) I'll use (5,2) y = -3/4 x + c 2 = -3/4 *5 + c c = 2 + 15/4 c = 23/4 So we have y = -3/4 x + 23/4 Now multiply the whole equation to get rid of fractions (multiply by 4) and rearrange. 4y = -3x + 23 3x + 4y -23 = 0 is the final answer.

Note: if you really must work with subscripts (like you might see in a textbook) then please make sure that you understand where the equations come from.

Condition for 2 Lines to be Perpendicular

I won't give reasons why here, but 2 lines are at right angles if their gradients multiply to -1.

Here's an examination question (May 2006 q11, worth 15 marks). Have a bash and then watch the clip to see how well you did.

The line l₁ passes through the points P(1, 2) and Q(11, 8).

1. Find an equation for l₁ in the form y = mx + c, where m and c are constants.
2. The line l₂ passes through the point R(10, 0) and is perpendicular to l₁. The lines l₁ and l₂ intersect at the point S. Calculate the coordinates of S.
3. Show that the length of RS is 3√5
4. Hence, or otherwise, find the exact area of the triangle PQR

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