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Revision:Coordinate Geometry

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Coordinate Geometry


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The distance between two points

The length of the line joining the points (x_1,y_1) and (x_2,y_2) is: \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.


Example

Find the distance between the points (5, 3) and (1, 4).

Distance = \sqrt{(1-5)^2 + (4-3)^2} = \sqrt{16 + 1} = \sqrt{17}.


The midpoint of a line joining two points

The midpoint of the line joining the points (x_1,y_1) and (x_2,y_2) is \left( \dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \right)


Example

Find the coordinates of the midpoint of the line joining (1, 2) and (3, 1).

Midpoint = \left( \frac{3+1}{2} , \frac{2+1}{2} \right) = (2, \frac{3}{2}).


The gradient of a line joining two points

The gradient m of the line joining the points (x_1,y_1) and (x_2,y_2) is: m = \dfrac{y_2-y_1}{x_2-x_1}.


Example

Find the gradient of the line joining the points (5, 3) and (1, 4).

Gradient = \dfrac{4-3}{1-5} = \dfrac{1}{-4} = -1/4.


The equation of a line using one point and the gradient

The equation of a line which has gradient m and which passes through the point (x_1,y_1) is:

y - y_1 = m(x - x_1).


Example

Find the equation of the line with gradient 2 passing through (1, 4).

y - 4 = 2(x - 1) \Rightarrow y - 4 = 2x - 2 \Rightarrow \underline{y = 2x + 2}.

Since m = \dfrac{y_2-y_1}{x_2-x_1}, the equation of a line passing through (x_1,y_1) and (x_2,y_2) can be written:

\dfrac{y-y_1}{x-x_1} = \dfrac{y_2-y_1}{x_2-x_1}.


Perpendicular Lines

We can work out the equation of a perpendicular line by using the relationship:

m_1m_2 = -1 \Rightarrow m_2 = \dfrac{-1}{m_1}
This means to find the gradient of a perpendicular line we take the reciprocal of the original line with it's sign changed.


Example

Find the equation of the line perpendicular to the line, which passes through the points (1, 3) and (4,9), and goes through the origin.
First, find the equation of the original line:

m = \dfrac{9-3}{4-1} \Rightarrow m = 2

To find the perpendicular line, we use m_1m_2 = -1, giving a gradient of -1/2. We then use the straight line formula to find the equation.

y - y_1 = m(x - x_1) \Rightarrow y - 0 = -\dfrac{1}{2}(x - 0) \Rightarrow y = -\dfrac{1}{2}x


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