James Gray

University of Edinburgh

Applicable Mathematics (Foundation) af0

2: Cartesian Geometry

In the 17th century Descartes has the idea of doing geometry using algebra.

Choose an origin and perpendicular co-ordinate axes.

Represent the point `P` (of the plane) by `(x,y)`.

2.1 Distance between two points (H pp 205-206)

Let `P = (x_1,y_1)` and `Q=(x_2,y_2)`.

Then

`PQ^2 = PR^2 + QR^2 = (x_2 - x_1)^2 +(y_2-y_1)^2`.

Exercises

Find the distance between `(1,2)` and `(4,7)`.
Find the distance between `(1,2)` and `(-3,7)`.
How far is the point `(1,2)` from the origin?
H p206 Q4.

2.2 Gradient of a straight line (H pp 1-7)

`l` is a straight line

Let `P=(x_1,y_1)` and `Q=(x_2,y_2)` be distinct points on `l`.

The gradient of `PQ = m_(PQ) = ("Change in" y)/("Change in" x) = (y_2 - y_1)/(x_2 - x_1)`.

If we had two other points `S` and `T` on the line `l` then `m_(ST) = m_(PQ)`.

Why?

Because the triangles `PQR` and `STU` are similar we have `QR = k times TU` and `PR = k times SU`.

So

`m_(PQ) = (QR)/(PR) = (k times TU)/(k times SU) = (TU)/(SU) = m_(ST).`

Therefore the gradient is the smae for any pair of points on the line `l`. So

gradient of `l = m_(PQ)`

for any pair of points `P` and `Q` on the line `l`.

Examples

Notes

For a horizontal line the gradient is `0` because
- gradient `= (y_2 - y_1)/(x_2 - x_1) = 0/(x_2 -x_1) = 0.`
For a vertical line the gradient is not defined.
Parallel lines have the same gradient.
Two lines with gradients `m` and `m'` are perpendicular prescisely when
- `mm' = -1`

Example

Example of `mm'=-1`

The triangle `STU` has vertices `S(-6,0)`, `T(-3,4)` and `U(5,-2)`. Show that it is a right-angled triangle.

Answer

`m_(ST) = (4-0)/(-3-(-6))=4/3 `
`m_(TU) = (-2-4)/(5-(-3)) = - 6/8 = - 3/4`
`m_(SU)=(-2-0)/(5-(-6))=- 2/11`

Hence `S hat(T) U = 90^(circ) since

`m_(ST) times m_(TU) = 4/3 times - 3/4 = -1`

2.3 Equation of a straight line

Let `l` be the line with gradient `m` and passing through the point `(a,b)`.

A point `Q(x,y)` lies on `l` precisely when `m = (y-b)/(x-a)`. That is when

`y-b = m(x-a)`

Example

Find the equation of the line through `(1,3)` with gradient `2`.

Where does this line cut the `x`-axis and the `y`-axis?

Show that the line does not pass through the point `(4,7)`.

Answer

Equation is `y-3 = 2(x-1)`; that is `y=2x+1`

When `x=0` then `y=1`. So it cuts the `y`-axis at `(0,1)`.
When `y=0` then `x=- 1/2`. So it cuts `x`-axis at `(- 1/2,0)`.

The values `x=4` and `y=7` do not satisfy `y=2x+1`, so `(4,7)` does not lie on the line.

Exercise

H p11 Q1(a)(c), 2(a) , 3

H p9 Q8(d)(f)

Note

`y-b = m(x-a)` can be rewritten

`y-b=mx-ma`
`y = mx +b -ma`
`y=mx+c`

where `c=b-ma` which is the `y`-intercept of the line (i.e. where it crosses the `y`-axis).

Alternative answer to the example above:

(Find the equation of the line through `(1,3)` with gradient `2`)

We know `m=2` so using other equation for line

`y=2x+c`

but we know that the line goes through `(1,3)` so

`3=2 times 1 + c \ \ \ \ \ \ "i.e."\ \ \ \ \ \ \ c=3-2=1`

So equation of line is `y=2x+1`.

Special Cases

Horizontal line through `(a,b)` has equation `y=b` (because `m=0`)
Verical line through `(a,b)` has equation `x=a`. (Here `m` is undefined.)

Remarks

Points `(x,y)` above the line `y=mx+c` satisfy `y>mx+c`.

Points `(x,y)` below the line `y=mx+c` satisfy `y<mx+c`.

2.4 Point of intersecion of two lines

Example

Find the point where the two lines `y=2x + 3` and `y = -3x +4` meet.

Answer

The point `(x,y)` of intersection satisfies both equations. Therefore

`y=2x+3 = -3x +4`

So `5x=1`. Hence `x=1/5` and `y = 2 times 1/5 +3 = 17/5`.

The equation of a line (again)

The equation of a line can also be written

`px +qy = r`

Example

Find the gradient of the line `2x+3y = 5`.

Answer

Rewrite the equation as `3y = -2x +5`. So `y=-2/3x +5/3`
Then gradient `= -2/3`.

The method of finding the intersection of two lines is solving simultaneous equations.

Example

Find the point of intersection of `3x+4y=5` and `2x-7y=13`.

Answer

Multiply the first equation by `7` to get `21x +28y=35` and multiply the second equation by `4` to get `8x-28y=52`. Now add the resulting equations to eliminate `y`. This gives `29x=87` and so `x=3`. Substitute `x=3` into the first equation in the question to find `y`. So `9+4y=5` and so `y=-1`. Therefore the point of intersection is `(3,-1)`.

2.5 Mid point of an interval

Let `P` have coordinates `(x_1,y_1)` and `Q` have coordinates `(x_2,y_2)`, and let `M` be the mid point of `PQ`

`x` coordinate of `M` is

`1/2(x_2-x_1) + x_1 = 1/2 x_2 - 1/2x_1+x_1`

`=1/2(x_1+x_2)`.

Similar for `y`.

So coordinates of `M` are `((x_1+x_2)/2 , (y_1+y_2)/2)`.

2.6 Applications (H pp 12-17)

Suppose we are given the poisitions of the vertices `A`, `B` and `C` of a triangle. Then we can now

find the gradient of any side
find the equation of any side
find the midpoint of any side

Median of a triangle

A median of a triangle is a line from a vertex to the mid-point of the opposite side.

Theorem

The three medians of a triangle are concurrent.

Altitude of a triangle

An altitude of a trianle is a line from a vertex perpendicular to the opposite side.

Theorem

The three altitudes of a triangle are concurrent.

Perpendicular bisector

The perpendicular bisector of the side of a triangle is the line perpendicular to the side and passing through the mid-point of the side.

Theorem

The three perpendicular bisectors of the sides of a triangle are concurrent.