Co-Ordinate Geometery!

[JafferAB]

Im currently working on this for my AS-Level Maths, and im completele stuck on finding the equation of a straight line! My book explains it about as well as a cow explains calculus to a baby!

Here for example is one of the questions...

Find the equation of...

(i) parralel to y = 2x and passing through (1,5);

all my book is telling is useless because i always end up with the question!!!!!!!

if someone could explain to me the steps needed to find the equation of this line, i would be very gratefull!!
thanks

[NotLKH]

Originally posted by JafferAB


(i) parralel to y = 2x and passing through (1,5);

Given Two Equations:

Eq1: Y = M₁X + B₁
Eq2: Y = M₂X + B₂

Eq2 is Parallel to Eq1 when:
M₂ = M₁

So, with your problem,

M₁ = 2
B₁ = 0

Now, this gives us a preliminary form for the family of all equations parallel to your original as:

Y = 2X + B₂ .

To Solve for B₂, we need a point that the new line will pass thru. Which you have kindly provided as being:
(1,5).

Assuming the form of this point follows the standard (X,Y) relation, we know

When X = 1, Y Must be 5.

So, lets plug these values into the generic family of parallel lines that we have so far:

5 = 2*1 + B₂

Which leads us to see that:

B₂ = 3.

Therefore the equation of a line parallel to y = 2x and passing through (1,5) is :

Y = 2X + 3

Now, just so long as I haven't made any errors, this should be the correct solution.

Does this Help any

[JafferAB]

wow, that shelped a treat i have been able to complete that section and if it helps any, you did get that right

now, as my bad forsight bargains! i need a bit more help on the same subject, but with perpendicular lines, e.g.

"Find the equations of the following lines..."

(i) perpendicular to y = 3x and passing through (0,0)

i know i need to find point on the other line, and i also know the gradient (m) of both lines is 3 (from what the y = mx+c equation states).

any help of the same nature if possible please

thanks again!

[NotLKH]

The Perpendicular is very similar to the Parallel, so Let me just rewrite my last response appropriately:

{BTW, I'm glad to Have Helped with the Parallel case. It seems easy Once you've done it a few times. }

Given Two Equations:

Eq1: Y = M₁X + B₁
Eq2: Y = M₂X + B₂

Eq2 is PERPENDICULAR to Eq1 when:
M₂ = (-1)/(M₁)

Or, in other words:

M₂ is the Negative Inverse of M₁

Of Course, if M₁ = 0, then you have a little problem. But in this instance we don't have to worry about that.

So, with your problem,

M₁ = 3
B₁ = 0

Now, this gives us a preliminary form for the family of all equations PERPENDICULAR to your original as:

Y = (-1/3)X + B₂ .

To Solve for B₂, we need a point that the new line will pass thru. Which you have kindly provided as being:
(0,0).

Assuming the form of this point follows the standard (X,Y) relation, we know

When X = 0, Y Must be 0.

So, lets plug these values into the generic family of PERPENDICULAR lines that we have so far:

0 = (-1/3)*0 + B₂

Which leads us to see that:

B₂ = 0.

Therefore the equation of a line PERPENDICULAR to y = 3x and passing through (0,0) is :

Y = (-1/3)X + 0, or more simply:
Y = (-1/3)X

Now, just so long as I haven't made any errors, this should be the correct solution.

Does this Help any

[JafferAB]

that helps alot, thank you

the actual answer was 3y + x = 0 but thats just a matter of moving the timesing both sides by 3 and moving the Y, so its technically the same

cheers again jafer

[A$$Bandit]

Ooooh, finally someone from the UK who's my age... have fun doing AS <cough>

[JafferAB]

hehe, AS levels arent too bad, loads of work to do though!

[SLH]

Wait until you get to maths at A-level, it was rather hard...