Re: Co-Ordinate Geometery!
Quote:
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
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Re: Re: Co-Ordinate Geometery!
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
http://www.vbforums.com/attachment.p...postid=1245329