# eq lines and surd

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• Dec 10th 2007, 10:05 AM
Chez_
eq lines and surd
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx
• Dec 10th 2007, 10:12 AM
topsquark
Quote:

Originally Posted by Chez_
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

We need more information. Where does BC cross AB? There are any number of lines perpendicular to AB.

Quote:

Originally Posted by Chez_
Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

In LaTeX code, [ math ] 6 - 3 \sqrt{3} [ /math ] (without typing the spaces between the [ ].

$\displaystyle 6 - 3\sqrt{3} = (6) + (-3)\sqrt{3}$

Is this what you are looking for?

-Dan
• Dec 10th 2007, 10:21 AM
janvdl
Quote:

Originally Posted by Chez_
have a picture of a line AB tht eq is y=4x-5 and passes through the point 9(2,3). The line BC is perp to AB and cuts the x axis at C. I need tofind the equation of line BC, which i have done, but i also need to find the x coordinate of C, but im not sure how do it? thnx

Also how could i express 3(in a square root thing)
--------
6-3(3 again in sqr root thing)
in the form of p+q 3(in sqroot)

Iv now with the help of you great people being able to complete my practise paper and practised and found some new tips lol thnx

$\displaystyle BC: f(x) = - \frac{x}{4} + D$

Somewhere BC intercepts AB

So the equations of the two lines become equal at some point.

$\displaystyle - \frac{x}{4} + D = 4x + 5$

Simplify that and you'll find that the D-value in $\displaystyle f(x)$ is given by the following function:

$\displaystyle D = \frac{17}{4} x + 5$

Where the $\displaystyle x$ is the $\displaystyle x$-axis value where the two lines cross. From there you'll find the D-value of f(x).

When you have that, set $\displaystyle f(x) = 0$

Then solve for $\displaystyle x$ and you'll find the C-value that you're looking for.
• Dec 10th 2007, 10:27 AM
topsquark
Quote:

Originally Posted by janvdl
$\displaystyle BC: f(x) = \frac{x}{4} + D$

That would be the line $\displaystyle f(x) = -\frac{x}{4} + D$

-Dan
• Dec 10th 2007, 10:30 AM
janvdl
Quote:

Originally Posted by topsquark
That would be the line $\displaystyle f(x) = -\frac{x}{4} + D$

-Dan

Hehe, that had to happen sometime. Thanks Topsquark. :D