# Thread: eq lines and surd

1. ## 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

2. 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.

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

3. 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.

4. 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

5. 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.