Cartesian equation form of a line

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Cartesian equation form of a line

IB Maths HL – Vectors, Cartesian equation form of a line

How can we find the Cartesian equation form of the line passing through the point $(-4,6)$ and is parallel to the vector $b= \begin{pmatrix} 6 \\ -8 \end{pmatrix}$ ?

Thanks
lora

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Joined: Wed Apr 10, 2013 7:36 pm

Re: Cartesian equation form of a line

IB Math HL – Cartesian equation form of a line

The Cartesian form for the equation of a straight line is :

$\frac{x- a_{1}}{ b_{1}}= \frac{y- a_{2}}{ b_{2}}$

where $a= \begin{pmatrix} a_{1} \\ a_{2} \end{pmatrix}$ is a position vector (a point on the line)

and $b= \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix}$ is a direction vector ( a vector parallel to the line)

Concerning your question we have that the Cartesian equation form is given by the following formula:

$\frac{x-(-4)}{6}= \frac{y-6}{-8}$

$\frac{x+4}{6}= \frac{y-6}{-8}$

Hope these help!!
lily

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Joined: Mon Jan 28, 2013 8:04 pm

Re: Cartesian equation form of a line

Thanks Lily!!
lora

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Joined: Wed Apr 10, 2013 7:36 pm