## IB Maths SL Vectors, Cartesian form of a line

Algebra, Functions and equations, Circular functions and trigonometry, Vectors, Statistics and probability, Calculus. IB Maths SL Revision Notes

### IB Maths SL Vectors, Cartesian form of a line

IB Mathematics SL – Vectors, Cartesian equation form of a line

How can we find the Cartesian equation form of the line passing through the point (2,4) and is parallel to the vector $b= \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ ?

Thanks
Jack

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Joined: Mon Jan 28, 2013 8:08 pm

### Re: IB Maths SL Vectors, Cartesian form of a line

IB Mathematics SL – Vectors, 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:

$\frac{x-2}{3}= \frac{y-4}{1}$

Hope these help!
miranda

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Joined: Mon Jan 28, 2013 8:03 pm