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A line passing through the points $A(-1,-4)$ and $B(1,-3)$ cuts the circle $(x-3)^2 + (y+2)^2 = 10$

How do write the equation on the line in vector, cartesian and parametric form?

I am new at maths so I'm not sure. I wrote it as

Vector $$ x=(-1/-4) + t(1/-3) $$ Parametric \begin{eqnarray} x&=&(-1)+t(1)\\ y&=&(-4+t(-3) \end{eqnarray}

but i do not know if i am right.

also the line intersects the circle at C and D. calculate the coordinates of C and D. how do you find C and D? can anyone help me please

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    No, these are not right. Your vector equation doesn't have a vector in it. Can you find a value for $t$ that gives the point $(1,-3)$ in the parametric equation? The formulae that you're working with have geometric meaning, you need to understand the geometric meaning in order to successfully write equations in these forms (it's dangerous to plug into multivariate formulae without knowing the meaning of the equations as they are so delicate). – 2017-01-09

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  1. Vector form:

$$\vec{r}=\vec r_0+\vec v$$

where $\vec r_0=\vec r_A$ (you can choose also $\vec r_0=\vec r_B$) and $\vec v=t(\vec B-\vec A)=t(2,1)$ is a vector parallel to the line passing through $A$ and $B$. So we get:

$$\vec r=(-1,-4)+t(2,1)$$

  1. Parametric form:

from the vector form we have: $$(x,y)=(x_0,y_0)+t(v_x,v_y)$$

hence \begin{cases}x=x_0+tv_x\\y=y_0+tv_y \end{cases}

\begin{cases}x=-1+2t\\y=-4+1t \end{cases}

  1. Cartesian form:

$${x-x_0\over v_x}={y-y_0\over v_y}$$

so

$${x+1\over 2}={y+4\over 1}$$

If you want to get the points $C$ an di $D$ you have to solve this system:

\begin{cases}(x-3)^2+(y+2)^2=2\\{x+1\over 2}=y+4\end{cases}

You can use the parametric form of the line as well.

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    so where does the 2 come from? in vector form wouldnt it be -2? so r→=(−1,−4)+t(-2,1) – 2017-01-09
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    $\vec B-\vec A=(1,-3)-(-1,-4)=(2,1)$ – 2017-01-09
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    my calculator said -2 so just wanted to ask. thank you :) – 2017-01-09
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    Pay attention to the double minus;-) – 2017-01-09
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    oh i had the equation the other way around lol to find the vector form for vector AB, do you use the same equation as for the equation line? – 2017-01-09
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    Oh yes you get the same equation line putting $\vec A-\vec B$ instead of $\vec B-\vec A$. – 2017-01-09
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    You are a star thank you so much :) – 2017-01-09
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    Ahahahah ;)...You're welcome! – 2017-01-09