# Thread: Need help with symmetric equation

1. ## Need help with symmetric equation

Write a symmetric equation and parametric equation for
a) Line passes through the point A (4,1) with direction vector m = [-3, 1]

b) passes through the point R(-6, 2) and S(4, -2)

2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form

*plz also explain them to me

2. b) passes through the point R(-6, 2) and S(4, -2)

First, we need a vector, v that will be parallel to the line. We thus let v be the vector that starts at point S and ends at point R.

v = <-6-4, 2-(-2)> = <-10, 4>.

Now that we have our direction vector v, all we need to do is choose of our two given points, R or S. Let's use R.

The equation of our line r = <-6, 2> + t<-10, 4> = <-6-10t, 2+4t>.

In parametric form:
$x = -6 - 10t$
$y = 2 + 4t$

In symmetric form:
$\frac{x+6}{-10} = \frac{y-2}{4}$.

3. Thanks for the explanation

now need part a and 2

4. a) Line passes through the point A (4,1) with direction vector m = [-3, 1]

We have a direction vector, m and a point to work with. This should not be difficult.

r = <4,1> + t<-3,1> = <4-3t, 1+t>

parametric form:
$x = 4 - 3t$
$y = 1 + t$

symmetric form:
$\frac{x-4}{-3} = \frac{y-1}{1}$

5. Thanks again

now need help with last question

6. 2) The equation of a line is 2x-3y+12=0. Write equation for line in parametric form.

What we are given here is an implicit equation of a line. To convert from an implicit representation to a parametric representation, follow my steps below:

We let $\vec r$ be the position vector, $$ and let $\vec n$ be the vector of coefficients, $<2, -3>$.

That is,
$\vec r = $
$\vec n = <2, -3>$

Now, we can rewrite 2x-3y+12=0 in terms of $\vec r$ and $\vec n$.

$\vec n \cdot \vec r = -12$

To write this in vector form, we must now do two things. A) we must find a point, $\vec{r_{0}}$ on the line and B) find a vector $\vec v$ parallel to the line.

To find any point on 2x-3y+12=0, we can simply just use the x-intercept. That is, we simultaneously solve for 2x-3y+12=0 and x=0.

$2x - 3y + 12 = 0$

$x = 0$

$-3y + 12 = 0$

$-3y = -12$

$3y = 12$

$y = \frac{12}{3}$ = 4

Therefore, we let point $\vec{r_{0}} = <0, 4>$.

Now, for part B that I mentioned above, find a vector $\vec v$ parallel to the line.

$\vec n = <2, -3>$

$\vec v = \vec{n_{\perp}} = <-3, -2>$

The general parametric equation in vector form is $\vec{r}(t) = \vec{r_{0}} + t \vec v$.

Thus, $\vec{r}(t) = <0, 4> + t<-3, -2>$.

Finally, in parametric non-vector form, we have:

x = -3t
y = 4 - 2t

Good luck!
-Andy

7. And to double-check my solution for question 2, let's substitute x=-3t and y=4-2t in 2x-3y+12 = 0:

$2x-3y+12 = 0$

$2(-3t) - 3(4-2t) + 12 = 0$

$-6t -12 + 6t + 12 = 0$

And we know we are correct...

-Andy

8. Thanks Andy

9. ## Vectors symmetric equations

This question is givin me trouble big tim

Que/

Find symmetric equations for the line

x=2+3t, y=4-4t, z=-1+6t

If anyone can solve this, itll be very greatful..
thanx

10. Originally Posted by K-D
This question is givin me trouble big tim

Que/

Find symmetric equations for the line

x=2+3t, y=4-4t, z=-1+6t

If anyone can solve this, itll be very greatful..
thanx
As abender has shown in the first reply you only have to solve each equation for t and set them equal:

$x=2+3t~\implies~\boxed{t=\dfrac{x-2}3}$
$y=4-4t~\implies~\boxed{t=\dfrac{y-4}4}$
$z=-1+6t~\implies~\boxed{t=\dfrac{z+1}6}$

Thus the line in symmetric form is:

$\dfrac{x-2}3 = \dfrac{y-4}4 = \dfrac{z+1}6$