Write an equation that passes through the points (4,-1) and (2,- 2).

How do you find the equation of a line with Guassian elimination given two points?

Paraphin [41]

Answer:

The solution is similar to the 2-point form of the equation for a line:

y = (y2 -y1)/(x2 -x1)·x + (y1) -(x1)(y2 -y1)/(x2 -x1)

Step-by-step explanation:

Using the two points, write two equations in the unknowns of the equation of the line.

For example, you can use the equation ...

y = mx + b

Then for the points (x1, y1) and (x2, y2) you have two equations in m and b:

b + (x1)m = (y1)

b + (x2)m = (y2)

The corresponding augmented matrix for this system is ...

$\left[\begin{array}{cc|c}1&x1&y1\\1&x2&y2\end{array}\right]$

____

The "b" variable can be eliminated by subtracting the first equation from the second. This puts a 0 in row 2 column 1 of the matrix, per Gaussian Elimination.

0 + (x2 -x1)m = (y2 -y1)

Dividing by the value in row 2 column 2 gives you the value of m:

m = (y2 -y1)/(x2 -x1)

This value can be substituted into either equation to find the value of b.

b = (y1) -(x1)(y2 -y1)/(x2 -x1) . . . . . substituting for m in the first equation

7 0

3 years ago

Find two vectors in R2 with Euclidian Norm 1 whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .