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1 year ago

Use slopes and y-intercepts to determine if the lines 5x-5y=-2 and -x+2y=4

Mathematics

1 answer:

HACTEHA [7]1 year ago

6 0

The given pair of lines are not perpendicular.

The line is a curve showing the shortest distance between 2 points.

5x - 5y = -2 - - - - - (1)
Transform the equation into standard form,
5x + 2 = 5y
y = 5x /5 + 2/5
y = x + 2/5

The slope of equation 1 is $m_1 = 1$ and intercept c = 2 / 5

Similarly
x + 2y = 4 - - - - - - - -(2)
Transform it into standard form
y = -x/2 + 4 /2
y = -x / 2 + 2

Slope of the equation 2 $m_2$ = -1 / 2 and intercept c = 2
Slope of line 1 * slope of line 2 = 1 * -1/2 = -1/2

Since the lines are not perpendicular because the pair of lines does not satisfy the property of perpendicular lines i.e
$m_1*m_2 = -1$

Thus, the given pair of lines are not perpendicular.

Learn more about lines here:

brainly.com/question/2696693

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Find two vectors in R2 with Euclidian Norm 1<br> 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 $x_2$ .

Remembering,

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

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

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