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svetoff [14.1K]

4 years ago

15

Find the equation of the line that passes through the point (−3, −4) and is parallel to the line that passes through the points

(−5, 1) and (7, −3).

1 answer:

Dimas [21]4 years ago

8 0

Answer:

$y = -\frac{1}{3}x - 5$

Step-by-step explanation:

To write the equation of a line use the point slope formula $y - y_1 = m(x-x_1)$ . Find m by using the slope formula with the points.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{1--3}{-5-7} =\frac{4}{-12} =-\frac{1}{3}$

Since the line is parallel, it will have the same slope.

Substitute m = -1/3 and (-3,-4).

$y --4 = -\frac{1}{3}(x --3)\\y + 4 = -\frac{1}{3}(x+3)\\y +4 = -\frac{1}{3}x - 1\\y = -\frac{1}{3}x - 5$

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Leona [35]

$\mathbf a=(1,1,1)^\top=1\,\mathbf i+1\,\mathbf j+1\,\mathbf k$
$\mathbf b=(-2,3,0)^\top=-2\,\mathbf i+3\,\mathbf j$
$\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\-2&3&0\end{vmatrix}=-3\,\mathbf i-2\,\mathbf j+5\,\mathbf k=(-3,-2,5)^\top$

Basically, you're looking for a matrix $\mathbf A$ such that

$\mathbf A(\mathbf a\times\mathbf b)=\mathbf 0$

i.e. a matrix $\mathbf A$ whose nullspace with basis vector $\mathbf a\times\mathbf b$ .

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$\mathrm{rank}\mathbf A+\underbrace{\mathrm{null}\mathbf A}_1=3\implies\mathrm{rank}\mathbf A=2$

One easy choice for a row would be $\begin{bmatrix}1&1&1\end{bmatrix}$ , since

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So,

$\underbrace{\begin{bmatrix}1&1&1\\0&1&\frac25\end{bmatrix}}_{\mathbf A}\begin{bmatrix}-3\\-2\\5\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

is one possible solution.

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