### Home > PC > Chapter 12 > Lesson 12.2.1 > Problem12-42

12-42.

Recall from Chapter 10 that the line through the points $A$ and $B$ can be described by the vector equation $\overrightarrow { O A } + t ( \overrightarrow { A B } )$ for $−∞ \lt t \lt ∞$.

1. Let $A = \left(2, 4\right)$ and $B = \left(1, −3\right)$. Since $\overrightarrow { O A }= 2\textbf{i} + 4\textbf{j}$ and $\overrightarrow { A B }= −\textbf{i} − 7\textbf{j}$ , verify that in this case $\overrightarrow { O A }+ t(\overrightarrow {AB})= 2\textbf{i} + 4\textbf{j} + t(−\textbf{i} − 7\textbf{j})$ .

Use substitution.

2. Rewrite this expression as $\left( \: \right)i + \left(\: \right)j$.

Use the distributive property, then combine like terms.

$2\textbf{i} + 4\textbf{j} + t\left(−\textbf{i} −7\textbf{j}\right)$
$= 2\textbf{i} + 4\textbf{j} −t\textbf{i} −7t\textbf{j}$
$= 2\textbf{ij} −t\textbf{i} + 4\textbf{j} −7t\textbf{j}$

3. Show that this line goes through the point $\left(0, −10\right)$.

If $\left(0, 10\right)$ is on the line, then $0\textbf{i} + 10\textbf{j}$ is possible.

$\left(2 − t\right)\textbf{i} + \left(4 − 7t\right)\textbf{j} = 0\textbf{i} + 10\textbf{j}$

$2 − t = 0$   $4 − 7t = 10$

Does t have the same value in both equations?
If it is on the line, then it should.