# How do you write the vector equation that passes through point (1,0) and parallel to <-2,-4>?

Dec 8, 2016

$\vec{r} = \left(\lambda + 1\right) \vec{i} - 2 \vec{j}$

#### Explanation:

the vector eqn. of a line is given by

$\vec{r} = \vec{a} + \lambda \vec{d}$

where

$\vec{r} \text{ }$is the general vector point on the line

$\vec{a} \text{ }$is a known vector point on the line

$\vec{d} \text{ }$is the direction of the line

$l a m \mathrm{da} \text{ }$ is a scalar

in this case $\vec{a} = \vec{i}$

$\vec{d} = 2 \vec{i} - 4 \vec{j}$or $\vec{i} - 2 \vec{j} \text{ }$in its simplest form

$\therefore \vec{r} = \vec{i} + \lambda \left(\vec{i} - 2 \vec{j}\right) = \left(\lambda + 1\right) \vec{i} - 2 \vec{j}$