Question

In standard form, the line parallel to the shown line and passing through the origin.

Answer 1

The equation of a line in standard form is written as:

`Ax+By=C`

Now, two lines are parallel if and only if their slopes are equal, that is:

`m_1=m_2`

The slope of a line is given by:

`m=\frac{y_2-y_1_{}}{x_2-x_1}`

where (x1,y1) and (x2,y2) are points through the line. From the graph we notice that the line passes through the points (0,3) and (5,-1) then its slope is:

`\begin{gathered} m=(-1-3)/(5-0) \\ m=-(4)/(5) \end{gathered}`

This means that the lin in the graph has a slope os -4/5 and hence the paralell line we are looking for has slope -4/5 as well.

Once we know the slope of the line we want we just need a point; the porblem states that the line passes through the origin, then it passes through the point (0,0). The equation of a line with slope m that passes throught the point (x1,y1) is given by:

`y-y_1=m(x-x_1)`

Plugging the values of the slope and the point we have that:

`\begin{gathered} y-0=-(4)/(5)(x-0) \\ y=-(4)/(5)x \end{gathered}`

Finally we just write the equation in standard form, therefore the equation of the line is:

`(4)/(5)x+y=0`