# If I had the scalar equation 4x-2y-1=0 how would I write it in the vector equation, parametric and symmetric form?

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### 1 Answer

You need to remember what is the parametric form of the equation such that:

`bar r= bar a + bar b*t`

You need to select a point that lies on the given line, hence, you need to select x and y coordinates to verify the given equation such that:

`x = 1/4 =gt 4*(1/4) - 2y - 1 = 0 =gt 1 - 2y -1 = 0 =gt -2y = 0 =gt y = 0`

Hence, the coordinates of the point that lies on the line are `(1/4,0)` .

You need to convert the scalar form of equation into slope intercept form such that:

`4x - 2y - 1 = 0 =gt -2y = -4x + 1`

`y = 2x - 1/2`

Notice that the slope of the line is the coefficient of x, hence m = 2 and the direction vector is (1,2).

Hence, you may write the vector equation of the line:

`bar r = lt1/4 , 0gt + lt1,2gt*t`

Once you have the vector equation, you may also write the parametric equations such that:

`x = 1/4 + t`

`y = 2t`

You may now determine the symmetric equations such that:

`(x - 1/4)/1 = (y - 0)/2 =gt x - 1/4 = y/2`

**Hence, evaluating the vector equation, parametric equations and symmetric equation yieldsbar `r = lt1/4 , 0gt + lt1,2gt*t ;x = 1/4 + t and y = 2t ; x - 1/4 = y/2.` **

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