# How to determine the equation of the line parallel to 3x - 2y + 4 = 0 and passing through (1,6)?

Mar 13, 2017

$y = \frac{3}{2} x + \frac{9}{2}$

#### Explanation:

color(orange)"Reminder "color(red)(bar(ul(|color(white)(2/2)color(black)("parallel lines have equal slope")color(white)(2/2)|)))

The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and b, the y-intercept.

$\text{Rearrange "3x-2y+4=0" into this form}$

$3 x \cancel{- 2 y} \cancel{+ 2 y} + 4 = 0 + 2 y$

$\Rightarrow 2 y = 3 x + 4$

divide ALL terms on both sides by 2

$\frac{\cancel{2} y}{\cancel{2}} = \frac{3}{2} x + \frac{4}{2}$

$\Rightarrow y = \frac{3}{2} x + 2 \leftarrow \text{ in form } y = m x + b$

$\Rightarrow \text{slope } = m = \frac{3}{2}$

The equation of a line in $\textcolor{b l u e}{\text{point-slope form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
m is slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

$\text{For parallel line " m=3/2" and } \left({x}_{1} , {y}_{1}\right) = \left(1 , 6\right)$

$\Rightarrow y - 6 = \frac{3}{2} \left(x - 1\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$

Distributing the bracket and simplifying gives the equation in an alternative form.

$y - 6 = \frac{3}{2} x - \frac{3}{2}$

$\Rightarrow y = \frac{3}{2} x - \frac{3}{2} + 6$

$\Rightarrow y = \frac{3}{2} x + \frac{9}{2} \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$
graph{(y-3/2x-2)(y-3/2x-9/2)=0 [-10, 10, -5, 5]}

Mar 13, 2017

$3 x - 2 y + 9 = 0.$

#### Explanation:

Recall that the eqn. of a line parallel to the given line

${l}_{1} : a x + b y + c = 0$ is of the Form ${l}_{2} : a x + b y + c ' = 0 , c ' \ne c .$

If we compare the slopes of the lines ${l}_{1} \mathmr{and} {l}_{2}$, we will find that

the result is quite obvious. If, in addition, #(x_0,y_0) in l_2, then,

$a {x}_{0} + b {y}_{0} + c ' = 0 , \text{ giving, } c ' = - a {x}_{0} - b {y}_{0.}$

$\therefore {l}_{2} : a x + b y = a {x}_{0} + b {y}_{0.}$

Accordingly, the eqn. of the reqd. line is given by,

$3 x - 2 y = 3 \left(1\right) - 2 \left(6\right) \Rightarrow 3 x - 2 y + 9 = 0.$