# How do you find the equation of a line that Contains point (-1, 2) and is parallel to x-2y=-3?

Jun 17, 2016

$\underline{\text{Every step shown. With practice a lot of these can be done in the head.}}$

$y = \frac{1}{2} x + 2 \frac{1}{2}$

#### Explanation:

Method:
Convert the given equation into standard form for ease of recognition.

Use the given information to determine the equation of the new line.
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$\textcolor{red}{\text{Converting the equation into standard form}}$

$\textcolor{g r e e n}{\text{Step 1}}$

Given: $\textcolor{b r o w n}{x - 2 y = - 3}$

Subtract $\textcolor{b l u e}{x}$ from both sides (isolates the term with $y$ in it)

" "color(brown)(x color(blue)(-x)-2y" "=" "-3color(blue)(-x))

$\text{ "0-2y" "=" } - x - 3$
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$\textcolor{g r e e n}{\text{Step 2}}$

Change the equation so that the $y$ term is positive

Multiply both sides by $\textcolor{b l u e}{- 1}$ giving:

$\text{ "+2y=+x+3" "->" } 2 y = x + 3$
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$\textcolor{g r e e n}{\text{Step 3}}$

Isolate $y$ so that it is on the left of = and everything else is on the right.

Divide both sides by $\textcolor{b l u e}{2}$

" "color(brown)(2/(color(blue)(2)) xxy=x/(color(blue)(2))+3/(color(blue)(2))

But $\frac{2}{2} = 1 \text{ and } 1 \times y = y$

$\text{ } \implies y = \frac{x}{2} + \frac{3}{2}$..............................Equation (1)
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$\textcolor{red}{\text{Building the new equation}}$

Standard form is $y = m x + c \text{ whare m is the gradient (slope)}$

From equation (1) the gradient is $\frac{1}{2}$

We are given that the new line passes through the point:
$\text{ } P \to \left(x , y\right) = \left(- 1 , 2\right)$

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Substituting into equation (1)

$\textcolor{b r o w n}{y = m x + c} \textcolor{b l u e}{\text{ "->" } 2 = \frac{1}{2} \left(- 1\right) + c}$

$\text{ } 2 = - \frac{1}{2} + c$

Add $\frac{1}{2}$ to both sides

$\text{ } 2 \frac{1}{2} = c$

Thus the new equation is:

$\text{ } \textcolor{m a \ge n t a}{y = \frac{1}{2} x + 2 \frac{1}{2}}$