# How do you write the equation of a line in slope intercept, point slope and standard form given Point: (-5, 4) Slope: m = 2?

Jun 2, 2018

Slope intercept: $y = 2 x + 14$
Point-slope: $y = 2 x + 14$
Standard form: $2 x - y = - 14$

#### Explanation:

Slope intercept equation is $y = m x + c$, where $m$ is the slope/gradient, $c$ is the y-intercept and $\left(x , y\right)$ is the format of the points/coordinates given $\left(- 5 , 4\right)$.

The point-slope is given as $y - {y}_{1} = m \left(x - {x}_{1}\right)$.

The standard form of an equation is $A x + B y = C$.

Now, for slope intercept, substitute all the values you have been given and solve for $c$, the y-intercept:

$y = m x + c$
$4 = 2 \left(- 5\right) + c$
$4 = - 10 + c$
$c = 14$

Then substitute only $m$ and $c$ to get the equation of the slope-intercept:

$y = 2 x + 14$

Now for the point-slope (which is just another method of finding the equation of the slope), you find that there's ${y}_{1}$ and ${x}_{1}$ in the formula, this corresponds to the $x$ and $y$ of the coordinates of the points you have been given, $\left(- 5 , 4\right)$. Where $\left({x}_{1} , {y}_{1}\right)$.

So, substitute the values of ${y}_{1}$, $x$, and ${x}_{1}$ and isolate $y$ to find the equation of the point-slope:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(4\right) = 2 \left(x - \left(- 5\right)\right)$
$y - 4 = 2 \left(x + 5\right)$
$y - 4 = 2 x + 10$
$y = 2 x + 10 + 4$
$y = 2 x + 14$

As you can see, both methods give the same results by substitution.

For the standard form, you use the format of the equation of the point-slope but rearrange it so that the $x$ and $y$ terms are on one side of the equal sign and the constants are on the other, $A x + B y = C$:

$y - 4 = 2 \left(x + 5\right)$
$y - 4 = 2 x + 10$
$- 2 x + y = 14$

This is almost in the format of $A x + B y = C$ but $A x$ cannot have a negative sign in front of it. To get rid of this negative, divide both sides of the equation by $- 1$:

$\frac{- 2 x + y}{-} 1 = \frac{14}{-} 1$
$2 x - y = - 14$

Which can also be solved by taking the results from one of the other equations and rearranged to follow the format of $A x + B y = C$.

Hope this helps.