How do you find the general form of the line passing through (-1,2) and (2,5)?

Jun 11, 2017

y=x+3

Explanation:

First, find the slope. To do this, plug in values for this equation.
$m = \frac{y 2 - y 1}{x 2 - x 1}$
m is the slope and the values are your original coords.
$m = \frac{5 - 2}{2 - - 1}$
$m = \frac{3}{3}$
$m = 1$
Now that we have the slope, we use it to find the y-intercept, and the slope-intercept form.
We use point-slope for this.
$y - y 1 = m \left(x - x 1\right)$
$y - 2 = 1 \left(x - - 1\right)$
$y - 2 = \left(x + 1\right)$
$y - 2 = x + 1$
$y = x + 3$
The slope is 1, and the y-intercept is 3. The slope-intercept form is "y=x+3", and the point-slope form is "y-2=1(x+1)"

Jun 11, 2017

$x - y + 3 = 0$

Explanation:

$\text{the equation of a line in "color(blue)"general form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{A x + B y + C = 0} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where A is a positive integer and B, C are integers.

$\text{to begin express the equation in "color(blue)"slope-intercept form}$

• y=mx+b

$\text{where m represents the slope and b, the y-intercept}$

$\text{to calculate m use the "color(blue)"gradient formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right) \text{ are 2 coordinate points}$

$\text{the points are } \left({x}_{1} , {y}_{1}\right) = \left(- 1 , 2\right) , \left({x}_{2} , {y}_{2}\right) = \left(2 , 5\right)$

$\Rightarrow m = \frac{5 - 2}{2 - \left(- 1\right)} = \frac{3}{3} = 1$

$\Rightarrow y = x + b \leftarrow \text{ is the partial equation}$

$\text{to find b use either of the 2 given points}$

$\text{using " (2,5)" then}$

$5 = 2 + b \Rightarrow b = 3$

$\Rightarrow y = x + 3 \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$

$\Rightarrow x - y + 3 = 0 \leftarrow \textcolor{red}{\text{ in general form}}$