# What is the slope-intercept form of the line passing through  (3,0)  and  (-4, 1) ?

$y = - \frac{1}{7} x + \frac{3}{7}$

#### Explanation:

The equation of straight line passing through the points $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(3 , 0\right)$ & $\left({x}_{2} , {y}_{2}\right) \setminus \equiv \left(- 4 , 1\right)$ is given as follows

$y - {y}_{1} = \setminus \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \left(x - {x}_{1}\right)$

$y - 0 = \setminus \frac{1 - 0}{- 4 - 3} \left(x - 3\right)$

$y = - \frac{1}{7} \left(x - 3\right)$

$y = - \frac{1}{7} x + \frac{3}{7}$

Above equation of line is the slope-intercept form: $y = m x + c$

Jul 18, 2018

$y = - \frac{1}{7} \cdot x + \frac{3}{7}$

#### Explanation:

After using slope formula for known 2 points,

$m = \frac{1 - 0}{- 4 - 3} = \frac{1}{- 7} = - \frac{1}{7}$

After using formula for known slope and a point,

$y - 0 = - \frac{1}{7} \cdot \left(x - 3\right)$

$y = - \frac{1}{7} \cdot x + \frac{3}{7}$

Jul 18, 2018

$y = - \frac{1}{7} x + \frac{3}{7}$

#### Explanation:

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

•color(white)(x)y=mx+b

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

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

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(3,0)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 4 , 1\right)$

$m = \frac{1 - 0}{- 4 - 3} = \frac{1}{- 7} = - \frac{1}{7}$

$y = - \frac{1}{7} x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find b substitute either of the 2 given points into}$
$\text{the partial equation}$

$\text{using "(3,0)" then}$

$0 = - \frac{3}{7} + b \Rightarrow b = 0 + \frac{3}{7} = \frac{3}{7}$

$y = - \frac{1}{7} x + \frac{3}{7} \leftarrow \textcolor{red}{\text{in slope-intercept form}}$