Question

What is an equation of the line that passes
through the points (-1, -6) and (6, 1)?

Answer 1

`(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{1}-\stackrel{y1}{(-6)}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-1)}}} \implies \cfrac{1 +6}{6 +1} \implies \cfrac{ 7 }{ 7 } \implies 1`

`\begin{array}ll \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{ 1}(x-\stackrel{x_1}{(-1)}) \implies y +6 = 1 ( x +1) \\\\\\ y+6=x+1\implies {\Large \begin{array}{llll} y=x-5 \end{array}}`

Answer 2

y = x - 5

Explanation:

The equation is y = mx + b

m = the slope

b = y-intercept

Slope = rise/run or (y2 - y1) / (x2 - x1)

Points (-1, -6) and (6, 1)

We see the y increase by 7, and the x increase by 7, so the slope is

m = 7/7 = 1

Y-intercept is located at (0, -5)

So, the equation is y = x - 5