Question

Find the equation of the line passing through the points of (-6, 15) and (4, 5)

Answer 1

To find the equation of the line passing through two points, you can use the point-slope form of a line. The slope of the line is given by the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the slope is m = (5 - 15) / (4 - (-6)) = -10/10 = -1.

The point-slope form of a line is y - y1 = m(x - x1), where (x1, y1) is one of the points on the line and m is the slope. Substituting in the values for m, x1, and y1, we get y - 15 = -1(x + 6). Simplifying this equation gives us y = -x + 9.

So, the equation of the line passing through the points (-6, 15) and (4, 5) is y = -x + 9.

Answer 2

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{5}-\stackrel{y1}{15}}}{\underset{\textit{\large run}} {\underset{x_2}{4}-\underset{x_1}{(-6)}}} \implies \cfrac{-10}{4 +6} \implies \cfrac{ -10 }{ 10 } \implies - 1`

`\begin{array}c \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}{15}=\stackrel{m}{- 1}(x-\stackrel{x_1}{(-6)}) \implies y -15 = - 1 ( x +6) \\\\\\ y-15=-x-6\implies {\Large \begin{array}{llll} y=-x+9 \end{array}}`