Question

Find the equation of a line passing through the points (2,6) and (-2,-10)

Answer 1

$\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{6})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-10-6}{-2-2}\implies \cfrac{-16}{-4}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-6=4(x-2) \\\\\\ y-6=4x-8\implies y=4x-2$

Answer 2

Answer:

y = 4x - 2.

Step-by-step explanation:

The slope is  difference in y values  / difference in x values

= (6 - -10) /  (2 - -2)

= 16 / 4

= 4.

Using the point-slope form of a line

y - y1 = m(x - x1)  where m = slope and (x1, y1) is a point on the line, we have:

y - 6 = 4(x - 2)

y = 4x - 8 + 6

y = 4x - 2.