Question

A line passes through the points (1, 4) and (3, –4). Which is the equation of the line?

Answer 1

$\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-4}{3-1}\implies \cfrac{-8}{2}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-4(x-1) \\\\\\ y-4=-4x+4\implies y=-4x+8$

Answer 2

Answer:

y=-4x+8

Step-by-step explanation:

First find the slope of the line using the slope formula.

-4-4/3-1=-4, The slope is -4.

Next, find the value of b by plugging in the value of the slope and either coordinate's x and y-values.  I will use the point (1,4).

y=mx+b

4 = -4(1)+b, Solve for b.

4 = -4+b

b = 8 The equation of the line is: y=-4x+8