Question

Find the equation of a line that passes through the points (1,-3) and (3,-4).

Answer 1

Answer: First, let's find the slope of the line:

slope = (change in y) / (change in x)

slope = (-4 - (-3)) / (3 - 1)

slope = -1/2

Now, let's choose one of the points, say (1,-3), and use the point-slope formula:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) are the coordinates of the point.

So, substituting in the values we get:

y - (-3) = (-1/2)(x - 1)

Simplifying this equation, we get:

y + 3 = (-1/2)x + 1/2

Subtracting 3 from both sides, we get:

y = (-1/2)x - 5/2

Therefore, the equation of the line that passes through the points (1,-3) and (3,-4) is y = (-1/2)x - 5/2.

Your welcome.

Explanation:

Answer 2

`(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{(-3)}}}{\underset{\textit{\large run}} {\underset{x_2}{3}-\underset{x_1}{1}}} \implies \cfrac{-4 +3}{2} \implies \cfrac{ -1 }{ 2 } \implies - \cfrac{ 1 }{ 2 }`

`\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}{(-3)}=\stackrel{m}{- \cfrac{ 1 }{ 2 }}(x-\stackrel{x_1}{1}) \implies y +3 = - \cfrac{ 1 }{ 2 } ( x -1) \\\\\\ y+3=- \cfrac{ 1 }{ 2 }x+\cfrac{1}{2}\implies y=- \cfrac{ 1 }{ 2 }x+\cfrac{1}{2}-3\implies {\Large \begin{array}{llll} y=- \cfrac{ 1 }{ 2 }x-\cfrac{5}{2} \end{array}}`