Respuesta :
[tex]\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}[/tex]
[tex]\bf \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-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}[/tex]
Answer:
x - 2y - 7 = 0
Step-by-step explanation:
To find the equation of a line that passes through (-5, -3) and (3,1)
All we need to do is to use the find the slope and then plug it into the straight line equation
straight line equation : y - [tex]y_{1}[/tex] = m (x - [tex]x_{1}[/tex])
[tex]x_{1}[/tex] = -5 [tex]y_{1}[/tex] = -3 [tex]x_{2}[/tex] = 3 [tex]y_{2}[/tex] = 1
m = slope = [tex]y_{2}[/tex] - [tex]y_{1}[/tex] / [tex]x_{2}[/tex] - [tex]x_{1}[/tex]
= 1 -(-3) / 3 -(-5)
=4/8
=1/2
m=1/2
So, we can now plug in our value into the equation;
y - [tex]y_{1}[/tex] = m (x - [tex]x_{1}[/tex])
y - (-5) = [tex]\frac{1}{2}[/tex] [x - (-3)]
y + 5 = [tex]\frac{1}{2}[/tex](x+3)
y + 5 = [tex]\frac{x + 3}{2}[/tex]
cross-multiply
2(y + 5) = x + 3
2y + 10 = x + 3
take 2y and 10 to the right-hand side of the equation;
x + 3 - 2y - 10 = 0
x - 2y - 7 = 0
Therefore the equation of the line is x - 2y - 7 = 0
