Question

Find an equation of the line that passes through the points (-5, -3) and (3, 1)

Answer 1

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 -  $y_{1}$   =   m (x  -  $x_{1}$)

$x_{1}$ = -5     $y_{1}$ = -3   $x_{2}$ = 3   $y_{2}$ = 1

 

m = slope =  $y_{2}$  -  $y_{1}$   /  $x_{2}$  -  $x_{1}$

                =   1 -(-3)  /   3 -(-5)

                 =4/8

                  =1/2

m=1/2

So, we can now plug in our value into  the equation;

y  -  $y_{1}$   =   m (x -  $x_{1}$)

y - (-5) = $\frac{1}{2}$ [x - (-3)]

y + 5 = $\frac{1}{2}$(x+3)

y +  5  =  $\frac{x + 3}{2}$

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

Answer 2

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{-3})\quad (\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-5)}}}\implies \cfrac{1+3}{3+5}\implies \cfrac{4}{8}\implies \cfrac{1}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{1}{2}}[x-\stackrel{x_1}{(-5)}]\implies y+3=\cfrac{1}{2}(x+5) \\\\\\ y+3=\cfrac{1}{2}x+\cfrac{5}{2}\implies y=\cfrac{1}{2}x+\cfrac{5}{2}-3\implies y = \cfrac{1}{2}x-\cfrac{1}{2}$