How do you write the equation of a line in slope intercept, point slope and standard form given (-3, -1) and (6, -4)?

Question

Hazel Anglin · Accepted Answer

The slope intercept form is #y=-1/3x-2#.

First determine the slope, #m#, using the equation: #m=(y_2-y_1)/(x_2-x_1)#, where #(x_1,y_1)# is one point and #(x_2,y_2)# is the other point. Either of the given points can be the first or second point. I'm going to use the first point as point 1, and the second point as point 2. #m=(-4-(-1))/(6-(-3))# Simplify. #m=(-4+1)/(6+3)# #m=-3/9=-1/3# The slope intercept form is: #y=mx+b#, where #b# is the y-intercept. We have slope, and #x# and #y# from one point. Insert the given values into the equation and solve for #b#. I'm going to use the first point #(-3,-1)#. #-1=-1/3xx-3+b# Simplify. #-1=-(-3)/3+b# #-1=1+b# Subtract #1# from both sides. #b=-2# Slope intercept form is #y=-1/3x-2#. graph{y=-1/3x-2 [-16.02, 16.02, -8.01, 8.01]}

Ethan Adkins · Answer

undefined To find the equation of the line passing through points (-3, -1) and (6, -4), follow these steps:

1. **Calculate the slope (m)**:
   $$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - (-1)}{6 - (-3)} = \frac{-3}{9} = -\frac{1}{3} $$

2. **Slope-intercept form (y = mx + b)**:
   Using the slope and one point, (-3, -1), substitute into the formula to find $b$:
   $$ -1 = -\frac{1}{3}(-3) + b $$
   Solve for $b$:
   $$ -1 = 1 + b $$
   $$ b = -2 $$
   Thus, the slope-intercept form is:
   $$ y = -\frac{1}{3}x - 2 $$

3. **Point-slope form (y - y1 = m(x - x1))**:
   Using the slope and one point, (-3, -1), substitute into the formula:
   $$ y - (-1) = -\frac{1}{3}(x - (-3)) $$
   Simplify to:
   $$ y + 1 = -\frac{1}{3}(x + 3) $$
   Thus, the point-slope form is:
   $$ y + 1 = -\frac{1}{3}x - 1 $$

4. **Standard form (Ax + By = C)**:
   Convert the slope-intercept form to standard form. Multiply everything by 3 to eliminate the fraction:
   $$ 3y = -x - 6 $$
   Rearrange to standard form:
   $$ x + 3y = -6 $$
   Thus, the standard form is:
   $$ x + 3y = -6 $$