How do you write an equation for a line passing through the points (5,-1) and (-3,-2)?

Question

Genesis Allen · Accepted Answer

The equation of the line in standard form is # x -8y = 13# The slope of the line passing through #(5,-1) and (-3,-2)# is #m= (y_2-y_1)/(x_2-x_1)= (-2+1)/(-3-5)=(-1)/(-8)=1/8# Let the equation of the line in slope-intercept form be #y=mx+c or y=1/8x+c# The point (5,-1) will satisfy the equation . So, # -1= 1/8*5+c or c= -1-5/8= -13/8# Hence the equation of the line in slope-intercept form is #y= 1/8x-13/8# The equation of the line in standard form is #y= 1/8x-13/8. or 8y =x-13 or x -8y = 13# {Ans]

Jace Anderson · Answer

undefined To write an equation for a line passing through the points (5, -1) and (-3, -2), you can use the point-slope form of a linear equation: $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is one of the given points.

1. First, calculate the slope ($ m $) using the formula $ m = \frac{y_2 - y_1}{x_2 - x_1} $, where $ (x_1, y_1) $ and $ (x_2, y_2) $ are the given points.
2. Then, choose one of the given points and substitute its coordinates into the point-slope form along with the calculated slope to get the equation of the line.

Using the given points:

$ (x_1, y_1) = (5, -1) $ and $ (x_2, y_2) = (-3, -2) $

1. Calculate the slope ($ m $):

$ m = \frac{-2 - (-1)}{-3 - 5} = \frac{-2 + 1}{-3 - 5} = \frac{-1}{-8} = \frac{1}{8} $

2. Choose one of the points, let's say (5, -1), and substitute into the point-slope form:

$ y - (-1) = \frac{1}{8}(x - 5) $

Simplify:

$ y + 1 = \frac{1}{8}(x - 5) $

Distribute:

$ y + 1 = \frac{1}{8}x - \frac{5}{8} $

Subtract 1 from both sides:

$ y = \frac{1}{8}x - \frac{5}{8} - 1 $

Simplify:

$ y = \frac{1}{8}x - \frac{5}{8} - \frac{8}{8} $

$ y = \frac{1}{8}x - \frac{13}{8} $

So, the equation of the line passing through the points (5, -1) and (-3, -2) is $ y = \frac{1}{8}x - \frac{13}{8} $.