How do you write the equation of a line in point slope form and slope intercept form given points (3, -8) (-2, 5)?

Question

Eli Assouline · Accepted Answer

Given the points #(3,-8)# and #(-2,5)#
both the point-slope form and the slope-intercept form require that we first determine the slope. The slope can be calculated as
#m = (Delta y)/(Delta x) = (5-(-8))/(-2-3) = - 13/5# Using the slope #m=-13/5# and the point #(3,-8)#
the slope-point form (#y-y_1 = m(x-x_1)#) is #y-(-8) = -13/5(x-3)#
or
#y+8 = -13/5(x-3)# The slope-point form can be converted into the slope-intercept form (#y=mx+b#) by some minor re-arranging of terms:
#y+8 = -13/5(x-3)# #rarr y = -13/5 x +39/5 -8# #rarr y = -13/5x -1/8#

Eva Adamson · Answer

undefined To write the equation of a line in point-slope form, use the formula: $ y - y₁ = m(x - x₁) $, where $ m $ is the slope and $ (x₁, y₁) $ is a point on the line.

To write the equation of a line in slope-intercept form, use the formula: $ y = mx + b $, where $ m $ is the slope and $ b $ is the y-intercept.

First, find the slope $ m $ using the given points.

$ m = \frac{y₂ - y₁}{x₂ - x₁} $

Substitute the points $ (3, -8) $ and $ (-2, 5) $:

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

$ m = \frac{13}{-5} $

$ m = -\frac{13}{5} $

Using point-slope form with the point $ (3, -8) $:

$ y - (-8) = -\frac{13}{5}(x - 3) $

$ y + 8 = -\frac{13}{5}(x - 3) $

$ y + 8 = -\frac{13}{5}x + \frac{39}{5} $

$ y = -\frac{13}{5}x + \frac{39}{5} - 8 $

$ y = -\frac{13}{5}x - \frac{1}{5} $

The equation in point-slope form is $ y + 8 = -\frac{13}{5}(x - 3) $ and in slope-intercept form is $ y = -\frac{13}{5}x - \frac{1}{5} $.