What is the equation of the line passing through #(12,-3)# and #(-8,2)#?

Question

Isaiah Anthony · Accepted Answer

The equation of the line is*** * #y=-1/4x# Slope of the line is #m=(y_2-y_1)/(x_2-x_1)=(2+3)/(-8-12)=5/-20=-1/4# The equation of the line passing through#(12,-3)#is #y-(-3)=-1/4(x-12)or y+3=-1/4x+3 or y=-1/4x#Check: in #(-8,2) 2= -1/4*(-8) or 2=2:.#The equation of the line is #y=-1/4x#[Ans]

Eva Adamson · Answer

undefined To find the equation of the line passing through two points $ (x_1, y_1) $ and $ (x_2, y_2) $, we can use the point-slope form of the equation, which is $ y - y_1 = m(x - x_1) $, where $ m $ is the slope of the line.

First, we find the slope $ m $ using the formula $ m = \frac{y_2 - y_1}{x_2 - x_1} $. Then, we choose one of the points and substitute its coordinates into the point-slope form along with the calculated slope to find the equation of the line.

Using the points $ (12, -3) $ and $ (-8, 2) $, the slope $ m $ is calculated as $ \frac{2 - (-3)}{-8 - 12} = \frac{5}{-20} = -\frac{1}{4} $.

Substituting the point $ (12, -3) $ and the slope $ -\frac{1}{4} $ into the point-slope form, we get the equation of the line as $ y - (-3) = -\frac{1}{4}(x - 12) $. Simplifying this equation gives us $ y + 3 = -\frac{1}{4}x + 3 $, which further simplifies to $ y = -\frac{1}{4}x $.