Write the equation of the line given points (-4,6) and (-8,10)

Accepted Solution

Answer:the equation of the line in slope-intercept form is: [tex]y=-x+2[/tex]Step-by-step explanation:First start by finding the slope of the segment that joins those two points using the general formula for the slope of the segments between two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] on the plane:[tex]Slope=\frac{y_2-y_1}{x_2-x_1}[/tex]Then for our case, calling (-4, 6) = [tex](x_1,y_1)[/tex] and (-8, 10) = [tex](x_2,y_2)[/tex], we have:[tex]Slope=\frac{y_2-y_1}{x_2-x_1}\\Slope=\frac{10-6}{-8-(-4)}\\Slope=\frac{4}{-8+4}\\Slope=\frac{4}{-4}\\Slope= -1[/tex]Now, knowing this, we can find the equation of the line by using the "point-slope" form of a line [of slope "m" and going through the point [tex](x_0,y_0)[/tex] that tells us: [tex]y-y_0=m(x-x_0)[/tex]We will be then using the found slope (-1) and for example one of the given points: (-4 , 6), thus:[tex]y-y_0=m(x-x_0)\\y-6=-1(x-(-4))\\y-6=-1(x+4)\\y-6=-x-4\\y=-x-4+6\\y=-x+2[/tex]