How do you write an equation of a line with points (-1,3), (0,8)?

First, we need to determine the slope of the line. The formula for find the slope of a line is:

#m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))#

Where #(color(blue)(x_1), color(blue)(y_1))# and #(color(red)(x_2), color(red)(y_2))# are two points on the line.

Substituting the values from the points in the problem gives:

#m = (color(red)(8) - color(blue)(3))/(color(red)(0) - color(blue)(-1)) = (color(red)(8) - color(blue)(3))/(color(red)(0) + color(blue)(1)) = 5/1 = 5#

Now, we can use the point-slope formula to write and equation for the line. The point-slope form of a linear equation is:

#(y - color(blue)(y_1)) = color(red)(m)(x - color(blue)(x_1))#

Where #(color(blue)(x_1), color(blue)(y_1))# is a point on the line and #color(red)(m)# is the slope.

Substituting the slope we calculated above and the values from the first point in the problem gives:

#(y - color(blue)(3)) = color(red)(5)(x - color(blue)(-1))#

#(y - color(blue)(3)) = color(red)(5)(x + color(blue)(1))#

We can also substitute the slope we calculated above and the values from the second point in the problem giving:

#(y - color(blue)(8)) = color(red)(5)(x - color(blue)(0))#

#(y - color(blue)(8)) = color(red)(5)x#

We can also convert this equation to slope-intercept form:

#y - color(blue)(8) = color(red)(5)x#

#y - color(blue)(8) + 8 = color(red)(5)x + 8#

#y - 0 = 5x + 8#

#y = 5x + 8#

#"the equation of a line in "color(blue)"slope-intercept form"# is.

#•color(white)(x)y=mx+b#

#"where m is the slope and b the y-intercept"#

#"to calculate m use the "color(blue)"gradient formula"#

#•color(white)(x)m=(y_2-y_1)/(x_2-x_1)#

#"let "(x_1,y_1)=(-1,3)" and "(x_2,y_2)=(0,8)#

#m=(8-3)/(0-(-1))=5/1=5#

#"note that "b=8" since point "(0,8)#

#y=5x+8larrcolor(red)"is equation of line"#