Question #26d13

The general form for the point-slope version of the equation of a line is:

#y=mx+b#

Here you are given the slope:

#-(5/2)#,

and the point of intersection with the y-axis, which is:

#(0,8)#

Let's calculate some values for #y = (-5/2)x + 8#, plug them in and check:

x | y

0 | 8
3 | 1/2
4 | -2

If I graph these points, I get:

graph{y=(-5/2)*x+8 [-20.75, 19.25, -8.08, 11.92]}

This is the graph the software gives me. It's a little confusing because 10 units on the graph are divided into only 4 parts instead of five.

Click on the graph to see the points of intersection with the x- and y-axes.

Slop (gradient) is given as #-5/2#

Given point #P_1->(x_1,y_1)=(-8,8)#

Gradient #->m=("change in y")/("change in x")#

So picking any hypothetical point #P_2-(x_2,y_2)#

We have:

#m=-5/2=("change in y")/("change in x") ->(y_2-y_1)/(x_2-x_1) = (y_2-8)/(x_2-(-8))#

We do not know the values of #y_2 and x_2# so write this as just:

#m=-5/2=(y-8)/(x+8)#

Cross multiply

#-5/2(x+8)=y-8#

Turn it round the other way:

#color(red)(y-8=-5/2(x+8) larr" Point slope form")#

#y-y_1=m(x-x_1)#

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

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#

#"where m is the slope and "(x_1,y_1)" a point on the line"#

#"here "m=-5/2" and "(x_1,y_1)=(-8,8)#

#rArry-8=-5/2(x+8)larrcolor(red)"in point-slope form"#