How do you write an equation in point slope form that passes through (1,4) and(4,9)?

We want this in the form #y=color(green)(m)x+color(purple)(c)#, where #m# is the gradient, and #c# is the y-intercept.

Gradient is the change in y over the change in x, or the rise over run; however you like to think about it.

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

So for our points #(color(blue)(1,4))# and #(color(red)(4,9)) #

#m=(color(red)9-color(blue)4)/(color(red)4-color(blue)1)#

#=5/3#

Now, by rearranging our equation for gradient, we can show that:

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

Here, #x# and #y# are general points on the line - we've just replaced #(color(red)(x_2,y_2))# with #(color(red)(x,y))#

Since we know #color(green)m#, and have a set of co-ordinates to sub in as #color(blue)((x_1, y_1)#, we can sub values in, then re-arrange to get to the #y=color(green)(m)x+color(purple)(c)# form.

#y-color(blue)(4)=color(green)(5/3)(x-color(blue)(1))#

#y-4=5/3x-5/3#
#y=5/3x-5/3+4#
#y=color(green)(5/3)x+color(purple)(7/3)#
as required.

An alternative method, after finding the gradient, is to substitute your gradient and a pair of points into #y=color(green)(m)x+color(purple)(c)#, rearrange to find #c#, then go back afterwards and write the equation. If you've been taught that way, its fine! But personally I prefer substituting straight into #y-color(blue)(y_1)=color(green)(m)(x-color(blue)(x_1))# since you don't have to go back afterwards. Up to you which method you use.

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

#•color(white)(x)y-y_1=m(x-x_1)#

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

#"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)=(1,4)" and "(x_2,y_2)=(4,9)#

#m=(9-4)/(4-1)=5/3#

#"use either of the 2 given points as point on line"#

#"using "(1,4)" then"#

#y-4=5/3(x-1)larrcolor(red)"in point-slope form"#