How do you find a standard form equation for the line with (1,3) and (5,9)?

We must use our knowledge of line graphs:

The general equation is: #y = mx +c #

Where #m# is our gradient...

To find #m#:

#m ="change in y" /"change in x" = (Deltay) / (Delta x) #

#=> m = (9-3)/(5-1) = 6/4 = 3/2 #

#=> y= 3/2 x + c #

Now to find #c#:

We can just substitute one of the points in:

# 3 = (3/2*1) +c #

Subtracting #3/2# from both sides...

#3-3/2 = c #

#c = 3/2 #

#=> y = 3/2x + 3/2 #

#=> y = 3/2 ( x+1 ) # if you understand factorisation...

#"the equation of a line in "color(blue)"standard form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))#

#"where A is a positive integer and B, C are integers"#

#"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(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))#

#"let "(x_1,y_1)=(1,3)" and "(x_2,y_2)=(5.9)#

#rArrm=(9-3)/(5-1)=6/4=3/2#

#rArry=3/2x+blarrcolor(blue)"is the partial equation"#

#"to find b use either of the two given points"#

#"substituting "(1,3)" into the partial equation"#

#3=3/2+brArrb=3/2#

#rArry=3/2x+3/2larrcolor(red)"in slope-intercept form"#

#"multiply through by 2"#

#2y=3x+3#

#rArr3x-2y=-3larrcolor(red)"in standard form"#