how to find th equation that passes though (1,3) and (8,5)

Question

is there a formula to find the answer>

Answer 1

yes, there are several methods
first you need the slope

slope = (5-3)/(8-1) = 2/7

general form: y = mx + b, where m is the slope and b is the y-intercept

y = (2/7)x + b
but the point (1,3) is on it, so

3 = (2/7)(1) + b
multiply by 7 to get rid of fractions
21 = 2 + 7b
7b = 19
b = 19/7

then y = (2/7)x + 19/7

check by using the point not used in the last step, it must satisfy my equation
for (8,5)
5 = (2/7)(8) + 2/7
5 = 16/7 + 19/7
5 = 35/7
5=5

my equation is correct

a "nicer" way is to take the slope and a point and use the point-slope form

so m = 2/7, point (1,3)
y - 3 = (2/7)(x - 1)
again, times 7
7y - 21 = 2x - 2
2x - 7y = -19

Your algebra skills should be good enough to see that my two versions are the same

Answer 2

I should not have called y = mx + b the "general form"
y = mx + b is called the slope y-intercept form

general form would be: 2x - 7y + 19 = 0
some texts call 2x-7y=-19 the standard form