Question

A line passes through (2, –1) and (8, 4).Write an equation for the line in point-slope form.
Rewrite the equation in standard form using integers.

Answer 1

(2,-1)(8,4)
slope = (4-(-1) / (8-2) = (4 + 1)/6 = 5/6

y - y1 = m(x - x1)
(8,4)...x1 = 8 and y1 = 4
slope(m) = 5/6
now we sub
y - 4 = 5/6(x - 8) <==point-slope form
y - 4 = 5/6x - 20/3
y = 5/6x - 20/3 + 4
y = 5/6x -20/3 + 12/3
y = 5/6x - 8/3
-5/6x + y = - 8/3...multiply by -6
5x - 6y = 16 <==standard form

Answer 2

Gradient of the line;
m=$\frac{change in y}{change in x}$
m=$\frac{4--1}{8-2}$
m=$\frac{5}{6}$
y=mx+c
y=$\frac{5}{6} x+c$
Replacing for x and y using point (2, -1)
-1=$\frac{5}{6} *2+c$
-1-5/3=c
c=-8/3
y=$\frac{5}{6} x- \frac{8}{3}$