Question

What is the equation of the line that passes through the points (1,2/3)

Answer 1

First find the slope
the slope/gradient formula is
(y₂ - y₁) = m(x₂ -x₁)
y₂ = -1                       y₁ = $\frac{2}{3}$
x₂ = -1                       x₁ = 1
Plug them in
(-1 - $\frac{2}{3}$) = m(-1 - 1)
$-\frac{5}{3}$ = m(-2)
Divide by -2
$\frac{5}{3}/-2$ = $\frac{-2m}{-2}$
-2 and -2 cancels out
$- \frac{5}{6}$ = m
So we found the slope, now apply the ones we know to the equation of the line which is:
          y = mx + c
Take any of the 2 coordinates
-1 = -$\frac{5}{6}$(-1) + c
-1 = $\frac{5}{6} + c$
Take the fraction to the other side
-1 - $\frac{5}6}$ = $\frac{5}{6} - \frac{5}{6} + c$
fraction cancels out
$- \frac{11}{6}$ = c

Equation of that line is:
y = $- \frac{5}{6}x - \frac{11}{6}$