A direct variation equation is one that requires y varies directly as x and looks like this in equation form:
![(y)/(x) =k](https://img.qammunity.org/2019/formulas/mathematics/high-school/3xy5zlaxdmdbvev4kumca64vw6v90u5dxd.png)
where k is the constant of variation. If we solve this for y, we have y = kx, which happens to be a linear function... a line. k here, then, serves as the slope. So what we are given as points on a direct variation function are actually points on a line. The equation for this requires that we find the slope and then rewrite the formula accordingly. First the slope:
![m(k)=(-4-(-3))/(-12-(-9))=(-4+3)/(-12+9)=(1)/(3)](https://img.qammunity.org/2019/formulas/mathematics/high-school/8ofgnt3muz91kkz4fp3muqv26uzyqf0uqm.png)
Now we need to write the equation by using one of the points' coordinates. I picked the first point that has an x coordinate of -9 and a y coordinate of -3. Fitting those into the slope-intercept form of a line,
![-3=(1)/(3)(-9)+b](https://img.qammunity.org/2019/formulas/mathematics/high-school/gegh42zlvixu4kesnd2cevguk73as0w8jr.png)
which simplifies to
-3 = -3 + b and b = 0. That means that the equation of direct variation is
or just
![y=(1)/(3)x](https://img.qammunity.org/2019/formulas/mathematics/high-school/n5aynks4swnhqkmz98b11xo20mzk28mdoz.png)