The Solution.
The equation of the line is given by the formula below:
![y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/csobd57zth7rh9k4hz9amldzpq2owf0z4j.png)
Where m is the slope of the line, given as
![m=(y_2-y_1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/78uaqhwt0aws3qfwxigaftpihnmb1gzxtp.png)
Picking 2 points in the line, we have
![(-2,2)\text{ and (4,6)}](https://img.qammunity.org/2023/formulas/mathematics/college/m2a23wexukh6qnbxzus30u8lg01xxgsntx.png)
That is,
![\begin{gathered} (x_1=-2,y_1=2) \\ (x_2=4,y_2=6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ygffk1k53pqo90ek8endkbswyub3gxt7za.png)
So, finding the slope,m, we have
![m=(6-2)/(4--2)=(4)/(4+2)=(4)/(6)=(2)/(3)](https://img.qammunity.org/2023/formulas/mathematics/college/hz8n9fw2jegh6ceyb2xsksnpyh35qq1b0r.png)
So, substituting into the formula for the equation of a line, we get
![y-2=(2)/(3)(x--2)](https://img.qammunity.org/2023/formulas/mathematics/college/edby7j5ypmb1gnlt5hx2ftuexz6t43v392.png)
![y-2=(2)/(3)(x+2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/fo5r569fmz61yogl4pjboy9ak4cuolllag.png)
Cross multiplying, we get
![3(y-2)=2(x+2)](https://img.qammunity.org/2023/formulas/mathematics/college/2k1y5bgr76crnaqtz084d32x2ld8luby8l.png)
![3y-6=2x+4](https://img.qammunity.org/2023/formulas/mathematics/college/qanfxvpxfzz09emtbsd9wpmjllqp7bxv5b.png)
![3y-2x=4+6](https://img.qammunity.org/2023/formulas/mathematics/college/xipxsshf77ed7wzvix4lkefnp3tokkmsm5.png)
![3y-2x=10](https://img.qammunity.org/2023/formulas/mathematics/college/548us7hg7p4jslwbf6kv2o2qd60wfzum4s.png)
![or\text{ 3y=2x+10}](https://img.qammunity.org/2023/formulas/mathematics/college/xtxb5vf2rqzqci78l8sz9yxj4bfogarrh3.png)
Therefore, the correct answer is 3y = 2x + 10 or 3y - 2x = 10