We know that two lines are parallel if and only if their slopes are equal. For this reason, we need to find the slope of line given by the equation 7x-2y-5=0 in order to get the slope of the equation we are looking for; to do this we solve the equation for y:
![\begin{gathered} 7x-2y-5=0 \\ 2y=7x-5 \\ y=(7)/(2)x-(5)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ra2jb6c4lbvmdwlrsiqvvppayx3u3mce9t.png)
Now, this equation is written in slope-intercept form:
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
Comparing it with the equation we found we conclude that the slope is 7/2 and hence the equation we are looking for will also have this slope.
Now that we know this, we have to remember that the equation of a line that passes through the point (x1,y1) and has slope m is given by:
![y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/csobd57zth7rh9k4hz9amldzpq2owf0z4j.png)
Plugging the slope we found and the point given we have:
![\begin{gathered} y-3=(7)/(2)(x+2) \\ y-3=(7)/(2)x+7 \\ y=(7)/(2)x+10 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oal9gowj863zg2fhin2omirxafq4rp4mje.png)
Therefore, the equation of the line we are looking for in slope intercept form is:
![y=(7)/(2)x+10](https://img.qammunity.org/2023/formulas/mathematics/college/3e2yte6ihg6imt2xs8sir5kd8myw8g26dr.png)
To write in general form we write it in the form Ax+By+C=0:
![\begin{gathered} y=(7)/(2)x+10 \\ 2y=7x+20 \\ 7x-2y+20=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/m54c8uga2hwpry9bij2w5ie0z6cay6sixe.png)
Therefore, the equation of the line in general form is:
![7x-2y+20=0](https://img.qammunity.org/2023/formulas/mathematics/college/i278qytdw7hqg1ek23jzk39ch7n5l3y4n2.png)