The easiest way to find the line parallel to 3x -2y =5 through (1,2), we must set 3x - 2y = 5 into slope-intercept form
![3x-2y =5\\-2y = -3x+5\\y = (3)/(2)x-(5)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/xhdk62320c0bwqri5gqacrsxn9dgencr7e.png)
In slope-intercept form, y = mx + b, m is the slope. For two lines to be parallel, their slopes must be equal.
Thus the new line that is parallel to 3x - 2y = 5 must also have a slope of (3/2)x
![y= (3)/(2) x+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/vwga4rbtz48s11lnrc0o161az5iufibt8o.png)
To find b, we must plug in (1,2)
![2 = (3)/(2)*1 + b\\b = (1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/i27davokdtwfhb38r4u49bznfeiz1wc9wh.png)
Thus the equation that is parallel to 3x-2y = 5 and passes through (1,2) is
![y = (3)/(2)x+(1)/(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/7h4cmjikbaewk5v0h4z4cifvjadsr7ge8q.png)
Hope that helps!