Answer:
[tex]\frac{1}{6} x+3y+\frac{13}{2}[/tex] or x + 18y + 39
Step-by-step explanation:
[tex]\frac{3}{2} y[/tex] = [tex]3x-6[/tex]
→ First multiply everything by 2 to get rid of the fraction
3y = 6x - 12
→ We want to find the equation that's parallel to this line and passes through (3 , 2). The first thing we know about parallel lines is that the gradient is the negative reciprocal so,
[tex]3y = -\frac{1}{6} x+c[/tex]
→ Now we substitute in the values (3 , 2)
[tex]6 =- \frac{1}{2}+c[/tex]
→ Add [tex]\frac{1}{2}[/tex] to both sides to isolate c
[tex]\frac{13}{2} =c[/tex]
So the equation of the line that is parallel to [tex]\frac{3}{2} y[/tex] = [tex]3x-6[/tex] is [tex]3y =- \frac{1}{6} x+\frac{13}{2}[/tex] but we are not finished we are asked to but the equation in the format
ax + by + c = 0 so,
[tex]3y =- \frac{1}{6} x+\frac{13}{2}[/tex]
Rearrange
[tex]\frac{1}{6} x+3y+\frac{13}{2}[/tex]
If the question want's the answer in whole numbers then multiply everything by 6
x + 18y + 39
