Find the equation of a line passes through the point `(1,3)` and parallel to the line `3x-5y+7=0`.

Explanation

Given line is
`3x-5y+7=0`
Its slope
`m_(1)=-("Coefficient of" x)/("Coefficient of" y)`
`=-(3)/(-5)=(3)/(5)`
Slope of parallel line
`m_(2)=m_(1)=(3)/(5)`
Now, the equation of a line passes through the point `(1,3)` and whose slope is `(3)/(5)`, is
`y-3=(3)/(5)(x-1)`
`implies5y-15=3x-3`
`implies3x-5y+12=0`