Question 112963
First find the slope of 3x-4y= 5


*[invoke converting_linear_equations "standard_to_slope-intercept", 3, -4, 5, 2, 1]


So the slope of 3x-4y= 5 is {{{3/4}}}



If you want to find the equation of line with a given a slope of {{{3/4}}} which goes through the point (-p, q), you can simply use the point-slope formula to find the equation:



---Point-Slope Formula---
{{{y-y[1]=m(x-x[1])}}} where {{{m}}} is the slope, and *[Tex \Large \left(x_{1},y_{1}\right)] is the given point


So lets use the Point-Slope Formula to find the equation of the line


{{{y-q=(3/4)(x-(-p))}}} Plug in {{{m=3/4}}}, {{{x[1]=-p}}}, and {{{y[1]=q}}} (these values are given)



{{{y-q=(3/4)x-(3/4)(-p)}}} Distribute {{{3/4}}}


{{{y-q=(3/4)x+(3/4)p}}} Multiply {{{-3/4}}} and {{{-p}}} to get {{{(3/4)p}}}


{{{y=(3/4)x+(3/4)p+q}}} Add q to  both sides to isolate y



So the equation parallel to  3x-4y= 5 and goes through (-p, q) is {{{y=(3/4)x+(3/4)p+q}}}