Question 121645


First convert the standard equation {{{3x+1y=8}}} into slope intercept form


*[invoke converting_linear_equations "standard_to_slope-intercept", 3, 1, 8, 2, 1]




Now let's find the equation of the line that is perpendicular to {{{y=-3x+8}}} which goes through (-5,-2)


*[invoke equation_parallel_or_perpendicular "perpendicular", "-3", "8", -5,-2]



So by using substitution, elimination, or graphing, we find that the two lines intersect at the point (2.5,0.5)


Now let's use the distance formula to find the distance between the two points (-5,-2) and (2.5,0.5)



Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(-5,-2\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(2.5,0.5\right)]


{{{d=sqrt((-5-2.5)^2+(-2-0.5)^2)}}} Plug in {{{x[1]=-5}}}, {{{x[2]=2.5}}}, {{{y[1]=-2}}}, {{{y[2]=0.5}}}


{{{d=sqrt((-7.5)^2+(-2.5)^2)}}} Evaluate {{{-5-2.5}}} to get -7.5. Evaluate {{{-2-0.5}}} to get -2.5. 


{{{d=sqrt(56.25+6.25)}}} Square each value


{{{d=sqrt(62.5)}}} Add


So the distance approximates to


{{{d=7.90569415042095}}}


which rounds to

{{{d=7.906}}}




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Answer:

So the distance between the point (-5,-2) and the line 3x+y=8 is approximately 7.906 units